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Let $P(x) = x^2-1$ be a polynomial, and let $a$ be a positive real number satisfying $$ P(P(P(a))) = 99. $$ The value of $a^2$ can be written as $m+\sqrt{n}$ , where $m$ and $n$ are positive integers, and $n$ is not divisible by the square of any prime. Find $m+n$ .
*Proposed by **HrishiP***
|
<details><summary>Solution</summary>First, we obtain $P(a)=a^2-1$ . Upon plugging in this value into the polynomial again, we obtain $$ P(P(a))=(a^2-1)^2-1=(a^2-1+1)(a^2-1-1)=a^2(a^2-2)=a^4-2a^2. $$ Finally, upon plugging in this value into the polynomial again, we obtain
\begin{align*}
P(P(P(a)))&=(a^4-2a^2)^2-1
&=(a^4-2a^2+1)(a^4-2a^2-1)
&=(a^2-1)^2((a^2-1)^2-2)
&=(a^2-1)^4-2(a^2-1)^2.
\end{align*}
Setting this equal to $99$ and letting $y=a^2-1$ , we get $y^4-2y^2=99$ . Adding $1$ to both sides of the equation gives us $$ y^4-2y^2+1=100 \implies (y^2-1)^2=100 \implies (a^4-2a^2)^2=100 \implies a^4-2a^2=\pm 10. $$ Next, by the quadratic formula, we obtain $$ a^2=\frac{2\pm \sqrt{4\pm40}}{2} = 1 \pm \sqrt{11}. $$ Since $a^2$ is positive, we have that $a^2=1 + \sqrt{11}$ , so the requested answer is $1+11=\boxed{012}$ .</details>**Remark.** This gives me DMC 10B #15 vibes, but I think that this one is more clever.
|
[
"alternatively you can just note that the answer is constant so we just need to find one value of $a$ that works\n\nwe have $P(P(a))=10$ so that $a^2-1=\\sqrt{11}\\rightarrow \\boxed{012}$ ",
"Wait yes, working backwards here is probably cleaner.",
"i did it a diff way $a^2=x$ so you get $x^4-4x^3+4x^2-100$ and then difference of squares ",
"Trivial? $$ P(P(P(a)))=99 \\implies P(P(a))=10 \\implies P(a)= \\sqrt{11} \\implies a^2=\\sqrt{11}+1 \\implies 11+1=\\boxed{012} $$ ",
"I thought this was a bit easy. We must have $P(P(a)) = 10$ since the negative option gives imaginary results, which means $P(a) = \\sqrt{11}$ , so $a^2 - 1= \\sqrt{11} \\implies a^2 = 1+\\sqrt{11} \\implies \\boxed{12}$ .",
"Whoops.\n\n<details><summary>Solution</summary>We have \n\\begin{align*}\nP(P(P(a))) &= 99 \nP(P(a^2-1)) &= 99 \nP((a^2-1)^2-1)&=99 \n((a^2-1)^2-1)^2-1&=99\n((a^2-1)^2-1)^2&=100\n\\end{align*}**Case 1:** $(a^2-1)^2-1=10$ We have\n\\begin{align*}\n (a^2-1)^2-1&=10 \n (a^2-1)^2&=11 \n\\end{align*}**Subcase 1:** $a^2-1=\\sqrt{11}$ We have $$ a^2-1=\\sqrt{11} \\implies a^2=1+\\sqrt{11} $$ **Subcase 2:** $a^2-1=-\\sqrt{11}$ We have $$ a^2-1=\\sqrt{11} \\implies a^2=1+\\sqrt{11} $$ **Case 2:** $(a^2-1)^2-1=-10$ We have $$ (a^2-1)^2=-9 $$ However, square roots can't be negative thus there are no solutions for this case. So our answer is $1+11=\\boxed{12}.$</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 DIME/2673239.json"
}
|
In $\triangle ABC$ with $AC>AB$ , let $D$ be the foot of the altitude from $A$ to side $\overline{BC}$ , and let $M$ be the midpoint of side $\overline{AC}$ . Let lines $AB$ and $DM$ intersect at a point $E$ . If $AC=8$ , $AE=5$ , and $EM=6$ , find the square of the area of $\triangle ABC$ .
*Proposed by **DeToasty3***
|
Interesting problem!
Trivially $AM=MC=DM=4$ when drawing the circumcircle of $\triangle ADC$ , so $D=2$ . Applying LoC on $\triangle AME$ we get $25=36+16-48 \cos(\angle AME) \implies \frac{9}{16} = \cos(\angle AME)$ Applying LoC now on $\triangle AMD$ we get $AD^2 = 32-32 (\tfrac{9}{16}) = 14 \implies AD=\sqrt{14}$ Doing Pythagoras now on $\triangle ADC$ we get $DC=5 \sqrt{2}$ For the final step in the solution, let $N$ be the midpoint of $AB$ , and $AN=BN=y$ . Stewart’s Theorem on $\triangle AME$ yields $$ 14(5-2y)+4(2y) = 10y(5-2y)+5(4y^2-14) \implies y=2 $$ $$ (2+2)^2 - (\sqrt{14})^2 = BD^2 \implies BD=\sqrt{2} $$ The final step is $$ \left(\frac{\sqrt{2} \cdot \sqrt{14}}{2} + \frac{\sqrt{14} \cdot 5 \sqrt{2}}{2} \right)^2 = (\sqrt{7}+5 \sqrt{7})^2 = \boxed{252} $$
|
[
"MD=4, DE=2, then do some area ratios and heron bash to get the answer",
"oops I fakesolved by using the fact that aime answers are only integers. :blush: ",
"<blockquote>oops I fakesolved by using the fact that aime answers are only integers. :blush:</blockquote>\n\nHuh? Now I'm curious as to how you did it.",
"So let $AB=x \\implies BE=5-x,$ and $DM=y \\implies DE=6-y.$ By herons formula, the area of $\\triangle{AEM}$ is $\\frac{15\\sqrt{7}}{4}.$ So now let the altitude of $M$ onto $AE$ be $H,$ we have $MH \\cdot 5=\\frac{15\\sqrt{7}}{2} \\implies MH=\\frac{3\\sqrt{7}}{2},$ and since the altitude from $C$ to $AE$ is $2$ times of that, it is $3\\sqrt{7}.$ So the square of the area of $\\triangle{ABC}$ is $\\frac{63x^2}{4},$ now you would guess $x$ is probably an integer cuz of aime and $x$ must be a multiple of $2,$ so if you draw a very accurate diagram, $x=4$ looks pretty reasonable, so the answer is $252.$ ",
"i just trigbashed",
"<blockquote>So let $AB=x \\implies BE=5-x,$ and $DM=y \\implies DE=6-y.$ By herons formula, the area of $\\triangle{AEM}$ is $\\frac{15\\sqrt{7}}{4}.$ So now let the altitude of $M$ onto $AE$ be $H,$ we have $MH \\cdot 5=\\frac{15\\sqrt{7}}{2} \\implies MH=\\frac{3\\sqrt{7}}{2},$ and since the altitude from $C$ to $AE$ is $2$ times of that, it is $3\\sqrt{7}.$ So the square of the area of $\\triangle{ABC}$ is $\\frac{63x^2}{4},$ now you would guess $x$ is probably an integer cuz of aime and $x$ must be a multiple of $2,$ so if you draw a very accurate diagram, $x=4$ looks pretty reasonable, so the answer is $252.$ </blockquote>\n\nI guessed 255 by measuring LOL",
"<details><summary>hint</summary>Trig bash to get $\\tan(\\angle{ABC})=\\sqrt{7}$ from tangent subtraction formula after solving for $\\cos(\\angle{ACB})=\\frac{5\\sqrt{2}}{8}$</details>",
"<blockquote>So let $AB=x \\implies BE=5-x,$ and $DM=y \\implies DE=6-y.$ By herons formula, the area of $\\triangle{AEM}$ is $\\frac{15\\sqrt{7}}{4}.$ So now let the altitude of $M$ onto $AE$ be $H,$ we have $MH \\cdot 5=\\frac{15\\sqrt{7}}{2} \\implies MH=\\frac{3\\sqrt{7}}{2},$ and since the altitude from $C$ to $AE$ is $2$ times of that, it is $3\\sqrt{7}.$ So the square of the area of $\\triangle{ABC}$ is $\\frac{63x^2}{4},$ now you would guess $x$ is probably an integer cuz of aime and $x$ must be a multiple of $2,$ so if you draw a very accurate diagram, $x=4$ looks pretty reasonable, so the answer is $252.$ </blockquote>\n\nummm thats epic. Actually in the test I sillied thinking $\\frac{\\sqrt{7}}{2\\sqrt{2}}=\\frac{\\sqrt{14}}{8}$ which result in the area being $\\frac{7\\sqrt{2}\\cdot \\sqrt{14}}{2}$ :oops_sign: ",
"I used trig + heron to get $CD = 5\\sqrt2, AD = \\sqrt{14}, CE = \\sqrt{79}$ and set up coordinates w/D at $(0,0)$ . I then got $E$ and was able to find point B based on E being the intersection point. It was nasty but it worked.",
"phythagorean+similar triangles plus extensions to some sides = solved",
"Bary kills this one.",
"I think the fact that the problem asks for the square of the area can help narrow the whole problem down to just finding one integer"
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1038,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 DIME/2673241.json"
}
|
Let $a_1,a_2,\ldots,a_6$ be a sequence of integers such that for all $1 \le i \le 5$ , $$ a_{i+1}=\frac{a_i}{3} \quad \text{or} \quad a_{i+1}={-}2a_i. $$ Find the number of possible positive values of $a_1+a_2+\cdots+a_6$ less than $1000$ .
*Proposed by **stayhomedomath***
|
AHHHH HOW IS EVERYONE SO SMART
For this problem I basically spent an hour bashing, first mapping a triangle that had all possible values of each set of $a_n$ terms for $1 \le n \le 6$ , of course setting the initial value to $x$ . Then, I spend EVEN LONGER finding all possible values of sums with making an EVEN LARGER triangle. In the end, I recognized all the numerators were multiples of $7$ , so the answer was simply $\lfloor \tfrac{1000}{7} \rfloor = \boxed{142}$
|
[
"the sequence 27, -54, -18, 36, 12, 4 and all its multiples gives 7x => 142, but i cant figure out how to explicitly prove",
"Note that ${-}2 \\equiv \\tfrac{1}{3} \\equiv 5 \\pmod{7}$ , so $$ a_1+a_2+\\cdots + a_6 \\equiv a_1(1+5+5^2+\\cdots+5^5) \\equiv 0 \\pmod{7}, $$ which means that $a_1+a_2+\\cdots+a_6$ must be a multiple of $7$ .**Remark (DeToasty3):** This problem and #11 have similar general approaches of <details><summary>doing this</summary>testing out values and then proving a claim using \"if and only if\"</details>.**Remark (stayhomedomath):** Both #9 and #11 are from me :)",
"<blockquote>the sequence 27, -54, -18, 36, 12, 4 and all its multiples gives 7x => 142, but i cant figure out how to explicitly prove</blockquote>\n\no I did the same",
"is there official sol?",
"Well the official solution is basically just find a sequence that generates all multiples of 7 and then do what I did to show that no other values work.",
"o lol $ $ ",
"<blockquote>o lol $ $ </blockquote>\n\nWe can choose the sequence being $9k,3k,k,-2k,4k,8k$ ",
"<blockquote><blockquote>o lol $ $ </blockquote>\n\nWe can choose the sequence being $9k,3k,k,-2k,4k,8k$ </blockquote>\n\nNow that I see the solution, this problem is beautiful! Great job proposing it @stayhomedomath!",
"I disliked the problem because the key realization is very much not obvious and seems \"out of the blue\". Too hard for a #9 in my opinion.",
"tbh “out of the blue” observations is what makes a problem hard and beautiful",
"...or annoyingly bashy.",
"<blockquote>...or annoyingly bashy.</blockquote>\n\ncope\n\n(this problem is not annoyingly bashy)",
"hi asdf . thanks for helping me Ratio someone on aops",
"How would one be able to come up with this construction?"
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1010,
"boxed": true,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 DIME/2673242.json"
}
|
Given a parallelogram $ABCD$ , let $\mathcal{P}$ be a plane such that the distance from vertex $A$ to $\mathcal{P}$ is $49$ , the distance from vertex $B$ to $\mathcal{P}$ is $25$ , and the distance from vertex $C$ to $\mathcal{P}$ is $36$ . Find the sum of all possible distances from vertex $D$ to $\mathcal{P}$ .
*Proposed by **HrishiP***
|
Since $ABCD$ must be a parallelogram, the plane $\mathcal{P}$ will always be fixed for some configuration of $ABCD$ . Thus, we fix the plane $\mathcal{P}$ and then proceed.
Let the notation $\{\mathcal{F} \}\{\mathcal{B} \} \in \mathcal{P}$ denote the set of all points relative to $\mathcal{P}$ such that $\mathcal{F}$ contains all points in the "front", and $\mathcal{B}$ has all points in the "back". There are multiple cases:**Case 1:** $\{A,B,C \}\{\} \in \mathcal{P}$
By parallelogram properties, the distance from $D$ to the plane is $(49-25)+36=60$ and is obvious by a diagram**Case 2:** $\{A,C \} \{B \} \in \mathcal{P}$ By parallelogram properties, the net distance between $A$ and $B$ is $49+25$ , so $D$ is $49+25+36=110$ units away from the plane**Case 3:** $\{A,B \} \{C \} \in \mathcal{P}$ By parallelogram properties, the net distance between $A$ and $B$ is $49-25$ , so the distance from $D$ to the plane is $36-(49-25)=12$ units away**Case 4:** $\{A \} \{B,C \} \in \mathcal{P}$ By parallelogram properties, the net difference between $C$ and $B$ is $36-25$ , so the distance from $D$ to the plane is $49-(36-25)=38$ units away
The answer is $$ 60+110+12+38=\boxed{220} $$
|
[
"let the z-coordinate for $A$ be $z_A$ , and similarly for the other points\n\nthen we have $|z_A|=49,|z_B|=25,|z_C|=36$ .\n\nnotice that because $ABCD$ is a parallelogram, we have $z_A+z_D=z_B+z_C\\rightarrow z_D=z_B+z_C-z_A$ .\n\nthe possible values for $z_B+z_C$ are $-61,-11,11,61$ and the possible values for $z_A$ are $-49,49$ so testing values from here gives $|z_D|=110,60,38,12\\rightarrow \\boxed{220}$ .",
"AMC 10A #17 moment",
"How \n\nOh I didn't do that one",
"<blockquote>AMC 10A #17 moment</blockquote>\n\n~~No, the other way around.~~",
"Did anyone notice how the answer is just $2\\cdot(49+36+25)$ ? It's very neat."
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1062,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 DIME/2673243.json"
}
|
Richard has an infinite row of empty boxes labeled $1, 2, 3, \ldots$ and an infinite supply of balls. Each minute, Richard finds the smallest positive integer $k$ such that box $k$ is empty. Then, Richard puts a ball into box $k$ , and if $k \geq 3$ , he removes one ball from each of boxes $1,2,\ldots,k-2$ . Find the smallest positive integer $n$ such that after $n$ minutes, both boxes $9$ and $10$ have at least one ball in them.
*Proposed by **vvluo** & **richy***
|
Simple recursion suffices. Let each marble be labeled $1,2,3 \cdots $ such that for every $n$ th minute, the marble with label $a$ is placed into a box such that there exists no marbles with labels $b<a$ . Let the function $f(c)$ denote the number of the marble that was placed into box $c$ first. We get $$ f(1)=1 $$ $$ f(2)=2 $$ $$ f(3)=3 $$ $$ f(4)=5 $$ $$ f(5)=8 $$ $$ f(6)=13 $$ It is clear $f(c)=f(c-1)+f(c-2)$ for $c>2$ . Therefore, $f(7)=21$ , $f(8)=34$ , $f(9)=55$ , and $f(10)=\boxed{89}$ . Note when marble $89$ is placed, box $9$ is guaranteed to have at least $1$ marble because boxes $1-8$ are only cleared.
|
[
"clearly the first time that box $10$ is filled is when we have $$ 1111111110\\rightarrow 0000000011 $$ which means that box $9$ will also be filled, so we just need to smallest positive integer $n$ such that after $n$ minutes box $10$ is filled.\n\nlet $f(n)$ be the smallest positive integer such that after $f(n)$ minutes box $n$ is filled. we want $f(10)$ .\n\nthen clearly we have $f(1)=1,f(2)=2$ .\n\nrecursion stuff gives $f(n)+f(n+1)=f(n+2)$ for $n\\geq 1$ so this is fibonacci so that $f(10)=\\boxed{089}$ .",
"i messed up with recursion here\n\nf(3) = 5\nf(4) = 9\nf(5) = 18\nf(6) = 34\n\ni thought difference of two consecutive is the next perfect square then use summation laws to find f(10) where did i go wrong",
"<blockquote>i messed up with recursion here\n\nf(3) = 5\nf(4) = 9\nf(5) = 18\nf(6) = 34\n\ni thought difference of two consecutive is the next perfect square then use summation laws to find f(10) where did i go wrong</blockquote>\n\n<details><summary>what</summary>$f(3)=3$ , $f(4)=5$ , $f(5)=8$ , $f(6)=13$</details>",
"<blockquote><blockquote>i messed up with recursion here\n\nf(3) = 5\nf(4) = 9\nf(5) = 18\nf(6) = 34\n\ni thought difference of two consecutive is the next perfect square then use summation laws to find f(10) where did i go wrong</blockquote>\n\n<details><summary>what</summary>$f(3)=3$ , $f(4)=5$ , $f(5)=8$ , $f(6)=13$</details></blockquote>\n\noops i know where i went wrong now",
"misread and thought it was remove all of them up to k-1 :(",
"<blockquote>misread and thought it was remove all of them up to k-1 :(</blockquote>\n\nsame",
"<blockquote>misread and thought it was remove all of them up to k-1 :(</blockquote>\n\nRead correctly and did this lots of times lol. Bashed up to 29 terms and noticed the pattern. Didn't bother proving it but hey, it worked."
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1056,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 DIME/2673244.json"
}
|
Let $a$ and $b$ be real numbers such that $$ \left(8^a+2^{b+7}\right)\left(2^{a+3}+8^{b-2}\right)=4^{a+b+2}. $$ The value of the product $ab$ can be written as $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
*Proposed by **stayhomedomath***
|
Everything was right until somehow I got $a=\tfrac{29}{4}$ ....
First off, make everything base $2$ to get $$ \left(2^{3a} + 2^{b+7} \right)\left(2^{a+3} + 2^{3b-6} \right) = 2^{2a+2b+4} $$ Clearly the RHS is a power of $2$ , which means both expressions in the LHS product are also powers of $2$ .
Now, suppose $2^c+2^d=2^e$ and WLOG let $2^c < 2^d$ . Clearly we can factor to get $2^c(1+2^{d-c}) = 2^e$ , so contradiction because $1+2^{d-c}$ is never a power of $2$ . Reverse the expression and we get the same result. Thus $2^c=2^d$ Now, back to the original problem, we get $2^{3a}=2^{b+7}$ and $2^{a+3} = 2^{3b-6}$ . Therefore, $$ 3a-b=7 $$ $$ 3a-9b=-27 $$ $$ 8b = 34 \implies b = \frac{17}{4} $$ $$ 3a=\frac{45}{4} \implies a = \frac{15}{4} $$ $$ ab=\frac{15 \cdot 17}{4 \cdot 4} = \frac{255}{16} \implies 255+16=\boxed{271} $$ If the following configuration had not worked, we would have ended up with $3$ systems which would have not been fun to solve...
|
[
"the LHS is $(2^{3a}+2^{b+7})(2^{a+3}+2^{3b-6})=4^{a+b+2}$ , this motivates us to set $3a=b+7$ and $a+3=3b-6$ giving $(a,b)=(\\frac{15}{4},\\frac{17}{4})$ . this works so the answer is $\\frac{255}{16}\\rightarrow \\boxed{271}$ .",
"wait i don't really understand the motivation for the above solution\n\nessentially the main idea is just <details><summary>this</summary>am-gm on the two terms of the product restrict stuff</details>\n\nwhich then tells you that $3a = b+7$ and $a+3 = 3b - 6$ **must** occur because equality :heart_eyes_cat:",
"@above the motivation is just to try it and see if it works lol. ",
"<blockquote>wait i don't really understand the motivation for the above solution\n\nessentially the main idea is just <details><summary>this</summary>am-gm on the two terms of the product restrict stuff</details>\n\nwhich then tells you that $3a = b+7$ and $a+3 = 3b - 6$ **must** occur because equality :heart_eyes_cat:</blockquote>\n\noops that works a lot better :blush:",
"Let $2^a = x$ and $2^b = y.$ Then we have $(x^3 + 128y)(8x + y^3/64) = 16x^2y^2$ but by AM-GM $$ (x^3 + 128y)(8x + y^3/64) \\ge 4 \\sqrt{128x^3y} \\times \\sqrt{xy^3/8} = 16x^2y^2 $$ Equality occurs only when $x^3 = 128y$ and $512x = y^3.$ So $3a = b+7$ and $9+a = 3b$ and solving this gives the answer.\n\nNote: the fact that there's only one choice of $ab$ essentially gives away the fact that this is an equality case of some inequality.",
"wait thats a bogus sol cuz the RHS isn't necessarily a power of 2. a and b aren't integers",
"[quote name=\"squareman\" url=\"/community/p23245021\"]\nLet $2^a = x$ and $2^b = y.$ Then we have $(x^3 + 128y)(8x + y^3/64) = 16x^2y^2$ but by AM-GM $$ (x^3 + 128y)(8x + y^3/64) \\ge 4 \\sqrt{128x^3y} \\times \\sqrt{xy^3/8} = 16x^2y^2 $$ Equality occurs only when $x^3 = 128y$ and $512x = y^3.$ So $3a = b+7$ and $9+a = 3b$ and solving this gives the answer.\n\nNote: the fact that there's only one choice of $ab$ essentially gives away the fact that this is an equality case of some inequality.\n</blockquote>\n\nHow does AM-GM tell us that $(x^3 + 128y)(8x + y^3/64) \\ge 4 \\sqrt{128x^3y} \\times \\sqrt{xy^3/8}$ ?",
"Simple solution: Rewrite as $(2^{3a}+2^{b+7})(2^{a+3}+2^{3b-6}) = 2^{2a+b+4}$ , so $2^{4a+3}+2^{3a+3b-6}+2^{a+b+10}+2^{4b+1} = 2^{2a+b+4}$ . Now I claim the only solution to $2^p+2^q+2^r+2^s=2^e$ occurs with $p=q=r=s$ . WLOG $p<q<r<s<e$ , taking $\\pmod {2^q}$ gives $2^p \\equiv 0 \\pmod {2^q}$ so $p = q$ . So on until you get $p = q = r = s$ , now solve to get $a = \\frac{15}{4}, b = \\frac{17}{4} \\implies \\frac{255}{16} \\implies \\boxed{271}$ .",
"a,b,c,d,e aren't necessarily integers so the above reasoning is flawed",
"o.\n\nthat is not good.",
"<blockquote>the LHS is $(2^{3a}+2^{b+7})(2^{a+3}+2^{3b-6})=4^{a+b+2}$ , this motivates us to set $3a=b+7$ and $a+3=3b-6$ giving $(a,b)=(\\frac{15}{4},\\frac{17}{4})$ . this works so the answer is $\\frac{255}{16}\\rightarrow \\boxed{271}$ .</blockquote>\n\nThis motivates us to set [Absolute Random Ness]",
"lol I just expanded and prayed each is equal. ",
"<blockquote>the LHS is $(2^{3a}+2^{b+7})(2^{a+3}+2^{3b-6})=4^{a+b+2}$ , this motivates us to set $3a=b+7$ and $a+3=3b-6$ giving $(a,b)=(\\frac{15}{4},\\frac{17}{4})$ . this works so the answer is $\\frac{255}{16}\\rightarrow \\boxed{271}$ .</blockquote>\n\ndang it I should've tried this problem",
"I did AM-GM and then matched $3a=a+3$ and $b+7=3b-6$ :clown: why isn't this good",
"why is this problem so sus :skull:"
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1050,
"boxed": false,
"end_of_proof": false,
"n_reply": 16,
"path": "Contest Collections/2022 Contests/2022 DIME/2673245.json"
}
|
A positive integer $n$ is called $\textit{un-two}$ if there does not exist an ordered triple of integers $(a,b,c)$ such that exactly two of $$ \dfrac{7a+b}{n},\;\dfrac{7b+c}{n},\;\dfrac{7c+a}{n} $$ are integers. Find the sum of all un-two positive integers.
*Proposed by **stayhomedomath***
|
If $n \equiv 0 \pmod{7}$ it's not hard to see that $n$ is not un-two. Otherwise, suppose we have $7a + b \equiv 0 \pmod{n}$ and $7b+c \equiv 0 \pmod{n}$ which means that $49a \equiv c \pmod{n}.$ We know $n$ is un-two if and only if $7c+a \equiv 344a \equiv 0 \pmod{n}$ for all $a$ so we just sum up all factors of $344$ to get $\boxed{660}.$
|
[
"<details><summary>Solution</summary>Set $a=1,b=-7,c=49$ to get $$ \\frac{7a+b}{n}=0,\\frac{7b+c}{n}=0,\\frac{7c+a}{n}=\\frac{344}{n}. $$ Then clearly we must have $n\\mid 344$ . <insert explanation that i got during the test but forgot> it is sufficient so we have $(15)(44)=660$ .</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1020,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 DIME/2673246.json"
}
|
A sequence of polynomials is defined by the recursion $P_1(x) = x+1$ and $$ P_{n}(x) = \frac{(P_{n-1}(x)+1)^5 - (P_{n-1}(-x)+1)^5}{2} $$ for all $n \geq 2$ . Find the remainder when $P_{2022}(1)$ is divided by $1000$ .
*Proposed by **treemath***
|
<details><summary>Author's/Official Solution</summary>The recursion is applying the second roots of unity filter to $(P_n(x)+1)^5$ that only returns terms with odd degree. For the first step, $P_2(x)$ is the odd-degree terms of $(x+2)^5$ , which is $x^5+40x^3+80x$ . For all subsequent steps, we again apply the second roots of unity filter to $(P_n(x)+1)^5$ that only returns terms with odd degree.
In general, if a polynomial $P(x)$ has only terms of odd degree, then $(P(x))^n$ will only have terms of odd degree if $n$ is odd and will only have terms of even degree if $n$ is even. The proof for this is left as an exercise to the reader. Since $P_n(x)$ will have all odd coefficients for all $n \ge 2$ , we can treat $P_n(x)$ as a single term with odd degree. Thus, by the Binomial Theorem, we get that $$ P_{n+1}(x)=(P_n(x))^5+10(P_n(x))^3+5(P_n(x)) $$ for all $n \geq 2$ .
Next, we see that $P_2(1)=121$ , which is equivalent to $1$ modulo $8$ . We see that $$ P_3(1) \equiv 1^5+10(1)^3+5(1)\equiv 0 \pmod{8}, $$ so $P_n(1)\equiv 0 \pmod{8}$ for all $n \geq 3$ .
Next, we see that $121 \equiv {-}4 \pmod{125}$ . Computing, we get $$ P_3(1) \equiv ({-}4)^5+10({-}4)^3+5({-}4) \equiv 66 \pmod{125}. $$ We can rewrite $66$ as $50+16$ . From here, we get that
\begin{align*}
P_4(1) &\equiv (50+16)^5+10(50+16)^3+5(50+16)
&\equiv 16^5 + 10(16)^3 + 5(16)
&\equiv 116 \pmod{125}.
\end{align*}
Note that $116=100+16$ , so for any $n \geq 5$ , we see that
\begin{align*}
P_n(1) &\equiv (100+16)^5+10(100+16)^3+5(100+16)
&\equiv 16^5 + 10(16)^3 + 5(16)
&\equiv 116 \pmod{125}.
\end{align*}
Thus, we have that $P_{2022}(1)$ is equivalent to $0$ modulo $8$ and $116$ modulo $125$ . By the Chinese Remainder Theorem, the requested answer is $\boxed{616}$ .</details>**Remark (DeToasty3).** While I really like the <details><summary>idea</summary>keeping in only the terms with an odd degree</details> of this problem, unfortunately the big mod bash afterwards kind of downgrades the quality of the problem for me by a bit.
|
[
"the only thing i was able to get on the test was $P_{n-1}(x)=-P_{n-1}(-x)$ for all $n\\geq 2$ giving $$ P_{n+1}(x)=\\frac{(P_n(x)+1)^5+(P_n(x)-1)^5}{2} $$ for all $n\\geq 2$ ",
"Try showing that $$ P_{n+1}(x)=(P_n(x))^5+10(P_n(x))^3+5(P_n(x)) $$ for all $n \\geq 2$ .\n\nNow, use CRT with mod $125$ and mod $8$ by using the explicit value of $P_2(1)$ . Mod $8$ should be easy, but good luck with mod $125$ ...\n\nEdit: I realized that you could still use CRT with the expression that you have...",
"<blockquote>Try showing that $$ P_{n+1}(x)=(P_n(x))^5+10(P_n(x))^3+5(P_n(x)) $$ for all $n \\geq 2$ .\n\n</blockquote>\n\nyeah i got this alternate form by expanding",
"hriship told me that its a roots of unity filter thing?? idk",
"huh what is answer for this?",
"is there official sol to this?",
"o, roots of unity filter, I have no idea what that is. ",
"isn't P_2(1)=122?, not 121?",
" $$ P_{2}(1) = \\frac{(P_{1}(1)+1)^5 - (P_{1}(-1)+1)^5}{2} = \\frac{(2+1)^5 - (0+1)^5}{2} = \\frac{243-1}{2} = 121. $$ ",
"ohhhhh wait I got the starting index of the recurrence wrong, thanks detoasty3",
"No problem.",
"<blockquote>The proof for this is left as an exercise to the reader. </blockquote>\n\n",
"wait you've gotta be kidding me it was just a crt bash after that oops ",
"what if you don't know roots of unity filter? Is there another sol?",
"huh i used engineers induction and went with 722 because $P_n(x) = -P_n(-x)$ and used recursion for $P_n(1)$ ",
"<blockquote>what if you don't know roots of unity filter? Is there another sol?</blockquote>\n\nWell, this roots of unity filter is like recognizing which terms add up and which terms cancel out. But here there's no special thing you have to know, it's more like just intuition. Oops I'm not sure how to explain this.\n\nSomeone else who got $722$ as the answer said that $P_2(1)=122$ , so maybe that's where you went wrong? I haven't actually tried computing it mod $1000$ with $122$ instead of $121$ , so I'm not actually sure.",
"<blockquote><blockquote>what if you don't know roots of unity filter? Is there another sol?</blockquote>\n\nWell, this roots of unity filter is like recognizing which terms add up and which terms cancel out. But here there's no special thing you have to know, it's more like just intuition. Oops I'm not sure how to explain this.\n\nSomeone else who got $722$ as the answer said that $P_2(1)=122$ , so maybe that's where you went wrong? I haven't actually tried computing it mod $1000$ with $122$ instead of $121$ , so I'm not actually sure.</blockquote>\n\nyep thats where i went wrong...\n\nrly need to work on not sillying",
"what I had an (easier to find) solution.\n\nEverything is $\\pmod {1000}$ . Basically you bash and get $P_2(1)=121$ , and then calculating $P_3(1)=816$ isn't that hard by CRT since $121$ is close to $125$ . Then calculating $P_4(1)=616$ is a good amount of computation but again it isn't that bad because one of the things you are raising to the power of 5 is 615 which will be $0\\pmod {125}$ anyway. Computing $P_5(1)=616$ is really similar to the previous computation. Now it just repeats and the answer is 616.",
"Oops manually computing each term mod 1000 works too...",
"Wait, never mind. Isn't it basically required to show that $P_{n}(x)={-}P_{n}({-}x)$ for all $n\\geq 2$ in order to progress? This is the crux of the problem, I believe. Whatever happens afterwards should hinge on this key claim.\n\n(BTW this is probably what HrishiP meant by roots of unity filter.)",
"wait i just realized that during the test i calculated to $P_4(1)=616$ and that i was trying to find a number so that n = n^5 + 10n^3 + 5n mod 1000 oops",
"Wait darn...",
"I disliked this problem too, since the trick was to \"compute terms\", which was both bashy and uninspired.",
"<blockquote>I disliked this problem too, since the trick was to \"compute terms\", which was both bashy and uninspired.</blockquote>\n\nbashy and uninspired is a very maa thing to do ",
"Here's a story about this problem. When the original proposer got to the part where $$ P_{n+1}(x)=(P_n(x))^5+10(P_n(x))^3+5(P_n(x)) $$ for all $n \\geq 2$ , he decided that due to the coefficients, the recursion would just be multiplying by $16$ each time, which would lead to a far less annoying follow-up. Unfortunately, no one noticed this flaw in the solution until three submissions in, where we received our first $616$ answer. I became suspicious and scanned through the solution, and sure enough, I managed to catch the mistake.\n\nWe apologize for not picking up on this earlier. In fact, we almost decided to replace this problem with something else, but the remainder staying constant after a few recursions was ultimately decent enough for us to keep it (at the time, at least). Now though, I wish there was something a lot cleaner to fill this spot, but again, it's too late to change an entire problem at this time, I guess.",
"I think that the CRT part of this problem is under-appreciated because there is a relatively clean general solution that works with higher modulo (like mod $10^5$ or $10^6$ ) that nobody has found yet.\n-----\nLet $a_n = P_n(1)$ . We have $a_2 = 121$ and $a_{n+1} = a_n^5 + 10a_n^3 + 5a_n$ for $n\\ge 2$ . We want to find $a_{2022}$ modulo $8$ and $125$ . \n\nMod $8$ is easy: we see that $a_2\\equiv 1\\pmod{8}$ , $a_3\\equiv 1+10+5\\equiv 0\\pmod{8}$ , and $a_n\\equiv 0\\pmod{8}$ for $n\\ge 3$ .\n--\nNow, we compute $a_{2022}\\pmod{125}$ . The key idea is to compute $a_n$ modulo increasing powers of $5$ , and using the residue of each power to compute the residue of the next power.\n\nFirst, we calculate mod $5$ . Note that $a_2\\equiv 1\\pmod{5}$ and $a_{n+1}\\equiv a_n^5\\pmod{5}$ for $n\\ge 2$ . This means $a_n\\equiv 1\\pmod{5}$ for $n\\ge 2$ .\n\nNow, let $a_n = 5b_n + 1$ for $n\\ge 2$ . By the Binomial Theorem, \n\\begin{align*}\na_{n+1} &\\equiv a_n^5 + 10a_n^3 + 5a_n \\pmod{25} \n5b_{n+1} + 1 &\\equiv (5b_n + 1)^5 + 10(5b_n + 1)^3 + 5(5b_n + 1) \\pmod{25} \n5b_{n+1} + 1 &\\equiv 1 + 10 + 5 \\pmod{25} \nb_{n+1} &\\equiv 3 \\pmod{5} \n\\end{align*}\nThus, $b_n \\equiv 3\\pmod{5}$ for $n\\ge 3$ .\n\nFinally, let $b_n = 5c_n + 3$ for $n\\ge 3$ . Again by the Binomial Theorem,\n\\begin{align*}\na_{n+1} &\\equiv a_n^5 + 10a_n^3 + 5a_n \\pmod{125} \n5b_{n+1} + 1 &\\equiv (5b_n + 1)^5 + 10(5b_n + 1)^3 + 5(5b_n + 1) \\pmod{125} \n5b_{n+1} + 1 &\\equiv 25b_n + 1 + 150b_n + 10 + 25b_n + 5 \\pmod{125} \n5b_{n+1} &\\equiv 200b_n + 15 \\pmod{125} \nb_{n+1} &\\equiv 40b_n + 3\\pmod{25} \n5c_{n+1}+3 &\\equiv 40(5c_n + 3) + 3\\pmod{25} \n5c_{n+1} &\\equiv 120 \\equiv 20\\pmod{25} \nc_{n+1} &\\equiv 4\\pmod{5}\n\\end{align*}\nThus, $c_n\\equiv 4\\pmod{5}$ for $n\\ge 4$ . This means $a_n \\equiv 5(5\\cdot 4 + 3) + 1 \\equiv 116\\pmod{125}$ for $n\\ge 4$ .\n\nNow, routine CRT computation gives $a_n\\equiv \\boxed{616}\\pmod{1000}$ for $n\\ge 4$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1070,
"boxed": true,
"end_of_proof": false,
"n_reply": 27,
"path": "Contest Collections/2022 Contests/2022 DIME/2673247.json"
}
|
A spinner has five sectors numbered ${-}1.25$ , ${-}1$ , $0$ , $1$ , and $1.25$ , each of which are equally likely to be spun. Ryan starts by writing the number $1$ on a blank piece of paper. Each minute, Ryan spins the spinner randomly and overwrites the number currently on the paper with the number multiplied by the number the spinner lands on. The expected value of the largest number Ryan ever writes on the paper can be written as $\tfrac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .
*Proposed by **treemath***
|
Cool problem!
<details><summary>Sol</summary>Let $a_k$ denote the $k^{\text{th}}$ term in our sequence of numbers, where the $0^{\text{th}}$ term is $1$ .
Now let's figure out the probability that $a_n$ is the last positive term in the sequence for a given $n$ . Following this positive we can have:
A $0$ . Probability: $\frac{1}{5}$ A negative followed by a $0$ . Probability: $\frac{2}{5} \cdot \frac{1}{5}$ A negative, a positive, and a $0$ . Probability: $\frac{2}{5} \cdot \frac{2}{5} \cdot \frac{1}{5}$ A negative, a positive, a positive, and a $0$ . Probability: $\frac{2}{5} \cdot \frac{2}{5} \cdot \frac{2}{5} \cdot \frac{1}{5}$
This sequence of a negative, positives, and $0$ continues infinitely. $\frac{\frac{1}{5}}{1-\frac{2}{5}} = \frac{1}{3}$ , so there's a $\frac{1}{3}$ probability that none of the terms following $a_n$ are positive.
We can compute the probability that $a_n$ is positive inductively.
Claim: The probability that $a_n$ is positive is $\frac{(\frac{4}{5})^n}{2}$ Base case: The probability that $a_1$ is positive is $\frac{\frac{4}{5}}{2} = \frac{2}{5}$ .
Inductive step: Given that the probability that $a_n$ is positive is $\frac{(\frac{4}{5})^n}{2}$ , we know that the probability that $a_n$ is negative is also $\frac{(\frac{4}{5})^n}{2}$ , since the probability that $a_n$ is not $0$ is $(\frac{4}{5})^n$ . From here, if $a_n$ is negative there is probability of $\frac{2}{5}$ that $a_{n+1}$ is positive, and if $a_n$ is positive there is probability of $\frac{2}{5}$ that $a_{n+1}$ is positive. Thus the probability that $a_{n+1}$ is positive is $\frac{2}{5} \cdot 2 \cdot \frac{(\frac{4}{5})^n}{2} = \frac{(\frac{4}{5})^{n+1}}{2}$ , completing our induction.
Combining the probability that $a_n$ is positive with the probability that all terms after $a_{n}$ are non-positive, we get that there is a probability of $\frac{(\frac{4}{5})^n}{2} \cdot \frac{1}{3}$ that $a_n$ is the last positive in our sequence. Note that this means that $a_n$ is also the greatest positive in our list, since magnitude cannot decrease unless we come across a $0$ .
Now we just need the expected value of $a_n$ , given that it is positive. Note that we can ignore if the spins leading up to $a_n$ are negative or positive, since we know that $a_n$ is positive. The $n$ terms that we take the product of to get $a_n$ have magnitude $\frac{5}{4}$ or $1$ with equal probability. The expected value of $a_n$ is thus $\frac{{n \choose n} \cdot (\frac{5}{4})^n + {n \choose {n-1}} \cdot (\frac{5}{4})^{n-1} + \ldots + {n \choose 0} \cdot (\frac{5}{4})^0}{2^n} = \frac{(\frac{5}{4} + 1)^n}{2^n} = (\frac{9}{8})^n$ .
Thus we can get our expected value by adding up the expected value of $a_n$ for all $n \geq 1$ : $\[ \sum_{n=1}^{\infty} (\frac{9}{8})^n \cdot \frac{(\frac{4}{5})^n}{2} \cdot \frac{1}{3} = \frac{3}{2}\]$ , but note that this discards the probability of our sequence being entirely non-positve after $a_0 = 1$ , which happens with a probability of $\frac{1}{3}$ , and thus contributes $1 \cdot \frac{1}{3}$ to the expected value. $\frac{3}{2} + \frac{1}{3} = \frac{11}{6} \implies \boxed{17}$</details>
|
[
"PROBABLY WRONG! but here goes\n\n<details><summary>eek</summary>Let $f(n)=\\text{E}[\\text{max}(\\text{largest number Ryan writes if he starts with the number }n,1)]$ . Then $$ f(1)=\\frac{f(-1.25)+f(-1)+f(0)+f(1)+f(1.25)}{5} $$ or $$ f(1)=\\frac{f(-1.25)+f(-1)+1+f(1.25)}{4}. $$ Notice that $f(1.25)=1.25f(1)$ (and $f(n)=nf(1)$ for all $n\\geq 1$ ), so we have $$ f(1)=\\frac{f(-1.25)+f(-1)}{4}+\\frac{4+5f(1)}{16}\\rightarrow 11f(1)-4=4(f(-1.25)+f(-1)). $$ Now consider $f(-1.25^n)$ for some nonnegative integer $n$ . We have $$ f(-1.25^n)=\\frac{1.25^{n+1}f(1)+1.25^nf(1)+1+f(-1.25^n)+f(-1.25^{n+1})}{5} $$ or $$ f(-1.25^n)=\\frac{1.25^{n+1}f(1)+1.25^nf(1)}{4} + \\frac{1}{4} + \\frac{f(-1.25^{n+1})}{4}. $$ Now, using the sum of infinite geometric series we get $$ f(-1.25^n)=\\frac{1}{3}+\\frac{9(1.25^n)}{11}f(1). $$ From earlier we had $$ 11f(1)-4=4(f(-1.25)+f(-1))=4\\left(\\frac{1}{3}+\\frac{9(1.25)}{11}f(1)+\\frac{1}{3}+\\frac{9(1)}{11}f(1)\\right) $$ so that $$ 11f(1)-4=\\frac{81}{11}f(1)+\\frac{8}{3}\\rightarrow f(1)=\\frac{11}{40}\\cdot \\frac{20}{3}=\\frac{11}{6}\\rightarrow \\boxed{017}. $$</details>\n\nhmmm my original ans was 22/9\n\nedit: ok that was kind of dumb :flushed:",
"<blockquote><details><summary>Part of your solution</summary>$$ \\frac{11}{40}\\cdot \\frac{20}{3}=\\frac{22}{3}\\rightarrow \\boxed{025}. $$</details></blockquote>\n\nDang, that very end though...",
"answer is 11/6 => 017 so i think i made a mistake somewhere\n\n@above oops.",
"BRUH IM SUCH AN IDIOT \nHWORAWJIIOASDJDISDJAIODJIASOD ",
"Let $p_0$ be the probability that Ryan never writes a number larger than $1.$ Now consider the probability $p_n$ that Ryan's largest number is $1.25^n$ for some positive integer $n.$ The number must first reach the magnitude of $1.25^n.$ The probability that this even occurs is $(2/3)^n$ (Think about the first $n$ rolls that are either $-1.25, 0, 1.25,$ each of these rolls must be the $-1.25$ or $1.25$ but not $0$ ). \n\nOnce this happens, there is probability $1/2$ that the number is positive, at which point there's $p_0$ probability that no number rolled is ever larger. There is probability $1/2$ that the number is negative, at which point there is $1/4$ probability that Ryan rolls a $-1$ before either of $-1.25, 1.25,$ or $0$ and gets the number $1.25^n,$ at which point there is again $p_0$ probability that no number rolled is ever larger. \n\nWe know that $$ \\frac{\\frac{2}{3}}{1-\\frac{2}{3}} \\cdot \\left( \\frac{1}{2} + \\frac{1}{2} \\cdot \\frac{1}{4} \\right) \\cdot p_0 + p_0 = 1, $$ solving for $p_0$ gives us $\\frac{4}{9}.$ Our answer is $$ \\frac{\\frac{2}{3} \\cdot \\frac{5}{4}}{1-\\frac{2}{3} \\cdot \\frac{5}{4}} \\cdot \\left( \\frac{1}{2} + \\frac{1}{2} \\cdot \\frac{1}{4} \\right) \\cdot \\frac{4}{9} + \\frac{4}{9} = \\frac{11}{6} \\implies \\boxed{17}. $$ Note: You could also solve for $p_0$ using states. ",
"Very COOL prolbem.\n",
"Bruh what is wrong with my sol??\n\nLet the first 0 occur on the $s+1$ step. Then, let $s_o$ be the last round that is positive. Note that the absolute value of this expression until it reaches 0 is increasing, which means that the maximum will be the last time that it is positive. The expected value of each round is $\\frac{9}{8} \\Longrightarrow (\\frac{9}{8})^{s_o}$ in total, and the probability of it being positive at $s_o$ is $\\frac{1}{2}$ by symmetry, and it being negative all after that is $(\\frac{1}{2})^{s-s_o}$ . Finally, there is a $(\\frac{4}{5})^s$ chance that the sequence does not reach 0 before $s+1$ , and a $\\frac{1}{5}$ chance that there is a 0 at step $s+1$ . This means that if we fix the variable $s$ and make $s_o$ the parameter, then the sum will be $\\sum_{j=1}^{s}{(\\frac{9}{8})^j*(\\frac{1}{2})^{s-j}*\\frac{1}{2}*(\\frac{4}{5})^s*\\frac{1}{5}}=\\frac{9}{50}*(\\frac{2}{5})^s*((\\frac{9}{4})^s-1)$ . Note that if $j=0$ , or it never becomes positive, the maximum it will eventually reach is 0, so we don't need to add anything. Now, summing this sum by varying $s$ , we get $\\sum_{s=1}^{\\infty}{\\frac{9}{50}*(\\frac{2}{5})^s*((\\frac{9}{4})^s-1)}=\\frac{3}{2}$ after simplifying. Note that $s=0$ yields a 0 as well, so we don't need to account. "
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
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{
"answer_score": 1120,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 DIME/2673248.json"
}
|
Let $\triangle ABC$ be acute with $\angle BAC = 45^{\circ}$ . Let $\overline{AD}$ be an altitude of $\triangle ABC$ , let $E$ be the midpoint of $\overline{BC}$ , and let $F$ be the midpoint of $\overline{AD}$ . Let $O$ be the center of the circumcircle of $\triangle ABC$ , let $K$ be the intersection of lines $DO$ and $EF$ , and let $L$ be the foot of the perpendicular from $O$ to line $AK$ . If $BL = 6$ and $CL = 8$ , find $AL^2$ .
*Proposed by **Awesome_guy***
|
<details><summary>Solution?</summary>Reflect $O$ over side $\overline{BC}$ to a point $P$ . Due to reflections, we have that $E$ is the midpoint of $\overline{OP}$ , and we are given that $E$ is also the midpoint of $\overline{BC}$ , so $BOCP$ is a parallelogram. We also have that $\angle BAC = 45^{\circ}$ , so $\angle BOC = 90^{\circ}$ . Since $BO=CO$ , we have that $BOCP$ is a square.**<span style="color:#f00">Claim:</span>** Points $A$ , $K$ , and $P$ are collinear.
*<span style="color:#00f">Proof:</span>* We have that $\angle APB = \angle OEB = 90^{\circ}$ , so $AD \parallel OP$ . Thus, $\triangle PKF \sim \triangle OKE$ . We also have that $AF:DF = OE:PE = 1:1$ , so $\triangle AKD \sim \triangle PKO$ by spiral similarity. It then follows that points $A$ , $K$ , and $P$ are collinear, as desired.
Now, we see that $\angle ALO = \angle PLO = 90^{\circ}$ , so $L$ lies on the circumcircle of $BOCP$ . By the Pythagorean Theorem, $$ BC^2=BL^2+CL^2=64+36=100 \implies BC=10. $$ By Ptolemy's Theorem on quadrilateral $BCLO$ ,
\begin{align*}
BC \cdot LO + CL \cdot BO &= BL \cdot CO
\implies 10 \cdot LO + 6 \cdot 5\sqrt{2} &= 8 \cdot 5\sqrt{2}
\implies LO &= \sqrt{2}.
\end{align*}
Finally, we have that $AO=5\sqrt{2}$ (radius of the circumcircle of $\triangle ABC$ ), $LO=\sqrt{2}$ , and $\angle ALO = 90^{\circ}$ , so by the Pythagorean Theorem, $$ AL^2=AO^2-LO^2=(5\sqrt{2})^2-(\sqrt{2})^2=50-2=\boxed{048}, $$ as requested.**Remark.** The idea of reflecting $O$ over side $\overline{BC}$ and the claim were both due to **ApraTrip**, and the proof of the claim was due to **richy**. The rest was written by me.</details>
Alternative formation: Prove that $AL^2=BL \cdot CL$ .
|
[
"it suffices to show that B, O, L, C concyclic and i half guessed this during the test but basically BLC is right by 6-8-10 (very cringe) and then BOC is 90 cuz BAC = 45 so B O L C concyclic. now it's simple by pythag and ptolemy => 48",
"official sol?",
"Guessed $50$ :wallbash: ",
"<blockquote>official sol?</blockquote>\n\n",
"SUS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! SUS!!!!!!!! SUSPICIOUS",
"Consider the circle $\\omega$ with diameter $BC,$ this passes through $O.$ Let $O'$ be the antipode. It's not hard to show $A,K,O'$ are collinear. So $L$ lies on $\\omega$ and also $LO_1$ bisects $\\angle BLC.$ So $\\angle BLA = \\angle ALC = 135^\\circ,$ but also $\\angle BAC = 45^\\circ.$ This is enough to show $\\triangle BLA \\sim \\triangle ALC$ so $AL$ is just the geometric mean of $BL$ and $CL.$ Our answer is $6 \\times 8 = \\boxed{48}.$ edit: compared to [SADGIME p14](https://artofproblemsolving.com/community/c2000465h2593532_hard_geo_p14)... I see that I did an oopsie with the difficulty of my mock :blush:",
"In my opinion, this was a very hard problem. Could not solve during the test.",
"<blockquote>In my opinion, this was a very hard problem. Could not solve during the test.</blockquote>\n\nI think this is much easier than the typical AIME p14, and could pass as a problem 11 or 12 (but not every problem in the final 5 range has to be placed absolutely perfectly).",
"<blockquote><blockquote>In my opinion, this was a very hard problem. Could not solve during the test.</blockquote>\n\nI think this is much easier than the typical AIME p14, and could pass as a problem 11 or 12 (but not every problem in the final 5 range has to be placed absolutely perfectly).</blockquote>\n\nI disagree. The key to solving this is to reflect O over BC, and that was a very nontrivial insight. I had first considered constructing the midpoints of AC and AB, but that got nowhere.",
"reflecting $O$ over $BC$ is not necessarily necessary",
"<blockquote>reflecting $O$ over $BC$ is not necessarily necessary</blockquote>\n\n..but very difficult without :)",
"<blockquote><blockquote>In my opinion, this was a very hard problem. Could not solve during the test.</blockquote>\n\nI think this is much easier than the typical AIME p14, and could pass as a problem 11 or 12 (but not every problem in the final 5 range has to be placed absolutely perfectly).</blockquote>\n\ni doubt it, the proof of the key claim (BOLC cyclic) is pretty hard and highly nontrivial\nhowever you can just cheese it by drawing an accurate diagram and assuming its true ",
"There's a typo in the solutions document. In line six of the solution, it says $\\triangle PKF \\sim \\triangle OKE$ ; however, I believe the correct formulation is $\\triangle DKF \\sim \\triangle OKE$ , instead.\n\nBy the way, a few comments on difficulty: I was not able to solve this during the test, but I don't believe that this is a particularly representative problem for the AIME -- mostly because many past AIME problems with such a flavor usually involved more nontrivial steps, instead of one key observation. Thus it's difficult to directly assign the problem an opinion to whether it is fit for its position.\n\nA note on motivation as well: though it feels obvious to me that such a claim of the concyclic would have existed, in hindsight, I think the true motivation for constructing the point $P$ comes from the line $\\overline{AK}$ .\n\nIn particular, it's pretty common to extend lines of the form $\\overline{AO}$ , where $O$ is the intersection of some two arbitrary lines. The $\\overline{OE}$ and $\\overline{AD}$ parallel exemplifies this unbelievably well.",
"Because $DO \\cap EF$ is named $K$ , you guess $AK$ is a symmedian and then you're done. This actually works for many problems!!\n\nThe proof isn't hard though. If you reflect $D$ and $O$ across $F$ and $E$ respectively, you get the points $A$ and $G$ where $G$ is the intersection of the tangents because $\\angle{BAC} = 45^{\\circ}$ . This means $AG$ , $FE$ , $DO$ are concurrent and $AG$ is the symmedian."
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1076,
"boxed": false,
"end_of_proof": false,
"n_reply": 15,
"path": "Contest Collections/2022 Contests/2022 DIME/2673249.json"
}
|
For positive integers $n$ , let $f(n)$ denote the number of integers $1 \leq a \leq 130$ for which there exists some integer $b$ such that $a^b-n$ is divisible by $131$ , and let $g(n)$ denote the sum of all such $a$ . Find the remainder when $$ \sum_{n = 1}^{130} [f(n) \cdot g(n)] $$ is divided by $131$ .
*Proposed by **ApraTrip***
|
<details><summary>Author's Solution</summary>Let all sums in this solution be taken modulo $131$ , and let $\text{ord}_{131}(x)$ be the order of $x$ modulo $131$ .
We claim that there exists an integer $s$ such that $r^{s} \equiv x \pmod{131}$ if and only if $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ . Let $g$ be a primitive root $\pmod{131}$ . Notice that when $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ , then $\tfrac{130}{\text{ord}_{131}(r)} \mid \tfrac{130}{\text{ord}_{131}(x)}$ . Thus, since $r \equiv g^{\tfrac{130a}{\text{ord}_{131}(r)}}$ and $x \equiv g^{\tfrac{130b}{\text{ord}_{131}(x)}}$ for some $a$ and $b$ where $\gcd(a, 130) = 1$ and $\gcd(b, 130) = 1$ , if we let $c \equiv \tfrac{b}{a} \pmod{130}$ and $d = \frac{\tfrac{130}{\text{ord}_{131}(x)}}{\tfrac{130}{\text{ord}_{131}(r)}}$ , then $$ r^{cd} = g^{\frac{130ac}{\text{ord}_{131}(x)}} = g^{\frac{130b}{\text{ord}_{131}(x)}} = x. $$ Thus, there exists an integer $s$ such that $r^{s} \equiv x \pmod{131}$ if $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ .
Now, notice that if $r^{s} \equiv x \pmod{131}$ , then $$ x^{\text{ord}_{131}(r)} \equiv r^{s \cdot \text{ord}_{131}(r)} \equiv 1 \pmod{131}, $$ which implies that $\text{ord}_{131}(x) \mid \gcd(\text{ord}_{131}(r), 130) = \text{ord}_{131}(r)$ , as $\text{ord}_{131}(r) \mid 130$ . Thus, our claim is true.
Thus, $f(x)$ is the number of integers $1 \le r \le 130$ such that $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ , and $g(x)$ is the sum of all integers $1 \le r \le 130$ such that $\text{ord}_{131}(x) \mid \text{ord}_{131}(r)$ . We will deal with both functions separately, and put them together at the end. $f(x)$ : It's well known that there are $\varphi(n)$ numbers $\pmod{p}$ with order $n$ where $p$ is prime and $n \mid p-1.$ Thus,
\begin{align*}
f(x) &= \sum_{\substack{\text{ord}_{131}(x) \mid d, d \mid 130}} \varphi(d) = \sum_{d \mid \tfrac{130}{\text{ord}_{131}(x)}} [\varphi(\text{ord}_{131}(x)) \cdot \varphi(d)]
&= \varphi(\text{ord}_{131}(x)) \cdot \sum_{d \mid \tfrac{130}{\text{ord}_{131}(x)}} \varphi(d) = \frac{130\varphi(\text{ord}_{131}(x))}{\text{ord}_{131}(x)},
\end{align*}
where the second equality follows as $\varphi(x)$ is multiplicative and any two factors of $130$ which multiply to be a factor of $130$ are relatively prime (as $130$ isn't divisible by any perfect powers). $g(x)$ : It's known that for all primes $p$ such that that $p-1$ isn't divisible by any perfect powers, the sum of all numbers with order $d$ (for some $d \mid p-1$ ) is $(-1)^{\text{the number of prime factors of } d}$ (see 2021 DIME #14). Thus, in order to find $g(x)$ , we'll do casework on the number of factors of $\text{ord}_{131}(x)$ , while utilizing the fact that $130$ has three prime factors.**Case 1:** $\text{ord}_{131}(x)$ has $0$ prime factors.
In this case, there would be $1$ multiple of $\text{ord}_{131}(x)$ with $0$ factors that divides $130$ , $3$ multiples of $\text{ord}_{131}(x)$ with one factor that divide $130$ , $3$ multiples of $\text{ord}_{131}(x)$ with two factors that divide $130$ , and $1$ multiple of $\text{ord}_{131}(x)$ with three factors that divides $130$ . Thus, $g(x) = 1-3+3-1 = 0$ when $\text{ord}_{131}(x)$ has $0$ prime factors.**Case 2:** $\text{ord}_{131}(x)$ has $1$ prime factor.
In this case, there would be $1$ multiple of $\text{ord}_{131}(x)$ with one factor that divides $130$ , $2$ multiples of $\text{ord}_{131}(x)$ with two factors that divide $130$ , and $1$ multiple of $\text{ord}_{131}(x)$ with three factors that divides $130$ . Thus, $g(x) = -1+2-1 = 0$ when $\text{ord}_{131}(x)$ has $1$ prime factor.**Case 3:** $\text{ord}_{131}(x)$ has $2$ prime factors.
In this case, there would be $1$ multiple of $\text{ord}_{131}(x)$ with two factors that divides $130$ , and $1$ multiple of $\text{ord}_{131}(x)$ with three factors that divides $130$ . Thus, $g(x) = 1-1 = 0$ when $\text{ord}_{131}(x)$ has $2$ prime factors. At this point, it might seem that $g(x)$ is $0$ for all $x$ (which would make this a very trolly problem), but this is not true.**Case 4:** $\text{ord}_{131}(x)$ has $3$ prime factors.
In this case, there would be $1$ multiple of $\text{ord}_{131}(x)$ with three factors that divides $130$ . Thus, $g(x) = -1$ when $\text{ord}_{131}(x)$ has three prime factors.
Thus, $g(x) = 0$ when $x$ is not a primitive root, and $g(x) = -1$ when $x$ is a primitive root. Thus, our sum becomes
\begin{align*}
\sum_{\text{ord}_{131}(x) = 130} [f(x) \cdot g(x)] &= - \sum_{\text{ord}_{131}(x) = 130} f(x)
&= - \sum_{\text{ord}_{131} (x) = 130} \frac{130 \varphi(130)}{130}
&= -\varphi(130)^2
&= -2304.
\end{align*}
Thus, the requested remainder when our sum is divided by $131$ is $\boxed{054}$ .</details>
If you have any questions about this solution or if you find any issues with it, please send a private message to **ApraTrip** (with the subject "Regarding 2022 DIME #15"), as I do not know what is going on here... :maybe:
|
[
"any solutions to this?",
"i think its smt with primitive roots",
"sol pls?",
"neither do I @above\n\nyou guys rlly should've made a DJMO that included this problem",
"yeah thats a rly hard order problem\n\n",
"<blockquote>neither do I @above\n\nyou guys rlly should've made a DJMO that included this problem</blockquote>\n\nBut then we would have to make 5 other olympiad problems... However, I agree that this problem is insane, I wouldn't expect it on a real AIME. ",
"While I do think that this is a bit too intensive for AIME, I also don't really think it's in the style of an Olympiad problem.\n\nThat being said, hopefully the scary length of the official solution and the notation stuff doesn't immediately scare people away. Assuming that orders and primitive roots are allowed on AIME (which at least 2021 DIME #14 used) then I think that it's some really clever use of ideas. However, the entirety of 2021 DIME #14 being summed up in one sentence in this solution makes me feel pretty intimidated.\n\nReminds me of that Season 2 OTIE #15 but I honestly think that one was even less reasonable... :wacko:\n\n(I'm saying all this from observing stuff; I'm definitely not at the level to solve most AIME #15s anyway...)\n\n(Wow I just contradicted myself because I said earlier that it's some really clever use of ideas even though I don't fully understand it...)\n\nHelp someone please share your own solution to this problem.",
"aime 2019 #14 used orders but tbh that is nothing compared to this",
"<blockquote>aime 2019 #14 used orders</blockquote>\n\nOh lol right\n\n<blockquote>but tbh that is nothing compared to this</blockquote>\n\nIndeed",
"<blockquote>Went back through solving the problem to understand what could be done to write a clearer solution. I think the solution becomes a lot easier to write if one states at the beginning: \"There exists a primitive root $q$ modulo 131. Then all of the numbers from 1 to 130 can be rewritten as $1, q, q^2, \\ldots, q^{129}$ \". Then we can basically go through all of the work just by referring to the exponents, without having to worry about order notation.</blockquote>",
"**Fun Fact.** This problem and 2021 DIME #14 are the only two DIME problems which ask for a remainder when divided by something other than $1000$ . This may seem like something that AIME would never do (count me in as one who thought that too), but here are some past AIME problems which do this:\n\n<details><summary>Spoilers</summary>\n\n- 1983 AIME #6 - remainder upon division by $49$ .\n- 2004 AIME I #1 - sum of all remainders upon division by $37$ (not really the same but whatever).\n</details>\n\nOops, actually it is pretty unusual. I'm guessing that it's because division by some $n$ less than $1000$ would automatically eliminate any answer choice greater than or equal to $n$ .",
"<blockquote>While I do think that this is a bit too intensive for AIME, I also don't really think it's in the style of an Olympiad problem.\n\nThat being said, hopefully the scary length of the official solution and the notation stuff doesn't immediately scare people away. Assuming that orders and primitive roots are allowed on AIME (which at least 2021 DIME #14 used) then I think that it's some really clever use of ideas. However, the entirety of 2021 DIME #14 being summed up in one sentence in this solution makes me feel pretty intimidated. </blockquote>\n\nYea this problem was definitely unsolvable on a real test it gives me 28-30 OMO vibes :P",
"lesgo I actually solved this and didn't silly ! (took me >1 hour though)\n\nHowever, I took this mock test untimed so probably couldn't have done so if this were the real AIME\n\nwould write a solution but i'm too lazy and it would take like 2 hours",
"someone named pi 28715871825 is reading",
"<blockquote>someone named pi 28715871825 is reading</blockquote>\n\n?????????",
"<blockquote><blockquote>someone named pi 28715871825 is reading</blockquote>\n\n?????????</blockquote>\n\nhi",
"<blockquote>\nhi</blockquote>\n\n*triple speed dual mini wave*",
"<blockquote><blockquote><blockquote>someone named pi 28715871825 is reading</blockquote>\n\n?????????</blockquote>\n\nhi</blockquote>\n\nwho is that",
"<blockquote><blockquote>\nhi</blockquote>\n\n*triple speed dual mini wave*</blockquote>\n\nOH NO",
"I felt like the problem should have just asked for the sum of $f(n)$ (and removing the mod 131 for answer extraction, since it wouldn't be necessary with this change). The $g(n)$ stuff from DIME #14 2020-2021 should have stayed out of this and its own separate problem. But if you were to change 131 to a prime $p$ such that $p-1$ is the product of distinct primes, then \\[\\sum_{n=1}^{p-1}f(n) = \\prod_{q|(p-1)}(q^2 - q + 1)\\]",
"New Version: (less complicated than it needs to be while keeping it separate from the other DIME problem)\n\nLet $a$ and $n$ be integers such that $1 \\leq a, n \\leq 130$ . Suppose $X$ is the number of pairs $(a, n)$ for which there exists an integer $b$ such that $a^b-n$ is divisible by $131$ . Find the remainder when $X$ is divided by $1000$ .\n\n<details><summary>Solution</summary>Let $g$ be a primitive root modulo $131$ , which exists, since $131$ is prime. Recall that $g^{130} \\equiv 1 \\pmod{131}$ and $\\{g^1, g^2, g^3, ..., g^{130}\\} \\equiv \\{1, 2, 3, ..., 130\\}$ by the essential properties of a primitive root. Thus, $g^c \\pmod{131}$ is uniquely determined by $c \\pmod{130}$ .\n\nLet $a \\equiv g^i \\pmod{131}$ and $n \\equiv g^j \\pmod{131}$ for some integers $i$ and $j$ such that $1 \\leq i, j \\leq 130$ . Then, for there to exist an integer $b$ such that $a^b \\equiv n \\pmod{131}$ , it follows that $g^{ib} \\equiv g^j \\pmod{131}$ or $ib \\equiv j \\pmod{130}$ . The prime factorization of $130$ is $2 \\cdot 5 \\cdot 13$ , which is a product of distinct primes.\n\nLet $p$ be an arbitrary prime that divides $130$ . It is required for $ib \\equiv j \\pmod{p}$ . By Chinese Remainder Theorem, the pair $(i, j) \\pmod{p}$ can be chosen separately for each prime $p$ that divides $130$ , and then the congruences can be merged to determine the actual values of $i$ and $j$ .\n\nClearly, $b$ cannot exist if $i$ is divisible by $p$ and $j$ is not divisible by $p$ .\n\n\n- If $p \\mid i$ , then $p \\mid j$ must follow. Clearly, $b$ exists when this happens.\n- If $p \\nmid i$ , then all values of $j$ will allow $b$ to exist. This can be seen by taking a value of $b$ such that $b \\equiv i^{-1}j \\pmod{p}$ .\n\nThere is $1$ pair $(i, j) \\pmod{p}$ in the first case and $(p-1) \\cdot p$ pairs $(i, j) \\pmod{p}$ pairs in the second case. So there are $p^2 - p + 1$ ways to determine $(i, j) \\pmod{p}$ .\n\nSince $(i, j) \\pmod{p}$ can be determined independently for each prime $p$ that divides $130$ , it follows that \\[X = \\prod_{p \\mid 130}(p^2 - p + 1) = (2^2 - 2 + 1)(5^2 - 5 + 1)(13^2 - 13 + 1) = 3 \\cdot 21 \\cdot 157 = 9891.\\] The requested remainder is $891$ .</details>\n\n",
"<blockquote>New Version: (less complicated than it needs to be while keeping it separate from the other DIME problem)\n\nLet $a$ and $n$ be integers such that $1 \\leq a, n \\leq 130$ . Suppose $X$ is the number of pairs $(a, n)$ for which there exists an integer $b$ such that $a^b-n$ is divisible by $131$ . Find the remainder when $X$ is divided by $1000$ .\n\n<details><summary>Solution</summary>Let $g$ be a primitive root modulo $131$ , which exists, since $131$ is prime. Recall that $g^{130} \\equiv 1 \\pmod{131}$ and $\\{g^1, g^2, g^3, ..., g^{130}\\} \\equiv \\{1, 2, 3, ..., 130\\}$ by the essential properties of a primitive root. Thus, $g^c \\pmod{131}$ is uniquely determined by $c \\pmod{130}$ .\nLet $a \\equiv g^i \\pmod{131}$ and $n \\equiv g^j \\pmod{131}$ for some integers $i$ and $j$ such that $1 \\leq i, j \\leq 130$ . Then, for there to exist an integer $b$ such that $a^b \\equiv n \\pmod{131}$ , it follows that $g^{ib} \\equiv g^j \\pmod{131}$ or $ib \\equiv j \\pmod{130}$ . The prime factorization of $130$ is $2 \\cdot 5 \\cdot 13$ , which is a product of distinct primes.\n\nLet $p$ be an arbitrary prime that divides $130$ . It is required for $ib \\equiv j \\pmod{p}$ . By Chinese Remainder Theorem, the pair $(i, j) \\pmod{p}$ can be chosen separately for each prime $p$ that divides $130$ , and then the congruences can be merged to determine the actual values of $i$ and $j$ .\n\nClearly, $b$ c\nannot exist if $i$ is divisible by $p$ and $j$ is not divisible by $p$ .\n\n\n- If $p \\mid i$ , then $p \\mid j$ must follow. Clearly, $b$ exists when this happens.\n- If $p \\nmid i$ , then all values of $j$ will allow $b$ to exist. This can be seen by taking a value of $b$ such that $b \\equiv i^{-1}j \\pmod{p}$ .\n\nThere is $1$ pair $(i, j) \\pmod{p}$ in the first case and $(p-1) \\cdot p$ pairs $(i, j) \\pmod{p}$ pairs in the second case. So there are $p^2 - p + 1$ ways to determine $(i, j) \\pmod{p}$ .\n\nSince $(i, j) \\pmod{p}$ can be determined independently for each prime $p$ that divides $130$ , it follows that \\[X = \\prod_{p \\mid 130}(p^2 - p + 1) = (2^2 - 2 + 1)(5^2 - 5 + 1)(13^2 - 13 + 1) = 3 \\cdot 21 \\cdot 157 = 9891.\\] The requested remainder is $891$ .</details></blockquote>\n\nThis is really XYZIME #15.",
"Oh right oops"
] |
[
"origin:aops",
"2022 Contests",
"2022 DIME"
] |
{
"answer_score": 1226,
"boxed": false,
"end_of_proof": false,
"n_reply": 24,
"path": "Contest Collections/2022 Contests/2022 DIME/2673252.json"
}
|
Let $f:\mathbb{N}^*\rightarrow \mathbb{N}^*$ be a function such that $\frac{x^3+3x^2f(y)}{x+f(y)}+\frac{y^3+3y^2f(x)}{y+f(x)}=\frac{(x+y)^3}{f(x+y)},~(\forall)x,y\in\mathbb{N}^*.$ $a)$ Prove that $f(1)=1.$ $b)$ Find function $f.$
|
Giả sử tồn tại hàm số thỏa mãn $$ \frac{x^3+3x^2f(y)}{x+f(y)}+\frac{y^3+3y^2f(x)}{y+f(x)}=\frac{(x+y)^3}{f(x+y)},\forall x,y\in \mathbb{Z^+}.(1) $$ Từ $(1)$ cho $x=y=1$ ta được $$ \frac{2(1+3f(1))}{1+f(1)}=\frac{8}{f(2)} $$ Hay ta được $$ f(2)=\frac{4(1+f(1))}{1+3f(1)}. $$ Suy ra $(1+f(1))|(4+4f(1))$ hay $(1+3f(1))|8$ suy ra $f(1)=1$ nên suy ra $f(2)=2.$ Ta sẽ chứng minh rằng $f(x)=x,\forall x\in \mathbb{Z^+}.$ Thật vật với $x=1,2$ thì khẳng định hiển nhiên đúng.Giả sử khẳng định đúng đến $x=k,k\in \mathbb{Z^+}.$ Thì ta chứng minh rằng $k+1$ cũng đúng.Từ $(1)$ thay $x=k,y=1$ ta được $$ \frac{k^3+3x^2}{k+1}+\frac{1+3k}{k+1}=\frac{(k+1)^3}{f(k+1)},\forall k\in \mathbb{Z^+}. $$ Hay ta được $$ f(k+1)=k+1,\forall k\in \mathbb{Z^+}. $$ Vậy theo quy nạp ta có tất cả hàm số cần tìm là $$ f(x)=x,\forall x\in \mathbb{Z^+}. $$
|
[
"Let $a=f(1)$ . $$ (x,y)=(1,1) \\rightarrow \\dfrac{2(1+3a)}{1+a}=\\dfrac{8}{f(2)} \\rightarrow 1+3a\\; |\\; 4a+4 \\rightarrow f(1)=a=1. $$ Using induction with the relation $(x,1)$ we deduce that $f(x)=x$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] |
{
"answer_score": 48,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810677.json"
}
|
$a)$ Prove that $2x^3-3x^2+1\geq 0,~(\forall)x\geq0.$ $b)$ Let $x,y,z\geq 0$ such that $\frac{2}{1+x^3}+\frac{2}{1+y^3}+\frac{2}{1+z^3}=3.$ Prove that $\frac{1-x}{1-x+x^2}+\frac{1-y}{1-y+y^2}+\frac{1-z}{1-z+z^2}\geq 0.$
|
$a)$ The expression factorizes as $(x-1)^2(2x+1)$ which is clearly positive for $x\geq 0$ . Moreover, equality occurs for $x=1$ . $b)$ Using the result from point $a)$ , we get $1-x^3\leq \frac{3}{2}(1-x^2)$ , thus $$ 3=\sum_{cyc}\frac{2}{1+x^3}\implies 0=\sum_{cyc}\frac{1-x^3}{1+x^3}\leq \frac{3}{2}\sum_{cyc}\frac{1-x^2}{1+x^3}=\frac{3}{2}\sum_{cyc}\frac{1-x}{1-x+x^2} $$ which is exactly what we wanted to prove. Equality occurs for $x=y=z=1$ (which does verify the constraint over the variables).
|
[
"Let $x,y,z\\geq 0$ such that $\\frac{2}{1+x^3}+\\frac{2}{1+y^3}+\\frac{2}{1+z^3}=3.$ Prove that $$ \\frac{2-x}{2-x+x^2}+\\frac{2-y}{2-y+y^2}+\\frac{2-z}{2-z+z^2}\\leq \\frac{3}{2} $$ \n"
] |
[
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] |
{
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810679.json"
}
|
$a)$ Solve over the positive integers $3^x=x+2.$ $b)$ Find pairs $(x,y)\in\mathbb{N}\times\mathbb{N}$ such that $(x+3^y)$ and $(y+3^x)$ are consecutive.
|
Part a) $x=1$ only solution. For $x>1, LHS>RHS$ Part b)
If $(x+3^y)$ and $(y+3^x)$ are consecutive $\Longrightarrow x+3^y+1= y+3^x$ Case $x=y$ $\Longrightarrow x+3^x+1= x+3^x \Longrightarrow 1= 0$ impossible
Case $y>x$ Let $y=x+n\Longrightarrow x+3^n3^x+1=x+n+3^x\Longrightarrow 3^x=\frac{n-1}{3^n-1}<1$ Contradiction
Case $x>y$ Let $x=y+n\Longrightarrow y+n+3^y+1=y+3^n3^y\Longrightarrow3^y=\frac{n+1}{3^n-1}$ $n=1\Longrightarrow 3^y=\frac{2}{2}\Longrightarrow 3^y=1\Rightarrow x=1, y=0$ is the only solution if $0\in Z$ .
If $n>1, 3^y=\frac{n+1}{3^n-1}<1$ . Contradiction.
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] |
{
"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810681.json"
}
|
Let $e$ be the identity of monoid $(M,\cdot)$ and $a\in M$ an invertible element. Prove that
[list=a]
[*]The set $M_a:=\{x\in M:ax^2a=e\}$ is nonempty;
[*]If $b\in M_a$ is invertible, then $b^{-1}\in M_a$ if and only if $a^4=e$ ;
[*]If $(M_a,\cdot)$ is a monoid, then $x^2=e$ for all $x\in M_a.$ [/list]
*Mathematical Gazette*
|
Notice that the set can be redefined as $M_a=\{x\in M : x^2=a^{-2}\}$ . $a)$ Observe that $a^{-1}\in M_a$ . $b)$ Since $b\in M_a$ , we have $b^2=a^{-2}$ , thus $$ b^{-1}\in M_a\iff b^{-2}=a^{-2}\iff e=b^2b^{-2}=a^{-2}a^{-2}=a^{-4}\iff a^4=e. $$ $c)$ Let $f$ be the identity of $M_a$ . $f\in M_a$ , thus $a^{-2}=f^2=f$ , implying that $a^{-4}=(a^{-2})^2=f^2=a^{-2}$ , and finally $a^{-2}=e$ . From the definition of $M_a$ , the result follows.
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] |
{
"answer_score": 32,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810712.json"
}
|
Let $(G,\cdot)$ be a group and $H\neq G$ be a subgroup so that $x^2=y^2$ for all $x,y\in G\setminus H.$ Show that $(H,\cdot)$ is an Abelian group.
|
Let $x \in G \setminus H.$ For any $a \in H,$ we have $ax \notin H$ and so $(ax)^2=x^2,$ which gives $axa=x$ hence $$ a=xa^{-1}x^{-1}, \ \ \ \ \ \forall a \in H. $$ So if $a,b \in H,$ then $$ xb^{-1}a^{-1}x^{-1}=x(ab)^{-1}x^{-1}=ab=(xa^{-1}x^{-1})(xb^{-1}x^{-1})=xa^{-1}b^{-1}x^{-1}, $$ which gives $b^{-1}a^{-1}=a^{-1}b^{-1}$ and so $ab=ba,$ i.e. $H$ is abelian.
The condition given in the problem is strong, and so I think we can say more about $G.$ For example, an obvious one is that $x^4=1$ for all $x \notin H$ because if $x \notin H,$ then $x^{-1} \notin H$ and so $x^2=x^{-2},$ which gives $x^4=1.$
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] |
{
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810713.json"
}
|
Find all values of $n\in\mathbb{N}^*$ for which \[I_n:=\int_0^\pi\cos(x)\cdot\cos(2x)\cdot\ldots\cdot\cos(nx) \ dx=0.\]
|
[
"[url=https://artofproblemsolving.com/community/c7h1646205p10391941] here [ \\url ]\n\nhttps://artofproblemsolving.com/community/c7h1646205p10391941"
] |
[
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] |
{
"answer_score": 0,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810715.json"
}
|
|
Let $I\subseteq \mathbb{R}$ be an open interval and $f:I\to\mathbb{R}$ a strictly monotonous function. Prove that for all $c\in I$ there exist $a,b\in I$ such that $c\in (a,b)$ and \[\int_a^bf(x) \ dx=f(c)\cdot (b-a).\]
|
Let $I=(k,l)$ and suppose $f$ is strictly decreasing (otherwise, look at the function $-f$ ). Consider some real number (which we will define later) $y_0\in (c,l)$ and let $g:(k,c)\rightarrow \mathbb{R}$ with $$ g(x)=\int_x^{y_0}f(t)dt - f(c)(y_0-x). $$ Observe that $g(c)<0$ since from the monotonicity of $f$ , $\int_c^{y_0}f(t)dt<f(c)(y_0-c)$ . Now $$ g(k)=\int_k^{y_0}f(t)dt - f(c)(y_0-k)=\int_k^c f(t)dt - f(c)(c-k)+\int_c^{y_0}f(t)dt - f(c)(y_0-c). $$ Letting $h(x)=\int_c^x f(t)dt - f(c)(x-c)$ , one can see that $\lim_{x\rightarrow c^{+}}h(x)=0$ , and since $\int_k^c f(t)dt - f(c)(c-k)>0$ (from the fact that $f$ is decreasing), there exists some $y_0\in(c,l)$ such that $g(k)>0$ . But now, $g$ is continuous (both the integral and the linear function are continuous), the Darboux property (MVT) implying that there exists some $r\in(k,c)$ such that $g(r)=0$ , which is exactly what we wanted to prove.
|
[
"Assume $f$ is strictly increasing (otherwise work with $-f$ ). Since $I$ is open, there exists $\\delta>0$ such that $(c-\\delta, c+\\delta) \\subseteq I$ . Fix such a $\\delta$ and consider the function (for $x>c$ )\n\\[\ng(x) = \\int_{c-\\delta}^{x} f(t)\\mathrm{d}t.\n\\]\nSince $f$ is strictly increasing, $\\int_{c-\\delta}^{c} f(t) \\mathrm{d}t < f(c)\\cdot \\delta$ , and there exists $\\varepsilon>0$ such that $\\int_{c-\\delta}^{c} f(t) \\mathrm{d}t + \\varepsilon< f(c)\\cdot \\delta$ .\n Next, there exists $\\delta_0 \\in (0, \\delta)$ such that $x\\in (c, c+\\delta_0) \\implies \\int_{c}^{x} f(t)\\mathrm{d}t < (x-c)f(c+\\delta) < {\\varepsilon}$ . So, we get that for a fixed $b \\in (c, c+\\delta_0)$ ,\n\\[\n g(b) = \\int_{c-\\delta}^{c} f(t)\\mathrm{d} t + \\int_{c}^{b} f(t)\\mathrm{d} t < \\int_{c-\\delta}^{c} f(t)\\mathrm{d} t + \\varepsilon < f(c)\\cdot \\delta < f(c)(b-(c-\\delta)).\n\\]\nNow consider the function\n\\[\nh(x) = \\int_{x}^{b} f(t)\\mathrm{d}t - f(c)(b-x).\n\\]\nWe have $h(c-\\delta)<0$ (from $g(b)$ 's analysis above) and $h(c)>0$ , which means there exists $a \\in (c-\\delta, c)$ such that $h(a)=0$ , since $h$ is continuous. That is, $a<c<b$ , and\n\\[\n\\int_{a}^{b} f(t) \\mathrm{d}t=f(c)(b-a).\n\\]"
] |
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"origin:aops",
"2022 Contests",
"2022 District Olympiad"
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{
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810716.json"
}
|
Let $f,g:\mathbb{R}\to\mathbb{R}$ be functions which satisfy \[\inf_{x>a}f(x)=g(a)\text{ and }\sup_{x<a}g(x)=f(a),\]for all $a\in\mathbb{R}.$ Given that $f$ has Darboux's Property (intermediate value property), show that functions $f$ and $g$ are continuous and equal to each other.
*Mathematical Gazette*
|
<blockquote>Let $f,g:\mathbb{R}\to\mathbb{R}$ be functions which satisfy \[\inf_{x>a}f(x)=g(a)\text{ and }\sup_{x<a}g(x)=f(a),\]for all $a\in\mathbb{R}.$ Given that $f$ has Darboux's Property (intermediate value property), show that functions $f$ and $g$ are continuous and equal to each other.
*Mathematical Gazette*</blockquote>**<span style="color:#f00">Claim 01.</span>** $f,g$ are increasing
*Proof.* Indeed, suppose $a > b$ , then we have
\begin{align*}
f(a) &= \sup_{x < a} g(x) \ge \sup_{x < b} g(x) = f(b)
g(a) &= \inf_{x > a} f(x) \ge \inf_{x > b} f(x) = g(b)
\end{align*}
---------------------------
Indeed, as $f,g$ have Darboux's Property and are increasing, it follows that $f$ and $g$ are continuous.
Let $x > y$ . We know that $\inf_{x > y} f(x) = g(y)$ implies $f(x) \ge g(y)$ . Furthermore,
\begin{align*}
f(x) \ge f(y) \ \forall x > y &\implies g(y) = \inf_{x > y} f(x) \ge f(y)
f(x) \ge g(y) \ \forall y < x &\implies f(x) \ge \lim_{y \to x} g(y) = g(x)
\end{align*}
and thus we conclude that $f = g$ .
|
[
"https://ssmr.ro/",
"<blockquote>[ see here ](https://ssmr.ro/) </blockquote>\n\nRomanian maths problems are very nice",
"Just a question, what is the name of this competition, can you attach website or something because these questions are college level yet they are for grade 11, so I want to know. Also i search mathematical olympiad in Romania and it showed only the selection competition for IMO",
"<blockquote>Just a question, what is the name of this competition, can you attack website or something because these questions are college level yet they are for grade 11, so I want to know. Also i search mathematical olympiad in Romania and it showed only the selection competition for IMO</blockquote>\nThis is the district round of the Romanian Mathematical Olympiad. Romanian students study all of this in the 11th grade, so we don't consider it college level. I am not sure what you mean by \"attack the website\". Assuming you meant attach, the Romanian Mathematical Society's Website is here: [https://ssmr.ro](https://ssmr.ro) You can access the problems and solutions of the district round here: [https://ssmr.ro/onm2022](https://ssmr.ro/onm2022). It's not in English though.",
"One more questions, which olympiad is this? is this selection for International mathematics olympiad?? because IMO doesn't require calculus and all those things. Is this entirely different olympiad for grade 11 and 12, then? \noh wow then that means the exams in Romania is completely different when it comes to selection for IMO. So this kind of content are taught in class 11 and 12 all over the Romania. Can you attach the curriculum of Romania mathematics to have some idea.",
"<blockquote>One more questions, which olympiad is this? is this selection for International mathematics olympiad?? because IMO doesn't require calculus and all those things. Is this entirely different olympiad for grade 11 and 12, then?</blockquote>\nNo, as I said above, it is the *district phase* of the olympiad. This is one of the phases required to go to the national phase of the olympiad, after which you go to the TSTs which select the IMO team.",
"<blockquote>Can you attach the curriculum of Romania mathematics to have some idea.</blockquote>\nIf you ask about class exams (like Baccalaureate), the curriculum is like this:\n-- for grade 9: linear and quadratic functions, trigonometry, vectors \n-- for grade 10: injective, surjective, bijective functions, logarithms, complex numbers, and binomial coefficients \n-- for grade 11: basic analysis (like limits and derivatives), and basic linear algebra (matrix operations and determinants)\n-- for grade 12: integrals, and basic abstract algebra.\n(as you can see the geometry was almost completely removed from high school curriculum :()\n\nIn fact, the Romanian baccalaureate (at math) is much easier than IB or A levels, in my opinion.\n\nBut if you ask about math olympiad curriculum, the things are different. Probably the idea of the Ministry of Education is that math olympiad curriculum should be based on the class curriculum, but the math olympiad curriculum contains some other hard things (like Holder's inequality for grade 9, or polynomials for grade 10, Taylor formula with Lagrange remainder for grade 11, Lebesgue criterion for integrability at grade 12) . As a consequence, the olympiad problems at grades 11-12 are basically college math problems. \nSo in grades 11-12 (in Romania) you have to study college mathematics to get to IMO TSTs, this being one of the reasons why the Romanian IMO team does not have too many 12th grade students. \nAnyway, the high school math olympiad curriculum (in Romanian) is here: [https://ssmr.ro/files/onm2022/Programa_ONGM_(liceu)_2022.pdf](https://ssmr.ro/files/onm2022/Programa_ONGM_(liceu)_2022.pdf)\n\n",
"\n<details><summary>@#8</summary>[**Miquel-point**](aops.com/community/user/779730) · Today at 3:12 AM [(view)](aops.com/community/p24859462)<span style=\"color:transparent\">helo</span>\n<blockquote>Can you attach the curriculum of Romania mathematics to have some idea.</blockquote>\nIf you ask about class exams (like Baccalaureate), the curriculum is like this:\n-- for grade 9: linear and quadratic functions, trigonometry, vectors \n-- for grade 10: injective, surjective, bijective functions, logarithms, complex numbers, and binomial coefficients \n-- for grade 11: basic analysis (like limits and derivatives), and basic linear algebra (matrix operations and determinants)\n-- for grade 12: integrals, and basic abstract algebra.\n(as you can see the geometry was almost completely removed from high school curriculum :()\n\nIn fact, the Romanian baccalaureate (at math) is much easier than IB or A levels, in my opinion.\n\nBut if you ask about math olympiad curriculum, the things are different. Probably the idea of the Ministry of Education is that math olympiad curriculum should be based on the class curriculum, but the math olympiad curriculum contains some other hard things (like Holder's inequality for grade 9, or polynomials for grade 10, Taylor formula with Lagrange remainder for grade 11, Lebesgue criterion for integrability at grade 12) . As a consequence, the olympiad problems at grades 11-12 are basically college math problems. \nSo in grades 11-12 (in Romania) you have to study college mathematics to get to IMO TSTs, this being one of the reasons why the Romanian IMO team does not have too many 12th grade students. \nAnyway, the high school math olympiad curriculum (in Romanian) is here: [https://ssmr.ro/files/onm2022/Programa_ONGM_(liceu)_2022.pdf](https://ssmr.ro/files/onm2022/Programa_ONGM_(liceu)_2022.pdf)\n\n-----------\n<span style=\"color:#5b7083\">[aops]x[/aops] 8[color=transparent]hellloolo</span> [aops]Y[/aops] 0 <span style=\"color:transparent\">hellloolo</span></details>\nWhat texts are used by Romanian students to learn calculus/analysis for such olympiads ? \n\nOf course, a lot of it is solving previous year problems so can you link some resources for these hard 11-12 grade analysis problems that are used. \nThe link ssmr.ro doesn't open for me. It says cannot connect to the server.",
"<blockquote> \nThe link ssmr.ro doesn't open for me. It says cannot connect to the server.</blockquote>\nStrange. All the official statements and solutions (in Romanian) are on [ssmr.ro](ssmr.ro).\n<blockquote>\nOf course, a lot of it is solving previous year problems so can you link some resources for these hard 11-12 grade analysis problems that are used. \n</blockquote>\nYou can see the statements of the problems from district olympiad and from national olympiad on AoPS:\n[https://artofproblemsolving.com/community/c3368_district_olympiad](https://artofproblemsolving.com/community/c3368_district_olympiad)\n[https://artofproblemsolving.com/community/c3365_romania_national_olympiad](https://artofproblemsolving.com/community/c3365_romania_national_olympiad)\n<blockquote>\nWhat texts are used by Romanian students to learn calculus/analysis for such olympiads? \n</blockquote>\nI'm not in grades 11-12 yet. But my teacher (who always has students at the national olympiad in grades 11-12) told me that for such olympiads the best study resources are year 1 (of college) textbooks.\nAnother resources are 11-12 grade olympiad articles from Mathematical Gazette.",
"Suppose if f have the Darboux property and let a<b and f(a)<f(b). Then f([a,b]) is an interval and so contains [f(a),f(b)]. If u>∈[f(a),f(b)] then of course u∈f([a,b]) and thus there exists some k∈[a,b] such that f(k)=u, i.e. f has IVP.\n\nNow let f have the IVP and let a<b, x,y∈f([a,b]) and z>∈R with x<z<y. We claim z∈f([a,b]). Indeed we have x=f(x′) and y=f(y′) for some x′,y′∈[a,b]. W.L.O.G assume that x′<y′. Then by the IVP there is some c∈[x′,y′] such that f(c)=z, i.e. z∈f([a,b]) and therefore f([a,b]) is an interval.\nLet x>y\nBy darboux property inf where x>y f(x)=g(y)\nWhich gives us f(x) greater than equal to g(y) so that we can conclude f=g for different cases\n"
] |
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"origin:aops",
"2022 Contests",
"2022 District Olympiad"
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810717.json"
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|
Let $A,B\in\mathcal{M}_3(\mathbb{R})$ de matrices such that $A^2+B^2=O_3.$ Prove that $\det(aA+bB)=0$ for any real numbers $a$ and $b.$
|
Interesting problem $A^2=-B^2$ take the det then $det(A)^2=-det(B)^2$ ==> det(A)=det(B)=0 (1)
If a=0 or b=0 with (1) ==> det(aA+bB)=0
If a=/=0 and b=/=0 we must prove det(A+xB)=0 for any x in R $Q(x)=det(A+xB)=det(B)x^3 + px^2 + qx +det(A)= px^2 + qx$ where p,q reals numbers $Q(i)Q(-i)=|Q(i)|^2 = (-p+iq)(-p-iq) = p^2 +q^2$ (2)
Q(i)Q(-i)=det((A+iB)(A-iB))= $det(A^2+B^2+iBA-iAB)=i^3det(BA-AB)$ =-idet(BA-AB) (3)
(2) and (3) give p=q=det(BA-AB)=0
This prove for any x in R, det(A+xB)=0
Remark AB-BA is not invertible
|
[
"What does the $O_3$ mean?",
"zero matrix $ 3\\times 3$ ",
"<blockquote> $A^2=-B^2$ take the det then $det(A)^2=-det(B)^2$ ==> det(A)=det(B)=0 (1)\nIf a=0 or b=0 with (1) ==> det(aA+bB)=0\n\nIf a=/=0 and b=/=0 we must prove det(A+xB)=0 for any x in R $Q(x)=det(A+xB)=det(B)x^3 + px^2 + qx +det(A)= px^2 + qx$ where p,q reals numbers $Q(i)Q(-i)=|Q(i)|^2 = (-p+iq)(-p-iq) = p^2 +q^2$ (2) $Q(i)Q(-i)=det((A+iB)(A-iB))=det(A^2+B^2+iBA-iAB)=i^3det(BA-AB)=-idet(BA-AB)$ (3) \n\n(2) and (3) give p=q=det(BA-AB)=0 \n\nThis prove for any x in R, det(A+xB)=0\n\nRemark AB-BA is not invertible</blockquote>\n\nplease put your work in latex it makes it easier to read"
] |
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"origin:aops",
"2022 Contests",
"2022 District Olympiad"
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"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810718.json"
}
|
Let $(x_n)_{n\geq 1}$ be the sequence defined recursively as such: \[x_1=1, \ x_{n+1}=\frac{x_1}{n+1}+\frac{x_2}{n+2}+\cdots+\frac{x_n}{2n} \ \forall n\geq 1.\]Consider the sequence $(y_n)_{n\geq 1}$ such that $y_n=(x_1^2+x_2^2+\cdots x_n^2)/n$ for all $n\geq 1.$ Prove that
[list=a]
[*] $x_{n+1}^2<y_n/2$ and $y_{n+1}<(2n+1)/(2n+2)\cdot y_n$ for all $n\geq 1;$ [*] $\lim_{n\to\infty}x_n=0.$ [/list]
|
Here is yet another way of doing part b) without the help of the auxiliary sequence.
Firstly, we will prove that $(x_n)$ is decreasing. Notice that $$ x_{n+1}-x_n = \left(\sum_{k=1}^n \frac{x_k}{n+k}\right) - x_n = \left(\sum_{k=1}^{n-1} \frac{x_k}{n+k}\right) + \frac{x_n}{2n} - x_n = \sum_{k=1}^{n-1} x_k\left(\frac{1}{n+k} - \left(1-\frac{1}{2n}\right)\frac{1}{n-1+k}\right). $$ However, every term of the form $\frac{1}{n+k} - \left(1-\frac{1}{2n}\right)\frac{1}{n-1+k}$ is negative, proving our claim.
Next, we will use the well-known lemma, stating that if a function $f:[0,1]\rightarrow\mathbb{R}$ is Riemann integrable and $(x_n)$ is a convergent sequence with $\lim_{n\rightarrow \infty} x_n = x$ , then $$ \lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n x_k f\left(\frac{k}{n}\right) = x\int_0^1 f(x)dx. $$ Returning to our problem, quite clearly all terms of the sequence are positive, and since the sequence is decreasing, by MCT the sequence has a real limit $x$ . Passing to the limit in the recurrence relation, we get that $$ x=\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k=1}^n x_k \frac{1}{1+\frac{k}{n}} =x\int_0^1 \frac{dx}{1+x} = x\ln 2, $$ finally implying that $x=0$ .
|
[
"<blockquote>\n[list=a]\n[*] $x_{n+1}^2<y_n/2$ \n[/list]</blockquote>\n\nThis fails even for $n=2$ . Check your definition of $y_n$ ...",
"I am very sorry, I made a typo. It's supposed to say $y_n=(x_1^2+x_2^2+\\cdots+x_n^2)/n.$ Fixing it.",
"You can do b without using the auxiliary sequence $(y_n)_n$ .\nLet $a_k=\\sup\\{x_k,x_{k+1},...\\}$ . Pick $n>k$ , and notice that $x_n\\leq(\\max(x_1,x_2,...,x_{k-1})-a_k)\\sum_{i=1}^{k-1}\\frac{1}{n+i}+a_k\\sum_{i=1}^{n}\\frac{1}{n+i}$ . Passing this inequality by $n$ to the superior limit we get $\\varlimsup_{n\\to\\infty}x_n\\leq ln(2)a_k$ . Letting $k\\to\\infty$ we obtain $\\varlimsup_{n\\to\\infty}x_n\\leq ln(2)\\varlimsup_{n\\to\\infty}x_n$ , and this implies $\\varlimsup_{n\\to\\infty}x_n=0\\Rightarrow\\lim_{n\\to\\infty}x_n=0$ "
] |
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"2022 Contests",
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{
"answer_score": 26,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810720.json"
}
|
Let $A\in\mathcal{M}_n(\mathbb{C})$ where $n\geq 2.$ Prove that if $m=|\{\text{rank}(A^k)-\text{rank}(A^{k+1})":k\in\mathbb{N}^*\}|$ then $n+1\geq m(m+1)/2.$
|
Let $m = \{ d_1 , d_2, \dots, d_m\}$ we have $$ \frac{m(m+1)}{2}=1+2+\cdots + m \leq d_1 +d_2 +\cdots d_m $$ $$ \leq \text{rank}(A^{i_1})-\text{rank}(A^{i_1+1})+\text{rank}(A^{i_2})-\text{rank}(A^{i_2+1})+\cdots +\text{rank}(A^{i_m})-\text{rank}(A^{i_m+1}) $$ $$ \leq \text{rank}(A^{i_1})-\text{rank}(A^{i_m+1})\leq n $$ That is assuming $i_1\leq i_2 \leq \cdots \leq i_n$ correspond to $d_1, d_2, \dots , d_m$ in some order and noting the fact that for each $j$ , $\text{rank}(A^{i_j+1})\geq \text{rank}(A^{i_{j+1}})$ .
It remains to show the equality case can not hold. But the equality of the last inequality holds when $\text{rank}(A^{i_1})=n$ and $\text{rank}(A^{i_m+1})=0$ that requires $A$ to be singular and full rank which is not possible. The conclusion follows.
|
[] |
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"2022 Contests",
"2022 District Olympiad"
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"answer_score": 28,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2810731.json"
}
|
We call a set of $6$ points in the plane *splittable* if we if can denote its elements by $A,B,C,D,E$ and $F$ in such a way that $\triangle ABC$ and $\triangle DEF$ have the same centroid.
[list=a]
[*]Construct a splittable set.
[*]Show that any set of $7$ points has a subset of $6$ points which is *not* splittable.
[/list]
|
a). Let $ABC$ and $DEF$ are equilateral triangles with the same circumcircle. Then the center is the centroid.
b). Let $O$ be the origin and $\vec{v}_{P} $ the position vector of the point $P$ . Now if $ABC$ and $DEF$ have the same centroid $G$ then $$ \vec{v}_{A}+\vec{v}_{B}+\vec{v}_{C}=3\vec{v}_{G}=\vec{v}_{D}+\vec{v}_{E}+\vec{v}_{F}. $$ For the set $\{A,B,C,D,E,F\}$ call this property a split.
Let $\mathcal{P}$ be set of $n$ points in the plane with $n\geq 7$ . Then fixing five of the points, we cannot have two distinct points with the same split. Thus say we have $\{P,Q,R,S,T,U\}$ and $\{P,Q,R,S,T,V\}$ then if we split the first set into $PQR$ and $STU$ , the second set can be split only into $PQS$ and $RTU$ or $PQU$ and $RST$ . After some algebra for the first split of the second set, we get $\vec{UV}=2\vec{RS}$ and for the second split, we get $\vec{UV}=2\vec{UR}$ .
Consider $\mathcal{S} \subseteq \mathcal{P}$ such that $|\mathcal{S}|=7$ for which we have a split. Now consider $X, Y\in\mathcal{S}$ such that the distance $XY$ is the smallest. However, applying the split with the remaining members of $\mathcal{S}$ with $X$ and $Y$ separately (not both in the same splitting set) we get a contradiction to the minimality of $XY$ .
|
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"2022 Contests",
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"answer_score": 64,
"boxed": false,
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"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811494.json"
}
|
Determine all $x\in(0,3/4)$ which satisfy \[\log_x(1-x)+\log_2\frac{1-x}{x}=\frac{1}{(\log_2x)^2}.\]
|
I claim $x=1/2$ is the only possibility.
We have
\begin{align*}
&\frac{\log_2(1-x)}{\log_2 x} + \log_2(1-x) = \log_2 x + \frac{1}{(\log_2 x)^2}
&\iff \log_2(1-x)\left(\frac{1+\log_2 x}{\log_2 x}\right) = \frac{(\log_2 x+1)((\log_2 x)^2-\log_2 x+1)}{(\log_2 x)^2}.
\end{align*}
If $\log_2 x+1=0$ , that is if $x=1/2$ , we enjoy equality. Assuming this is not the case, we perform cancellations and get
\[
\log_2(1-x) = \frac{(\log_2 x)^2 -\log_2 x+1}{\log_2 x}.
\]
Now, set $x=2^t$ . We then get
\[
\log_2 (1-x) =\frac{t^2-t+1}{t} \implies 1-x = 2^{\frac{t^2-t+1}{t}} \iff f(t)\triangleq 2^t + 2^{\frac{t^2-t+1}{t}}=1.
\]
It is readily verified $t\mapsto (t^2-t+1)/t$ is increasing on $(-\infty,-1]$ and decreasing on $(-1,0)$ . Moreover, $f(-1)<1$ , it follows that such a $t$ , if exists, is in $(-1,0)$ . Notice, however, that as $x<3/4$ , we must have $2^{(t^2-t+1)/t}>1/4\iff (t^2-t+1)/t > -2$ . Notice, however, that setting $t=-s$ for $s>0$ and recalling $s+1/s\ge 2$ , we have
\[
\frac{t^2-t+1}{t} = t+\frac1t -1 = -1 - (s+1/s)\le -3,
\]
impossible. Hence, $x=1/2$ is the only solution.
|
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"boxed": false,
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"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811496.json"
}
|
Let $z_1,z_2$ and $z_3$ be complex numbers of modulus $1,$ such that $|z_i-z_j|\geq\sqrt{2}$ for all $i\neq j\in\{1,2,3\}.$ Prove that \[|z_1+z_2|+|z_2+z_3|+|z_3+z_2|\leq 3.\]*Mathematical Gazette*
|
Joy Ma Tara! :bruce: Let's look at the geometric interpretation of the problem:
<details><summary>New problem</summary>Let $ABC$ be a non obtuse triangle such that all the interior angles are atleast $45^o$ each. Let $D,E,F$ be the midpoints of sides $BC,CA,AB$ respectively. If $O$ is the circumcenter prove that $2OD+2OE+2OF \leq 3$ .</details>
Observe that $2OD=|z_1+z_2|$ , $2OE=|z_3+z_2|$ , $2OF=|z_1+z_3|$ by parallelogram law of vector addition. Also $OD+OE+OF = \cos A+\cos B+\cos C \stackrel{Jensen}{\leq}3\cos(\frac{A+B+C}{3}) = \frac{3}{2}$ as desired.
|
[
"This is simple geo. Just use Jensen after introducing antipodes of these points.",
"<blockquote>This is simple geo. Just use Jensen after introducing antipodes of these points.</blockquote>\nExactly; but before doing so, prove that the triangle determined by $z_1,z_2$ and $z_3$ is either acute or right, so Jensen can work.",
"this problem is pretty simple because the first one we can find out that the angle between $z_1,z_2$ is bigger than 90 and it can turn into a Trigonometric inequality and that one is pretty easy."
] |
[
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] |
{
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811498.json"
}
|
A positive integer $n\geq 4$ is called *interesting* if there exists a complex number $z$ such that $|z|=1$ and \[1+z+z^2+z^{n-1}+z^n=0.\] Find how many interesting numbers are smaller than $2022.$
|
<details><summary>storage</summary><span style="color:#f00">**Claim:-**</span>all $n$ of the form $5\rho-1$ , for some $\rho \in \mathbb{Z^{+}}$ works
<span style="color:#600">pf:-</span>since $|z|=1\implies z\cdot \overline{z}=1$
now taking conjugate of given equation we arise at $z^{n}+z^{n-1}+z^{n-2}+z+1=0$ subtract this with original equation to get: $z^{n-2}=z^2\implies z^{n-4}=1\implies z^{n-1}=z^3, z^{n}=z^{4}$ so plugging these back in original equation , equation re-transforms to $1+z+z^3+z^{4}+z^{5}$ , clearly we can observe $z\neq 1$ so we have $z^{5}=1\implies z$ is $5$ th root of unity
now this gives $n\equiv -1 \pmod 5$ hence all $n$ of form $5\rho-1$ works , for $4\le n\le 2022$ as claim follows $\blacksquare$ such number of n's are $\boxed{404}$</details>
|
[
"We observe: $1$ and $-1$ are not solutions of the equation.\nLet be $z_0=\\cos\\alpha+i\\sin\\alpha$ a solution of the equation, where $\\alpha\\in[0,2\\pi)$ .\nUsing $(\\cos\\alpha+i\\sin\\alpha)^k=\\cos (k\\alpha)+i\\sin (k\\alpha),\\;\\forall k\\in\\mathbb{N}$ , the equation becomes: $1+\\cos\\alpha+\\cos(2\\alpha)+\\cos((n-1)\\alpha)+\\cos(n\\alpha)+$ $+i[\\sin\\alpha+\\sin(2\\alpha)+\\sin((n-1)\\alpha)+\\sin(n\\alpha)]=0\\Longrightarrow$ $\\Longrightarrow \\begin{cases}\\cos\\alpha+\\cos(2\\alpha)+\\cos((n-1)\\alpha)+\\cos(n\\alpha)=-1;\\sin\\alpha+\\sin(2\\alpha)+\\sin((n-1)\\alpha)+\\sin(n\\alpha)=0.\\end{cases}$ $\\cos\\alpha+\\cos(2\\alpha)+\\cos((n-1)\\alpha)+\\cos(n\\alpha)=2\\cos\\dfrac{\\alpha}{2}\\cdot\\left(\\cos\\dfrac{3\\alpha}{2}+\\cos\\dfrac{(2n-1)\\alpha}{2}\\right)=$ $=4\\cos\\dfrac{\\alpha}{2}\\cdot\\cos\\dfrac{(n-2)\\alpha}{2}\\cdot\\cos\\dfrac{(n+1)\\alpha}{2}=-1\\quad(1)$ . $\\sin\\alpha+\\sin(2\\alpha)+\\sin((n-1)\\alpha)+\\sin(n\\alpha)=2\\cos\\dfrac{\\alpha}{2}\\cdot\\left(\\sin\\dfrac{3\\alpha}{2}+\\sin\\dfrac{(2n-1)\\alpha}{2}\\right)=$ $=4\\cos\\dfrac{\\alpha}{2}\\cdot\\cos\\dfrac{(n-2)\\alpha}{2}\\cdot\\sin\\dfrac{(n+1)\\alpha}{2}=0\\quad(2)$ . $\\cos\\dfrac{\\alpha}{2}\\cdot\\cos\\dfrac{(n-2)\\alpha}{2}\\ne0$ (otherwise, contradiction with (1)).\nResults: $\\sin\\dfrac{(n+1)\\alpha}{2}=0\\Longrightarrow \\alpha=\\dfrac{2m\\pi}{n+1}$ , where $m\\in\\mathbb{N},\\;m\\le n$ .\n\nReplacing $\\alpha$ in (1) we obtain: $4\\cos\\dfrac{m\\pi}{n+1}\\cdot\\cos\\dfrac{m(n-2)\\pi}{n+1}\\cdot\\cos(m\\pi)=4\\cos\\dfrac{m\\pi}{n+1}\\cdot\\cos\\dfrac{3m\\pi}{n+1}=-1$ .\nDenote $\\cos\\dfrac{m\\pi}{n+1}=t$ .\nThen: $\\cos\\dfrac{3m\\pi}{n+1}=4t^3-3t$ .\nThe equation (1) becomes: $16t^4-12t^2+1=0\\Longrightarrow t^2=\\dfrac{3\\pm\\sqrt5}{8}\\Longrightarrow t=\\pm\\dfrac{\\sqrt5\\pm1}{4}\\quad(3)$ (all possible combinations of signs $+$ and $-$ ; $4$ distinct values).\n\nFor $n=4$ , the given equation has solutions: $1+z+z^2+z^3+z^4=0\\Longleftrightarrow z^5=1$ and $z\\ne1$ , with the solutions $z_j=\\cos\\dfrac{2j\\pi}{5}+i\\sin\\dfrac{2j\\pi}{5},\\;j\\in\\{1,2,3,4\\}$ .\nUsing (3) results: $\\left\\lbrace\\cos\\dfrac{2j\\pi}{5}\\big|j\\in\\{1,2,3,4\\}\\right\\rbrace=\\left\\lbrace\\pm\\dfrac{\\sqrt5\\pm1}{4}\\right\\rbrace$ .\n\nReturn to the general case $n\\ge4$ :\nthe given equation has solutions with $|z|=1$ only if exist $m\\in\\mathbb{N}, m\\le n;\\;j\\in\\{1,2,3,4\\};\\;p\\in\\mathbb{N}$ such that $\\cos\\dfrac{m\\pi}{n+1}=\\cos\\dfrac{2j\\pi}{5}\\Longrightarrow \\dfrac{m\\pi}{n+1}=\\pm\\dfrac{2j\\pi}{5}+2p\\pi\\Longrightarrow$ $\\Longrightarrow (n+1)(10p\\pm2j)=5m\\Longrightarrow 5|n+1$ .\n\nRemains to prove: $\\forall n=5u-1,\\;u\\in\\mathbb{N},\\;\\exists m\\in\\mathbb{N}, m\\le n;\\;j\\in\\{1,2,3,4\\}$ such that $\\cos\\dfrac{m\\pi}{n+1}=\\cos\\dfrac{2j\\pi}{5}$ .\nWe observe: $m=2u\\le 5u-1=n,\\;j=1$ satisfy the condition.\n\nResults the set of interesting numbers smaller than $2022$ : $A=\\{4,9,\\dots,2014,2019\\}=\\{5u-1|u\\in\\mathbb{N},\\;1\\le u\\le404\\}$ and $|A|=404$ .\n",
"An easier solution than @above. Note that $\\overline{z} = \\frac{1}{z}$ . If we the conjugate of the equation, we will get: $$ 1+\\frac{1}{z}+\\frac{1}{z^2}+\\frac{1}{z^{n-1}} + \\frac{1}{z^n} = 0 \\Longrightarrow 1+z+z^{n-2}+z+1 = 0. $$ Together with the given equation we get $z^{n-2} = z^2,$ or $z^{n-4} = 1$ . Using this we can simplify the given expression to: $$ 1+z+z^2+z^3+z^4 = 0 \\Longrightarrow z^5=1. $$ Note that $z=1$ does not fit, so $z$ is a primitive fifth root of unity. Since $z^{n-4} = 1$ , it follows that $n \\equiv 4 \\pmod{5}$ . Obviously all such $n$ satisfy the problem, so the answer is $404$ .",
"Joy Kali! :bruce: Suppose $n$ is an *interesting* number. Then there exists uni-modular $z\\in \\mathbb{C}$ such that $1+z+z^2+z^{n-1}+z^n=0$ . Taking conjugate on both sides, and using that $z\\overline{z}=1$ , we have that $z^n+z^{n-1}+z^{n-2}+z+1=0$ . Thus $z^{n-2}=z^2$ and as $z\\neq 0$ , $z^{n-4}=1$ . Using this in parent equation we have $z^4+z^3+z^2+z+1=0$ that is $z$ is the 5th non real root of unity. As $5$ is prime, we must have $5|n-4$ , thus giving a necessary condition of *interesting* numbers.We note that this condition is sufficient. Thus the *interesting* numbers are $\\{4,9,14, \\cdots , 2019\\}$ and hence a total of $\\boxed{404}$ *interesting* numbers less than $2022$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] |
{
"answer_score": 1132,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811500.json"
}
|
Find all functions $f:\mathbb{R}\to\mathbb{R}$ which satisfy the following relationship for all $x,y\in\mathbb{R}:$ \[f(f(y-x)-xf(y))+f(x)=y\cdot(1-f(x)).\]
|
Let $P(x,y)$ denote the given assertion.
Claim: $f$ is surjective.
Proof: We note that \[f(f(y-x)-xf(y))=y(1-f(x))-f(x)\]
Noting that $f\equiv 1$ is not a solution, set $x$ such that $f(x)\ne1$ . Varying $y$ gives the desired result. $\blacksquare$ Claim: $f$ is injective.
Proof: $P(0,x): f(f(x))+f(0)=x(1-f(0))$ . If $f(a)=f(b)$ , then $a=b$ or $f(0)=1$ .
We will show $f(0)\ne 1$ . Suppose FTSOC $f(0)=1$ . Then $f(f(x))=-1$ for any $x$ . Since $f$ is surjective, $f\equiv -1$ , which is not a solution.
Thus, $a=b$ . $\blacksquare$ $P(x,-1): f(f(-x-1)-xf(-1))=1$ , so $f(-x-1)-xf(-1)=c$ for some constant $c$ . Set $x\to -x-1$ .
So $f(x)-(-x-1)f(-1)=c$ , which implies $f$ is linear.
Let $f(x)=ax+b$ . Note that $f(f(x))+b=x-xb\implies f(f(x))=x(1-b)-b$ . $f(f(x))=f(ax+b)=a^2x+(a+1)b$ . So $(a+1)b=-b$ .
Case 1: $a+1=-1$ .
Then $f(f(x))=4x-b$ . So $b=-3$ . Thus, $f(x)=-2x-3$ , which is not a solution.
Case 2: $b=0$ .
Then $f(x)=ax$ for some $a$ . But $f(f(x))=x\implies \boxed{f(x)=x}$ or $f(x)=-x$ . The former works but the latter doesn't.
|
[
"REDACTED",
"<blockquote>Denote $P(x,y)$ as usual be the assertion $P(0,x)\\leftrightarrow f(f(x))+f(0)=x(1-f(0)) \\leftrightarrow f(f(x)) = x(1-f(0)) - f(0)$ From here we get that $f(x)$ is bijective</blockquote>Not yet. What if $f(0)=1?$ ",
"Case 1: $f(0)\\neq 1$ $P(0,y)\\implies f(f(y)) = (1-f(0))y - f(0)\\implies f$ is injective $P(x,-1)\\implies f(f(-1-x) - xf(-1)) = -1\\implies f$ is linear\n\nPlugging back in, we get that the only linear solution is $f(x)\\equiv x$ Case 2: $f(0) = 1$ $P(0,y)\\implies f(f(y)) = -1$ $P(-1,-1)\\implies f(0) = -1$ contradiction. The only solution is $f(x)\\equiv x$ .",
"REDACTED",
"<blockquote>Case 2: $f(0) = 1$ $P(0,y)\\implies f(f(y)) = -1$ $P(-1,-1)\\implies f(-2) = -1$ $P(-1,0)\\implies f(-2) = 1$ contradiction, so the only solution is $f(x)\\equiv x$ .</blockquote>\nAm I being silly or doesn't $P(-1,-1)$ imply $f(1+f(-1))=-1?$ <blockquote>@OVlad:\nIf $f(0)=1$ , then $f(f(x)) = -2f(0)$ . Let $-2f(0)=a$ $P(x,0)\\leftrightarrow f(f(-x))+f(x)=0\\leftrightarrow a + f(x) = 0 \\leftrightarrow f(x)=$ const, then replace back to the problem to get contradiction</blockquote>\nAm I even sillier or doesn't $P(x,0)$ imply $f(f(-x)-x)+f(x)=0?$ ",
"@above sorry. I think it's fixed now.\n\n@below oops, now it should be fixed for real. The contradiction was much easier than I thought.",
"Let me post a full solution as well for convenience. The problem isn't hard, but you need to get a couple of details right.\n\nFirstly, $f\\equiv 1$ is not a solution, so there exists $c\\in\\mathbb{R}$ such that $f(c)\\neq 1.$ Then, by taking $P(c,y)$ we get that \\[f(\\text{something})=y\\cdot(1-f(c))-f(c).\\]Since $f(c)\\neq 1$ then by varying $y$ over the real numbers, $y(1-f(c))-f(c)$ will take all real values, so $f$ is surjective.\n\nNow, assuming that $f(0)=1,$ then $P(0,y)$ gives $f(f(y))=-1$ for all $y.$ Since $f$ is surjective, the latter is equivalent to $f(x)=-1$ for all $x.$ This is clearly not a solution, so we get that $f(0)\\neq 1.$ \n\nBy taking $P(0,y)$ and since $f(0)\\neq 1$ we can infer that $f$ is injective. Now, $P(x,-1)$ implies that $f(-x-1)-xf(-1)$ is constant for all $x,$ so $f$ is linear and with a couple more tweaks we get $f(x)=x.$ [rule]\nEdit: @CT17 I still don't understand why $P(-1,-1)\\implies f(-2) = -1$ ",
"*Ans:* $f(x)=x$ for all real $x$ .\n*Pf:* It is easy to see that the identity function works. Let $P(x,y)$ denote the given assertion, we have \\[P(0,x) \\implies f(f(x))+f(0)=x(1-f(0)).\\]\n<span style=\"color:#f00\">Case 1:</span> $f(0)=1.$ We have $f(f(x))=-1$ for all real $x$ . So, $f(-1)=f(f(f(0)))=f(f(1))=-1$ , but \\[P(-1,-1) \\implies f(0)=-1\\] a contradiction. \n<span style=\"color:#f00\">Case 2:</span> $f(0)\\ne 1.$ We have that $f$ is bijective by $P(0,x)$ . Let $f(t)=0$ for some real $t$ , we have \\[P(0,t) \\implies t=\\frac{2f(0)}{1-f(0)}\\] \\[P(t,t) \\implies f(f(0))=t\\] \\[P(0,0) \\implies f(f(0))=-f(0)\\] so \\[\\frac{2f(0)}{1-f(0)}=f(f(0))=-f(0) \\implies f(0)= 0 \\ \\textrm{or} \\ 3.\\]\n<span style=\"color:#0ff\">Case 2.1:</span> $f(0)=0.$ We have \\[P(-x,0)\\implies f(f(x))=-f(-x)\\] \\[P(0,x)\\implies f(f(x))=x\\] so $-f(-x)=x \\implies f(x)=x.$ <span style=\"color:#0ff\">Case 2.2:</span> $f(0)=3.$ We have $t=-3, f(-3)=0,$ \\[P(-x-3,-3)\\implies f(f(x))=2f(-x-3)-3\\]\\[P(0,x)\\implies f(f(x))=-2x-3\\] so \\[2f(-x-3)-3=-2x-3\\implies f(x)=x+3\\] which is not a solution upon checking. $\\blacksquare$ ",
" $P(0,0) : f(f(0)) + f(0) = 0$ . $P(0,y) : f(f(y)) + f(0) = y - yf(0) \\implies f(f(y)) = y - (y+1)f(0) \\implies f$ is bijective.\nAssume $a$ such that $f(a) = 0$ . $P(0,a) : (a+2)f(0) = a \\implies f(0) = \\frac{a}{a+2}$ . $P(a,a) : f(f(0)) = a \\implies a + \\frac{a}{a+2} = 0 \\implies a = 0 \\implies f(0) = 0$ . $P(0,y) : f(f(y)) = y$ $P(x,0) : f(f(-x)) + f(x) = 0 \\implies -x + f(x) = 0 \\implies f(x) = x$ .\nAnswers $: f(x) = x$ .",
"Let $P(x,y)$ denote the assertion $f(f(y-x)-xf(y))+f(x)=y(1-f(x)).$ If $f(0)=1,$ then $P(0,x)$ implies $f(f(x))=-1$ and also $f(1)=-1.$ So $P(f(1),f(1))$ implies $0=-2,$ false. If $f(0)\\neq 1,$ then $P(0,x)$ implies $f$ is bijective. And $P(-x-1,-1)$ implies $f$ is linear. The solution that works is the identity function. ",
"This problem was proposed at a math magazine issued by the Hellenic Mathematical Society. However, they made a *small* mistake at the problem statement. Here's the proposed version:\n\n<blockquote>Find all functions $f:\\mathbb{R}\\to\\mathbb{R}$ which satisfy the following relationship for all $x,y\\in\\mathbb{R}:$ \\[f(f(y-x)-xf(y)+f(x))=y\\cdot(1-f(x)).\\] </blockquote>\n\nI think this version is significantly harder than the original, and so I propose it here :)\n\n",
"A solution for the harder version: $f(f(y-x)-xf(y)+f(x))=y\\cdot(1-f(x)) (A)$ \nIt is obvious that $f(x)=1$ is not a solution, so there exist a number $u$ such that $f(u)$ is not equal to $1$ . For $x=u$ in $(A)$ we get that $f$ is surjective. \nFor $x=0$ in $(A)$ : $f(f(y)+f(0))=y(1-f(0)) (1) $ If $f(0)=1$ using $f$ is surjective we get from $(1)$ that $f(x)=0$ for all $x$ , contradiction. \nSo $f(0)\\neq 1$ and hence from $(1)$ $f$ is injective\nFor $y=0$ in $(A): f(f(-x)+f(x)-xf(0))=0 (2)$ For $x->-x$ in $(2)$ and using injectivity we get that $f(0)=0$ So $(1)$ gives $f(f(y))=y (3)$ For $y=0$ in $(A)$ and using injectivity: $f(-x)=-f(x) (4)$ Taking $y->-y$ in $(A)$ and adding with $(A)$ will give us (using $(4)$ and injectivity): $f(f(-y-x)-xf(-y)+f(x))=-f(f(y-x)-xf(y)+f(x))=f(-f(y-x)+xf(y)-f(x)) => f(x-y)+f(x+y)=2f(x) (5)$ For $x=y$ in $(5): f(2x)=2f(x)$ , so $(5)$ gives : $f(x-y)+f(x+y)=f(2x)$ , i.e. $f$ is additive\nUsing additivity in $(A)$ : $f(f(y-x))+f(-xf(y))+f(f(x))=y(1-f(x))$ and using $(3),(4)$ we get that: $f(xf(y))=yf(x) (6)$ For $y->f(y)$ in $(6)$ and using $(3)$ : $f(xy)=f(x)f(y)$ , hence $f$ is both additive and multiplicative.\nHence $f(x)=x$ for all $x$ or $f(x)=0$ for all $x$ from which only $f(x)=x$ satisfy $(A)$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 District Olympiad"
] |
{
"answer_score": 1178,
"boxed": false,
"end_of_proof": false,
"n_reply": 13,
"path": "Contest Collections/2022 Contests/2022 District Olympiad/2811501.json"
}
|
Charlotte is playing the hit new web number game, Primle. In this game, the objective is to guess a two-digit positive prime integer between $10$ and $99$ , called the *Primle*. For each guess, a digit is highlighted blue if it is in the *Primle*, but not in the correct place. A digit is highlighted orange if it is in the *Primle* and is in the correct place. Finally, a digit is left unhighlighted if it is not in the *Primle*. If Charlotte guesses $13$ and $47$ and is left with the following game board, what is the *Primle*? $$ \begin{array}{c}
\boxed{1} \,\, \boxed{3} [\smallskipamount]
\boxed{4}\,\, \fcolorbox{black}{blue}{\color{white}7}
\end{array} $$ *Proposed by Andrew Wu and Jason Wang*
|
<details><summary>Solution</summary>Because we are working with a two digit number and the digit $7$ is not the units digit, then it follows that the *Primle* has a ten's digit of $7$ . The prime numbers in the $70's$ are $71, 73,$ and $79$ . Because when Charlotte put the prime $13$ none of the digits were blue, $71$ and $73$ don't work. Therefore, the *Primle* is $\boxed{79}.$</details>
|
[
"<details><summary>Solution</summary>The blue $7$ tells us that the number is a prime with $7$ in the tens place. The uncolored numbers tell us that the units digit is not $1, 3$ , or $4$ . The only working prime number that satisfies these conditions can easily be checked as $\\fbox{79}$ .</details>",
"<details><summary>Solution</summary>From the second row, we know that the first digit of the prime has the be $7$ . So, the possible primes are $71, 73,$ and $79$ . But, we know that neither $1$ nor $3$ can be in the prime number from the information in the first row. Therefore, the *Primle* is $\\boxed{79}$ .</details>",
"<blockquote>Charlotte is playing the hit new web number game, Primle. In this game, the objective is to guess a two-digit positive prime integer between $10$ and $99$ , called the *Primle*. For each guess, a digit is highlighted blue if it is in the *Primle*, but not in the correct place. A digit is highlighted orange if it is in the *Primle* and is in the correct place. Finally, a digit is left unhighlighted if it is not in the *Primle*. If Charlotte guesses $13$ and $47$ and is left with the following game board, what is the *Primle*? $$ \\begin{array}{c} \n \\boxed{1} \\,\\, \\boxed{3} [\\smallskipamount]\n \\boxed{4}\\,\\, \\fcolorbox{black}{blue}{\\color{white}7}\n \\end{array} $$ *Proposed by Andrew Wu and Jason Wang*</blockquote>\nNice simple problem. \nInspired by [this](https://converged.yt/primel/)?",
"<details><summary>sol</summary>It starts with 7, but can't end with 1,3 so it has to be $\\boxed{79}$ .</details>",
"<details><summary>Solution</summary>The $7$ highlighted blue in the second guess tells us that the *Primle* has a tens digit of $7.$ \n\nThe two-digit primes that have a tens digit of $7$ are $71,73,79$ but the first three can' t be the *Primle* because they were left unhighlighted in the first guess.\n\nHence, the *Primle* must be $\\boxed{79}.$</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790580.json"
}
|
How many ways are there to fill in a $2\times 2$ square grid with the numbers $1,2,3,$ and $4$ such that the numbers in any two grid squares that share an edge have an absolute difference of at most $2$ ?
*Proposed by Andrew Wu*
|
<details><summary>Solution</summary>Note that the $1$ must be diagonally opposite the $4$ . Then, the $2$ and $3$ can be placed in any way.
Therefore, there are $4$ ways to place the $1$ , $1$ way to place the $4$ after that, and $2$ ways to place the $2$ and $3$ after that, for a total of $$ 4\cdot1\cdot2=\boxed{8}. $$</details>
|
[
"<details><summary>Solution</summary>By symmetry, we can place the $1$ in the top-left corner and then multiply our answer by $4$ for symmetry. This means that the $4$ must go diagonally across from the $1$ since the absolute difference is more than $2$ . Obviously, both orderings of the $2$ and $3$ will work as they are both within $2$ of $1$ and $4$ . So, the answer is $4 \\cdot 2$ or $\\fbox{8}$ .</details>",
"<details><summary>sol</summary>It is easy to see that $4$ and $1$ must be in opposite corners so they don't share an edge. We can choose the spot for the $1$ in $4$ ways, then $4$ is fixed. The other two numbers can be arranged in $2$ ways, so we get $4\\cdot 2 = \\boxed 8$ .</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1026,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790583.json"
}
|
The **Collaptz function** is defined as $$ C(n) = \begin{cases} 3n - 1 & n\textrm{~odd}, \frac{n}{2} & n\textrm{~even}.\end{cases} $$
We obtain the **Collaptz sequence** of a number by repeatedly applying the Collaptz function to that number. For example, the Collaptz sequence of $13$ begins with $13, 38, 19, 56, 28, \cdots$ and so on. Find the sum of the three smallest positive integers $n$ whose Collaptz sequences do not contain $1,$ or in other words, do not **collaptzse**.
*Proposed by Andrew Wu and Jason Wang*
|
<details><summary>Solution</summary>Hoping that the three smallest positive integers that do not collaptzse are reasonably small, we use brute force:**1:** Obviously, the Collaptz sequence of $1$ contains a $1,$ so $1$ collaptzses**2:** The first two terms of the Collaptz sequence of $2$ are $2,1$ so $2$ also collaptzses.**3:** The first five terms of the Collaptz sequence of $3$ are $3,8,4,2,1$ so $3$ also collaptzses.**4:** The first three terms of the Collaptz sequeuce of $4$ are $4,2,1$ so $4$ also collaptzses**5:** The first six terms of the Collaptz sequence of $5$ are $5,14,7,20,10,5$ so it repeats with period $5$ without having a $1$ in the first $5$ terms. Hence, $5$ does not collaptzse.**6:** The first two terms of the Collaptz sequence of $6$ are $6,3.$ Since $3$ collaptzses, $6$ must also collaptzse.**7:** The first four terms of the Collaptz sequeuce of $7$ are $7,20,10,5.$ Since $5$ does not collaptzse, $7$ also does not collaptzse.**8:** The first two terms of the Collaptz sequeunce of $8$ are $8,4.$ Since $4$ collaptzses, $8$ must also collaptzse.**9:** The first nine terms of the Collaptz sequence of $9$ are $9,26,13,38,19,56,28,14,7.$ Since $7$ does not collaptzse, $9$ also does not collaptzse. We can stop here since we have already found $3$ numbers which do not collaptzse.
Hence, the three smallest positive integers $n$ that do not collaptzse are $5,7,9$ which have sum $$ 5+7+9=\boxed{21}. $$</details>
|
[
"<details><summary>Solution</summary>We use brute force with some quick observations. Obviously $1$ to $4$ all do not work by testing. $5$ works because its sequence is $5, 14, 7, 20, 10, 5, ...$ . $6$ doesn't work because it leads into the sequence for $3$ . $7$ works because its sequence leads into the sequence for $5$ . $8$ obviously doesn't work. $9$ works because its sequence leads into the sequence for $7$ and thus $5$ from some quick bashing. Thus our answer is $5+7+9$ or $\\fbox{21}$ .</details>",
"<details><summary>Sol</summary>Note that $1,2,3,4$ trivially don't work. Then $n=5$ gives us the period collaptz sequence $5,14,7,20,10,5,14, \\cdots$ . We also know $6$ doesn't work because $6/2$ is $3$ and $3$ eventually becomes $1$ . But $7$ does work, since we get $7,20, 10, 5, \\cdots$ and it continues like the $n=5$ sequence. We can skip $8$ . $n=9$ gives $9,26,13,38, 19, 56, 28, 14, 7, \\cdots$ and since it reached $7$ , it will eventually become periodic. Our values are $5,7,9$ which have sum $\\boxed{21}.$</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1078,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790584.json"
}
|
Kara rolls a six-sided die, and if on that first roll she rolls an $n$ , she rolls the die $n-1$ more times. She then computes that the product of all her rolls, including the first, is $8$ . How many distinct sequences of rolls could Kara have rolled?
*Proposed by Andrew Wu*
|
<details><summary>Solution</summary>Since the product of all the rolls is $8,$ the first roll must be a factor of $8$ which can be $1,2,$ or $4$ ( $8$ is out of range).
Obviously, $1$ wouldn't work because then she would have $0$ more rolls, so the product couldn't be $8$ .
If she rolls a $2,$ she has $1$ more roll, so her second roll must be a $4$ .
If she rolls a $4,$ the product of her next $3$ rolls must be $2.$ Obviously, they must be some permutation of $(1,1,2)$ of which there are $3$ of them.
Hence, the total number of distinct sequences of rolls Kara could have rolled is $$ 1+3=\boxed{4}. $$</details>
|
[
"<details><summary>sol</summary>We know it must start with a factor of $8$ , so it has to be $1,2,4$ . $1$ clearly doesn't work because we roll the die $0$ times. For $2$ , we have 1 roll so that has to be $4$ . If she rolls a $4$ , she has 3 rolls to get a product of $2$ . This has to be $2,1,1$ in some order, which we can order in 3 ways. We get $1+3=\\boxed{4}$ .</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1036,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790586.json"
}
|
Cat and Claire are having a conversation about Cat's favorite number.
Cat says, "My favorite number is a two-digit positive integer with distinct nonzero digits, $\overline{AB}$ , such that $A$ and $B$ are both factors of $\overline{AB}$ ."
Claire says, "I don't know your favorite number yet, but I do know that among four of the numbers that might be your favorite number, you could start with any one of them, add a second, subtract a third, and get the fourth!"
Cat says, "That's cool, and my favorite number is among those four numbers! Also, the square of my number is the product of two of the other numbers among the four you mentioned!"
Claire says, "Now I know your favorite number!"
What is Cat's favorite number?
*Proposed by Andrew Wu*
|
<details><summary>Solution</summary>All two-digit positive integers with distinct nonzero digits $\overline{AB}$ , such that $A$ and $B$ are both factors of $\overline{AB}$ can be listed pretty easily: $$ 12,15,24,36,48. $$ (For $A\geq5,$ for $A$ and $B$ to both divide $\overline{AB},$ we must have $A=B,$ contradiction.)
Then, note that $12+48-36=24$ or $12+48-24-36,$ so the set of four numbers mentioned is $\{12,24,36,48\}.$
Finally, one can easily find (by guess and check) that $$ 24^2 = 12 \cdot 48, $$ so Cat's favorite number is $\boxed{24}.$</details>
|
[
"<details><summary>Solution</summary>We must have $A|10A+B$ , so $A|B$ . From here, we can see that the only possibilities are $12, 15, 24, 36, 48$ . We have $12 + 48 - 36 = 24$ , so our four numbers are $12, 48, 36, 24$ . We have $12 \\cdot 48 = 24^2$ , so the answer is $\\boxed{24}$ .</details>\n",
"<details><summary>Sol</summary>We can list the numbers with the first property. We get $12, 15, 24, 36, 48.$ Notice that $12,24,36,48$ satisfy the second property mentioned as well, and this can be motivated by the fact that they are all multiples of $12$ . Then $24^2 = 12 \\cdot 48$ , so her favorite number is $\\boxed{24}.$</details>",
"seems like an amc10 problem 10"
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1034,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790588.json"
}
|
Carissa is crossing a very, very, very wide street, and did not properly check both ways before doing so. (Don't be like Carissa!) She initially begins walking at $2$ feet per second. Suddenly, she hears a car approaching, and begins running, eventually making it safely to the other side, half a minute after she began crossing. Given that Carissa always runs $n$ times as fast as she walks and that she spent $n$ times as much time running as she did walking, and given that the street is $260$ feet wide, find Carissa's running speed, in feet per second.
*Proposed by Andrew Wu*
|
<details><summary>Sol</summary>Let $a$ be the number of seconds she spent walking. Then $a+an=30$ and $2a+2an^2=260$ or $a+an^2=130$ . From the first equation, we can divide by $n+1$ to get $a=\frac{30}{n+1}$ . We can substitute to get $\frac{30}{n+1}+\frac{30n^2}{n+1}=130$ . Multiplying by $\frac{n+1}{10}$ , we get $3n^2+3=13n+13$ . This becomes $3n^2-13n-10=0$ and factors to $(n-5)(3n+2)$ , so $n=5$ . We get $a=5$ , so her running speed is $2n =\boxed{10}$ .</details>
|
[
"<details><summary>Solution</summary>Let's say Carissa walks at $2$ ft/sec for $x$ seconds. She then travels a distance of $2x$ feet, and after that travels at $2n$ ft/sec for $nx$ seconds. She thus travels a total distance of $2x+2n^2x = 260$ , so $x+n^2x = 130$ . We also have $x + nx = 30$ . Dividing gives $\\frac{1+n^2}{1+n} = \\frac{13}{3}$ , or $3+3n^2 = 13 + 13n$ . This means $n = 5$ , for an answer of $2 \\cdot 5 = \\boxed{10}$ ft/sec.</details>",
"<details><summary>Solution</summary>Let the time she spent walking(in seconds) be $t$ .\nThen, our equation becomes $$ 2t+2tn^2=260\\Rightarrow t(n^2+1)=130 $$ and $$ t+tn=30\\Rightarrow t(n+1)=30 $$ There are many possibilities, but let's divide the two equations first. $$ \\frac{n^2+1}{n+1}=\\frac{13}{3}\\Rightarrow3n^2+3=13n+13\\Rightarrow3n^2-13n-10=0 $$ Factoring gives us $$ (3n+2)(n-5)=0 $$ $n$ can't be negative, so it is $5$ . We need her running speed, so $2\\cdot5=\\boxed{10\\frac{\\text{ft}}{\\text{s}}}$ .</details>",
"Remember to put units. For all we know, $10$ could mean $10$ knots. (Who gives a measure of rate a name like \"knots\"? I know it has to do with actual knots, but it's just weird.)",
"<details><summary>Solution</summary>Let $x$ be the time (in seconds) Carissa spent walking. Then, she spent $nx$ seconds running.\n\nWe can then set up the system of equations $$ x+nx=30, $$ $$ 2x+2n^2x=260. $$ Simplifying the second equation, we have $$ x+n^2x=130. $$ Dividing this by the first equation of the system gives $$ \\frac{n^2+1}{n+1}=\\frac{13}{3} \\rightarrow 3n^2-13n-10=0 \\rightarrow n = 5, -\\frac{2}{3}. $$ Obviously, $n>0,$ so $n=5.$ Hence, Carrisa's running speed is $$ 2n = 2\\cdot5=\\boxed{10} $$ feet per second.</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1028,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790589.json"
}
|
Given that six-digit positive integer $\overline{ABCDEF}$ has distinct digits $A,$ $B,$ $C,$ $D,$ $E,$ $F$ between $1$ and $8$ , inclusive, and that it is divisible by $99$ , find the maximum possible value of $\overline{ABCDEF}$ .
*Proposed by Andrew Milas*
|
<details><summary>Sol</summary>It will always be divisible by $9$ because the sum of the digits is $36$ . For $11$ , we need alternate digits to have a sum that differs by a multiple of $11$ . We can try to set the first few digits in order from $8$ and find a way to place the rest of the digits to satisfy the $11$ divisibility condition. If we have $876543\_\_$ , then the sum is currently $18$ for odds and $15$ for evens, so there's no way to do this. If we have $87654\_\_\_$ , then the current sum for odd digits is $18$ and even digits have $12$ . We either need to close the gap of 6 or extend it to 11, but neither of these are possible. If we have $8765\_\_\_\_$ , then the current sum for odds is $14$ and evens is $12$ . Again, we either need to close the gap of $2$ or extend it to 11. To extend, we need a difference of $9$ but $1+2+3+4=10$ so this is not possible. To close the gap, we need even digits to have a sum 2 higher than the odd digits. The sum of the even digits will be $\frac{10+2}{2}=6$ and the odd digits will have $10-6=4$ . To maximize the number, the digit after $5$ is $3$ , because it can't be $4$ and the next digit is $4$ . Then we get $1$ and $2$ . Our answer is $\boxed{87653412}$ .</details>
|
[
"The problem asks for a six-digit positive integer. ",
"<details><summary>Solution</summary>Since we want to maximize $\\overline{ABCDEF}$ , I let $\\overline{ABCD}=9876$ . Since the sum of these numbers is $\\equiv 3 \\pmod 9$ , $E+F$ needs to be $\\equiv 6 \\mod 9$ . Since we have already used $9,8,7,$ and $6$ , $E+F$ has to be equal to $6$ . Using the $11$ divisibility trick, the largest possible value of $\\overline{ABCDEF}$ is $\\boxed{987624}$ . This works because $\\frac{987624}{99}=9976$ .</details>",
"The digits are 1-8 @above.",
"<details><summary>Click to expand</summary>Answer must be $857241$ .</details>",
"Expanding, ABCDEF gives $10^{5}A+10^{4}B+10^{3}C+10^{2}D+10{E}+{F}$ This can be written as $99(k)+10A+B+10C+D+10E+F$ so, $(\\overline{AB}+\\overline{CD}+\\overline{EF}))=99$ or $198$ since, all digits are distinct and cannot be 0,9 $max(\\overline{AB}+\\overline{CD}+\\overline{EF})=87+65+43=195$ hence, $\\overline{AB}+\\overline{CD}+\\overline{EF}=98$ Since we want the largest number, $AB=87$ $CD=12$ $EF=00$ \nso, $\\overline{ABCDEF}=871200$ which works, $\\frac{871200}{99}=8800$ EDIT: I know that $max(\\overline{AB}+\\overline{CD}+\\overline{EF})=87+65+43=195$ is not true since it can be more but i dont think it matters because any number for which $AB\\neq87$ will be less that $871200$ ",
"How about 888822? I was looking for the solution in the form <u>8888ab</u>, since <u>88888a</u> didn't work.\nDepending on the interpretation of \"has distinct digits\". Is it all digits are distinct or there are some distinct digits.",
"\"Has distinct digits $A, B, C, D, E, F$ \" means that all the digits are distinct."
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1054,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790590.json"
}
|
Triangle $ABC$ has sidelengths $AB=1$ , $BC=\sqrt{3}$ , and $AC=2$ . Points $D,E$ , and $F$ are chosen on $AB, BC$ , and $AC$ respectively, such that $\angle EDF = \angle DFA = 90^{\circ}$ . Given that the maximum possible value of $[DEF]^2$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $\gcd (a, b) = 1$ , find $a + b$ . (Here $[DEF]$ denotes the area of triangle $DEF$ .)
*Proposed by Vismay Sharan*
|
Let $BD=x,ED=2x,AD=(1-x),DF=\frac{\sqrt{3}(1-x)}{2}$ The area will be noted by $\frac{\sqrt{3}}{2}x(1-x)$ , which is maximum value is got when $x=\frac{1}{2}$ , this time the area is $\frac{\sqrt{3}}{8}$ , the desired answer is $\frac{3}{64}$ leads to $\boxed{67}$
|
[
"Diagram!\n\n(points not necessarily optimally chosen to satisfy problem statement.)\n\n[asy]\nunitsize(1.5inch);\npair A = dir(180);\npair B = dir(120);\npair C = dir(0);\ndraw(A--B--C--A);\npair D = (3A + 5B)/(3 + 5);\npair F = foot(D, A, C);\npair E = extension(D, rotate(90, D)*F, C, B); \ndraw(D--E--F--cycle, blue+dashed);\nfill(D--E--F--cycle, lightcyan + opacity(0.05));\nmarkscalefactor=0.007;\ndraw(rightanglemark(D, F, A), red);\ndraw(rightanglemark(F, D, E), red);\ndot(\" $A$ \", A, dir(A));\ndot(\" $B$ \", B, dir(B));\ndot(\" $C$ \", C, dir(C));\ndot(\" $D$ \", D, dir(D));\ndot(\" $F$ \", F, S);\ndot(\" $E$ \", E, NE);\n[/asy]"
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1012,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790592.json"
}
|
Suppose that $P(x)$ is a monic quadratic polynomial satisfying $aP(a) = 20P(20) = 22P(22)$ for some integer $a\neq 20, 22$ . Find the minimum possible positive value of $P(0)$ .
*Proposed by Andrew Wu*
(Note: wording changed from original to specify that $a \neq 20, 22$ .)
|
The conditions of the problem allow us to define
\[ XP(X) = (X-a)(X-20)(X-22) + C. \]
Since $X$ divides the RHS, then $C = 440a$ . So,
\begin{align*}
P(X) = \dfrac{1}{X} \left[(X-a)(X-20)(X-22) + 400a\right] = X^2 - (42+a)X + (440+42a).
\end{align*}
Henceforth, we want to minimize the value of $P(0) = 440+42a$ which is $\boxed{20}$ and occurs at $a = -10$ .
|
[
"Suppose $P(x) = x^2 + dx + e$ , then let $q(x) = xP(x)$ be a monic cubic of the form $x^3 + dx^2 + ex$ , then it is given that $q(20) = q(22) = q(a) = C$ for some $C$ . Translate down by $C$ , then we get that $q(x) - C$ has roots $20, 22, a$ meaning that $q(x) - C = (x-20)(x-22)(x-a)$ . \n\nThis becomes $x^3 + dx^2 + ex - C = (x-20)(x-22)(x-a)$ . But $P(0) = e$ . By Vieta's, then $e$ is the sum of the roots taken $2$ at a time. Hence $e = 20(22) + 20a + 22a = 440 + 42a$ . This has maximum positive value at $a = -10$ with $e = 20$ . \n\nSo the answer is $P(0) = 20$ . "
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1010,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790593.json"
}
|
How many ways are there to choose distinct positive integers $a, b, c, d$ dividing $15^6$ such that none of $a, b, c,$ or $d$ divide each other? (Order does not matter.)
*Proposed by Miles Yamner and Andrew Wu*
(Note: wording changed from original to clarify)
|
We have $15^6 = 3^65^6$ . Let $a = 3^{x_1}5^{y_1}$ , $b = 3^{x_2}5^{y_2}$ , $c = 3^{x_3}5^{y_3}$ , $d = 3^{x_4}5^{y_4}$ .
Note that if any two $x_i$ or $y_i$ are equal, then the condition is violated (one number must divide the other). So they are all distinct, and the problem is equivalent to choosing $\{x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4\}$ , each nonnegative integers at most $6$ , such that that for all $i,j$ if $x_i < x_j$ then $y_i > y_j$ .
The number of ways to choose our variables so all $x_i$ and $y_i$ are distinct is just $(\binom{7}{4})^2 = 35^2$ . Given this set, since order does not matter, then exchange variables and assume WLOG that $x_1 < x_2 < x_3 < x_4$ . This uniquely forces $y_1 > y_2 > y_3 > y_4$ , and there is exactly $1$ way to rearrange the $y$ exponents in forming $a, b, c, d$ to satisfy this.
In total, the answer is just $35^2 * 1 = 1225$ .
|
[
"<blockquote>We have $15^6 = 3^65^6$ . Let $a = 3^{x_1}5^{y_1}$ , $b = 3^{x_2}5^{y_2}$ , $c = 3^{x_3}5^{y_3}$ , $d = 3^{x_4}5^{y_4}$ .\nIn total, the answer is just $35^2 * 1 = 1225$ .</blockquote>\n\nNot quite right. Keep in mind that ***<u>none of $a, b, c,$ or $d$ divide each other.</u>***\nThe following cases should be excluded.\nIf $a= 3^05^0=1$ , then $a$ divides $b, c$ and $d$ .\nIf $a=3^1$ , then $a$ divides $b=3^{x_2}, x_2 \\in \\{2, 3, 4, 5, 6\\}$ .\nIf $a=5^1$ , then $a$ divides $b=5^{x_2}, x_2 \\in \\{2, 3, 4, 5, 6\\}$ . $\\cdots \\cdots$ ",
"What? I’m not sure what your objection is. \n\nIn any case we coded the problem and 1225 is certainly correct, and this is the intended solution. Maybe you’re trying to say that they need to take the exponent = 0 case into account, but they do so. ",
"Very nice problem!\n\nThe bijection for the exponents to choosing 4 numbers from 0-6 is very nice. "
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 42,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/2790594.json"
}
|
Georgina calls a $992$ -element subset $A$ of the set $S = \{1, 2, 3, \ldots , 1984\}$ a **halfthink set** if
- the sum of the elements in $A$ is equal to half of the sum of the elements in $S$ , and
- exactly one pair of elements in $A$ differs by $1$ .
She notices that for some values of $n$ , with $n$ a positive integer between $1$ and $1983$ , inclusive, there are no halfthink sets containing both $n$ and $n+1$ . Find the last three digits of the product of all possible values of $n$ .
*Proposed by Andrew Wu and Jason Wang*
(Note: wording changed from original to specify what $n$ can be.)
|
<blockquote>Georgina calls a $992$ -element subset $A$ of the set $S = \{1, 2, 3, \ldots , 1984\}$ a **halfthink set** if
- the sum of the elements in $A$ is equal to half of the sum of the elements in $S$ , and
- exactly one pair of elements in $A$ differs by $1$ .
She notices that for some values of $n$ , with $n$ a positive integer between $1$ and $1983$ , inclusive, there are no halfthink sets containing both $n$ and $n+1$ . Find the last three digits of the product of all possible values of $n$ .
*Proposed by Andrew Wu and Jason Wang*
(Note: wording changed from original to specify what $n$ can be.)</blockquote>
<details><summary>Click to expand</summary>All numbers except $989, 991, 993, 995$ have halfthink sets.
Compare with a halfthink set with no consecutive terms: $$ \{1, 3, 5, ..., 989, 991, 994, 996, ..., 1984\} $$ We can get four cases based on if $n$ is even or odd and if $n$ is $992$ or below, or greater than $992$ .
If $n$ is even and $n \leq 992$ , take our no consecutive term set, replace $n-1$ with $n$ , then replace $994$ with $993$ . All good. This even works for $n=992$ .
If $n$ is even and $n > 992$ , take our no consecutive term set, replace $n+2$ with $n+1$ , then replace $991$ with $992$ . All good.
If $n$ is odd and $n < 992$ , take our no consecutive term set, replace $n+2$ with $n+1$ , then replace $991$ with $992$ . However, these two replacements interfere when $n = 989, 991$ , we'll have to deal with these later. The rest are all good.
If $n$ is odd and $n > 992$ , take our no consecutive term set, replace $n-1$ with $n$ , then replace $994$ with $993$ . However, these two replacements interfere when $n = 993, 995$ , we'll have to deal with these later. The rest are all good.
Now, take a look at $n = 989$ . Then we must have $989$ and $990$ in our sequence. To get the maximum value to the left of $989$ , we must take $1, 3, 5, ..., 987$ . To get the maximum value to the right, we must take $992, 994, ..., 1984$ . However, there is one extra value, so we must remove one, and to satisfy the sum constraint, we must remove the value $991$ , which doesn't exist in our sequence.
We can use the same type of reasoning for $989, 991, 993, 995$ , which all don't work. Therefore our answer is $$ (989)(991)(993)(995) \equiv (11)(9)(7)(5) \equiv 465 \pmod {1000}. $$</details>
|
[
"I think this works? If yes, very funny problem. (Though highly doubting this does).\n\n<details><summary>Sol</summary>Consider $n=1000$ . Notice that if we have all the evens or odds below $1000$ , thats $499$ numbers each (we don't include 999 because only one pair of elements differ by 1), and all the evens and odds above $1001$ , we have $491$ odds and $491$ evens. This is the worst case scenario, and this gives us $499+2+491=992$ elements, meaning we need one of these \"all evens/odds\" cases. We can casework and find that none sum to half the elements in $S$ , so $n=1000$ gives no halfthink sets. Then the last 3 digits of the product is $000$ .</details>",
"Maybe we can form a bijection between $\\{a_n\\} \\Longleftrightarrow \\{1985 - a_n\\}$ so that we cannot have $1985-a_n$ if we have $a_n$ where $a_n$ is an element of $A$ that satisfies this? I don't know if it is true (likely not). ",
"No one has solved the problem yet! \n\n<details><summary>Hint</summary>What sets satisfying the first condition have no adjacent integers?</details>",
"By pigeonhole on the sets $\\{1, 2\\}, \\{3, 4\\} ... \\{2k-1, 2k\\}$ (where $2k = n$ ), exactly one integer must be in these $n$ sets. For each of the sets $\\{2m-1, 2m\\}$ , if $2m$ is chosen, then the rest of the integers from the latter sets must be uniquely chosen (the even ones). Since $1+3+...1983 = 1984\\cdot992$ and $1+ 2+ ... 1984 = 1985\\cdot992$ , there must be $992$ evens, specifically the last $992$ . I believe this is the only set if no pairs of elements differ by 1. If exactly one consecutive $\\{n, n+1\\}$ , we may case work into $n=2m$ and $n=2m-1$ . \n\nFor the latter case, \nsubcase 1.1: $n \\le 991$ the numbers $\\{2k-1, 2k\\}$ will be in the set, which from the case where there are no consecutive, will be $\\{2k-1, 2k+1\\}$ . This sum is reduced by 1 so there must be 993 evens. However, this does not work when $n = 989$ since we cannot generate more evens and same for the reason with $n = 991$ . \nsubcase 1.2: $n \\ge 993$ all numbers within this range would work since we can change $n=994$ to $993$ (the sum increases by 1). But this won't work when $n = 993$ or when $n = 995$ . \n\nsubcase 2.1: $n \\le 990$ then all would work since we can change $991$ to $992$ . \nsubcase 2.2: $n \\ge 992$ then all would work since we now have $2k, 2k+1$ , which reduces the sum by 1 but we can account for this by letting the term $991$ be $992$ . \n\nAnd the product is congruent to 465 mod 1000"
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Let $ABC$ be a triangle with $AB = 5$ , $BC = 7$ , and $CA = 8$ , and let $D$ be a point on arc $\widehat{BC}$ of its circumcircle $\Omega$ . Suppose that the angle bisectors of $\angle ADB$ and $\angle ADC$ meet $AB$ and $AC$ at $E$ and $F$ , respectively, and that $EF$ and $BC$ meet at $G$ . Line $GD$ meets $\Omega$ at $T$ . If the maximum possible value of $AT^2$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $\gcd (a,b) = 1$ , find $a + b$ .
*Proposed by Andrew Wu*
|
With **jeteagle.**
The main claim is that $T$ is fixed as the midpoint of arc $\widehat{BC}$ or equivalently $AT$ is the angle bisector of $\angle{BAC}$ . To prove this, we will show that $EF, TD, BC$ are concurrent at $G$ where $T$ is the midpoint of arc $\widehat{BC}$ and $D$ is a variable point on $\widehat{BC}$ .
Let $M$ and $N$ be the midpoints of arcs $\widehat{AB}$ and $\widehat{AC}$ respectively. Let the incenter of $\triangle{ABC}$ be $I$ . By Pascal's Theorem on $ABNDMC$ we see that $E$ , $I$ , $F$ are collinear. If $TD$ and $BC$ intersect at $G$ , then Pascal's Theorem on $TDMCBA$ yields that $G$ , $E$ , and $I$ are collinear. Since $E$ , $I$ , $F$ are collinear, it follows that $G$ lies on line $EF$ proving the claim.
Now for the computational part. Noting that $\angle{BAC} = 60^\circ$ , we have $\angle{BTC} = 120^\circ.$ Since $BC = 7$ and $BT = CT$ , we find that $BT = \frac{7}{\sqrt{3}}$ . By Ptolemy's Theorem, we have $5 \cdot BT + 8 \cdot BT = AT \cdot BC$ which yields $AT = \frac{13}{\sqrt{3}}$ . Thus, $AT^2 = \frac{169}{3}$ so the answer is $169+3 = \boxed{172}.$
|
[
"I am being forced to learn asymptote for the first time :P so I decided I might as well practice on my own problems haha\n\n<details><summary>diagram (possible spoilers!)</summary>[asy]\nunitsize(1.5inch);\npair B = dir(210);\npair C = dir(330);\npair A = dir(134.5736);\nfilldraw(circumcircle(A,B,C), opacity(0.3)+lightcyan, blue);\ndraw(A--B--C--cycle);\npair D = dir(295);\ndraw(A--D, dashed);\ndraw(B--D--C);\npair Z = bisectorpoint(A, D, B);\npair Y = bisectorpoint(A, D, C);\npair E = extension(A, B, D, Z);\npair F = extension(A, C, D, Y);\ndraw(E--D--F, red+dashed);\npair G = extension(E, F, B, C);\ndraw(C--G);\ndraw(E--F--G, green);\npair O=circumcenter(A,B,C);\npair T = -D+2*foot(O,D,G);\ndraw(T--D--G, green);\ndraw(A--T, purple);\ndot(\" $A$ \", A, dir(A));\ndot(\" $B$ \", B, SW);\ndot(\" $C$ \", C, SE);\ndot(\" $D$ \", D, dir(D));\ndot(\" $E$ \", E, dir(E));\ndot(\" $F$ \", F, NE);\ndot(\" $G$ \", G, dir(G));\ndot(\" $T$ \", T, dir(T));\n[/asy]</details>"
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**p1** How many two-digit positive integers with distinct digits satisfy the conditions that
1) neither digit is $0$ , and
2) the units digit is a multiple of the tens digit?**p2** Mirabel has $47$ candies to pass out to a class with $n$ students, where $10\le n < 20$ . After distributing the candy as evenly as possible, she has some candies left over. Find the smallest integer $k$ such that Mirabel could have had $k$ leftover candies.**p3** Callie picks two distinct numbers from $\{1, 2, 3, 4, 5\}$ at random. The probability that the sum of the numbers she picked is greater than the sum of the numbers she didn’t pick is $p$ . $p$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $gcd (a, b) = 1$ . Find $a + b$ .
|
<details><summary>A2 (with sol)</summary>Essentially it means we divide $47-k$ candies to $n$ students evenly, or in other words the largest number less than or equal to $47$ which has a number between $10$ and $19$ inclusive as a factor.
We can try each $k$ starting from $0$ . $k=0$ doesn't work because $47$ is a prime $k=1$ doesn't work because the only factors are $1, 2, 23, 46$ , none of which are between $9$ and $20$ . $k=2$ works because $45$ candies can be divided to $15$ students (each student getting 3 candies), thus $\boxed{2}$ is our answer.</details>
|
[
"<details><summary>A1 (with sol)</summary>We can tell that there's not many integers (and it's the first problem) meaning we can directly do casework: **(i)** When the tens digit is 1, the units digit can be any integer from $2$ to $9$ , which is <u>$8$</u> numbers. **(ii)** When the tens digit is 2, the units digit can be $4, 6, \\text{or } 8$ , which is <u>$3$</u>.**(iii)** When it's 3, the units digit can be $6 \\text{ or } 9$ , which is <u>$2$</u>.**(iv)** Lastly when it's 4 the units digit can be $8$ , which is <u>$1$</u> number. Adding all of those gets us $\\boxed{14}$ numbers in total.</details>",
"<details><summary>A3 (with sol)</summary>The sum of these 5 numbers is $15$ so if the sum of two numbers is greater than the sum of the numbers she didn't pick, that sum can only be $8$ or $9$ ( $9$ is the largest possible sum of two numbers). If the sum is $8$ , it can only be $3$ and $5$ . If the sum is $9$ , it's $4$ and $5$ . $2$ cases out of $\\binom{5}{2} = 10$ different pairs of two numbers is $\\frac{1}{5}$ , getting an answer of $\\boxed{6}$ .</details>",
"Is this a hs comp???!!!",
"Wait, this is a hs school round?\n\n<details><summary>s1</summary>We wish to find two digit positive integers with DISTINCT digits that satisfy the condition that none may be $0$ and the units digit is a multiple of the tens digit. Hence, there is a total possible $9\ncdot 8=72$ combinations of numbers that satisfy the first condition. When the tens digit in $1$ , then the numbers $2-9$ can be the unit digit, hence $8$ possible numbers. A tens digit of $2$ yields $3$ numbers, a tens digit of $3$ yields $2$ numbers, a tens digit of $4$ yields $1$ number, and no tens digit after that is possible. So, there are a total of $14$ numbers</details>\n<details><summary>s2</summary>Suppose we wish to find the smallest remainder of $\\dfrac{47}{n}$ , with $n$ more than or equal to $10$ and less than or equal to $20$ . The smallest remainder is $0$ , yet this cannot be attained, for $47$ is prime. The second smallest remainder, $1$ , cannot be obtained, for $46$ is divisible by both $2$ and $23$ , two numbers not acceptable. However, a remainder of $2$ works, for $45$ is divisble by both $15$ and $3$ , hence $2$ is the smallest value of $k$ .</details>\n<details><summary>s3</summary>There are $10$ total combinations that are possible here for we overcount each pair once, hence a division of $2$ is necessary here. \na, b, (a+b) 15-(a+b)\n1, 2, 3, 12 \n1, 3, 4, 11\n1, 4, 5, 10 \n1, 5, 6, 9\n2, 3, 5, 10\n2, 4, 6, 9\n2, 5, 7, 8\n3, 4, 7, 8\n3, 5, 8, 7\n4, 5, 9, 6\nWe can observe that there are only two iterations in which her sum is greater than the sum of the other three numbers. We can see that $p$ will be $\\dfrac{1}{5}$ , with $a+b$ in the problem being $6$ .</details>"
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**p4** Define the sequence ${a_n}$ as follows:
1) $a_1 = -1$ , and
2) for all $n \ge 2$ , $a_n = 1 + 2 + . . . + n - (n + 1)$ .
For example, $a_3 = 1+2+3-4 = 2$ . Find the largest possible value of $k$ such that $a_k+a_{k+1} = a_{k+2}$ .**p5** The taxicab distance between two points $(a, b)$ and $(c, d)$ on the coordinate plane is $|c-a|+|d-b|$ . Given that the taxicab distance between points $A$ and $B$ is $8$ and that the length of $AB$ is $k$ , find the minimum possible value of $k^2$ .**p6** For any two-digit positive integer $\overline{AB}$ , let $f(\overline{AB}) = \overline{AB}-A\cdot B$ , or in other words, the result of subtracting the product of its digits from the integer itself. For example, $f(\overline{72}) = 72-7\cdot 2 = 58$ . Find the maximum possible $n$ such that there exist distinct two-digit integers $ \overline{XY}$ and $\overline{WZ}$ such that $f(\overline{XY} ) = f(\overline{WZ}) = n$ .
|
<details><summary>A6 (with sol)</summary>$f(\overline{AB}) = 10A+B-A\cdot B = (A-1)(10-B) + 10$ , so essentially, we want $(A-1)(10-B) = (C-1)(10-D)$ where the two numbers are $\overline{AB}$ and $\overline{CD}$ If we want this to be large intuitively we should make A as large and B as small as possible . So we start from $A=9$ and $B=0$ . This doesnt work as there is no other way to form a product of 90. $91$ doesn't work because again, there is no other way to form this product. Similarly 92 doesn't work. However, 93, does - the product is 56 and we can also form it by C=8 D=2. So $n$ is $56+10 = \boxed{66}$ A=8 B=2 C=9 D=3</details>
|
[
"<details><summary>A4 (with sol)</summary>For all $n \\le 2$ , $a_n = \\frac{n(n+1)}{2}-(n+1) = \\frac{n^2-n-2}{2} = \\frac{(n+1)(n-2)}{2}$ . \nSo the problem wants the largest possible value of $k$ where $$ \\frac{(n+1)(n-2)}{2} + \\frac{(n+2)(n-1)}{2} = \\frac{(n+3)(n)}{2} $$ Solving this yields $\\boxed{4}$ .</details>"
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"answer_score": 1018,
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**p7** Cindy cuts regular hexagon $ABCDEF$ out of a sheet of paper. She folds $B$ over $AC$ , resulting in a pentagon. Then, she folds $A$ over $CF$ , resulting in a quadrilateral. The area of $ABCDEF$ is $k$ times the area of the resulting folded shape. Find $k$ .**p8** Call a sequence $\{a_n\} = a_1, a_2, a_3, . . .$ of positive integers *Fib-o’nacci* if it satisfies $a_n = a_{n-1}+a_{n-2}$ for all $n \ge 3$ . Suppose that $m$ is the largest even positive integer such that exactly one *Fib-o’nacci* sequence satisfies $a_5 = m$ , and suppose that $n$ is the largest odd positive integer such that exactly one *Fib-o’nacci* sequence satisfies $a_5 = n$ . Find $mn$ .**p9** Compute the number of ways there are to pick three non-empty subsets $A$ , $B$ , and $C$ of $\{1, 2, 3, 4, 5, 6\}$ , such that $|A| = |B| = |C|$ and the following property holds: $$ A \cap B \cap C = A \cap B = B \cap C = C \cap A. $$
|
<details><summary>A8 (with sol)</summary>Note that $a_5$ represented in terms of $a_1$ and $a_2$ is $2a_1 + 3a_2$ . If there is exactly one, $m$ and $n$ can't be very large, so one quick sol is to just try each number. For example for $m$ we can start from... $8$ (only 2*1 + 3*2), which works. We don't need to try everything, we can jump to another number like.. $12$ , which also works (only 2*3 + 3*2). Let's go a bit further, to $14$ . This obvsly doesn't work- it an be $2*1 + 3*4$ or $2*4 + 3*2$ . Thus $m=12$ . Similar process finds $n=12$ , so our answer is $\boxed{109}$ .</details>
|
[
"<details><summary>A7 (with sol)</summary>Very simple problem - if we fold B over AC the pentagon is $ACDEF$ , and if we fold A over CF the quadrilateral is $FEDC$ which is half of $ABCDEF$ , so $k=\\boxed{2}$ .</details>"
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**p10**Kathy has two positive real numbers, $a$ and $b$ . She mistakenly writes $$ \log (a + b) = \log (a) + \log( b), $$ but miraculously, she finds that for her combination of $a$ and $b$ , the equality holds. If $a = 2022b$ , then $b = \frac{p}{q}$ , for positive integers $p, q$ where $gcd(p, q) = 1$ . Find $p + q$ .**p11** Points $X$ and $Y$ lie on sides $AB$ and $BC$ of triangle $ABC$ , respectively. Ray $\overrightarrow{XY}$ is extended to point $Z$ such that $A, C$ , and $Z$ are collinear, in that order. If triangle $ ABZ$ is isosceles and triangle $CYZ$ is equilateral, then the possible values of $\angle ZXB$ lie in the interval $I = (a^o, b^o)$ , such that $0 \le a, b \le 360$ and such that $a$ is as large as possible and $b$ is as small as possible. Find $a + b$ .**p12** Let $S = \{(a, b) | 0 \le a, b \le 3, a$ and $b$ are integers $\}$ . In other words, $S$ is the set of points in the coordinate plane with integer coordinates between $0$ and $3$ , inclusive. Prair selects four distinct points in $S$ , for each selected point, she draws lines with slope $1$ and slope $-1$ passing through that point. Given that each point in $S$ lies on at least one line Prair drew, how many ways could she have selected those four points?
|
<details><summary>answer p10</summary>$\frac{p}{q} = \frac{2023}{2022}$ . Since they are relatively prime, $4045$ .</details>
|
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**p13** Let $ABCD$ be a square. Points $E$ and $F$ lie outside of $ABCD$ such that $ABE$ and $CBF$ are equilateral triangles. If $G$ is the centroid of triangle $DEF$ , then find $\angle AGC$ , in degrees.**p14**The silent reading $s(n)$ of a positive integer $n$ is the number obtained by dropping the zeros not at the end of the number. For example, $s(1070030) = 1730$ . Find the largest $n < 10000$ such that $s(n)$ divides $n$ and $n\ne s(n)$ .**p15** Let $ABCDEFGH$ be a regular octagon with side length $12$ . There exists a region $R$ inside the octagon such that for each point $X$ in $R$ , exactly three of the rays $AX$ , $BX$ , $CX$ , $DX$ , $GX$ , and $HX$ intersect segment $EF$ . If the area of region $R$ can be expressed as $a -b\sqrt{c}$ for positive integers $a, b, c$ with $c$ squarefree, find $a + b + c$ .
|
<details><summary>13</summary>the answer should be $120^{\circ}$ Since $BE=BF;DE=DF$ and $\triangle{DEF}$ is equilateral triangle, it is clear that the centroid of the triangle lies on $BD$ $\angle{DAE}=150^{\circ},\angle{DAG}=75^{\circ}$ , it îs clear that $\triangle{AGD}\cong \triangle{CGD}$ , $\angle{AGC}=2\angle{AGD}=2*60^{\circ}=120^{\circ}$</details>
|
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**p16** Madelyn is being paid $\$ 50 $/hour to find useful *Non-Functional Trios*, where a Non-Functional Trio is defined as an ordered triple of distinct real numbers $ (a, b, c) $, and a Non- Functional Trio is *useful* if $ (a, b) $, $ (b, c) $, and $ (c, a) $ are collinear in the Cartesian plane. Currently, she’s working on the case $ a+b+c = 2022 $. Find the number of useful Non-Functional Trios $ (a, b, c) $ such that $ a + b + c = 2022 $.**p17** Let $ p(x) = x^2 - k $, where $ k $ is an integer strictly less than $ 250 $. Find the largest possible value of k such that there exist distinct integers $ a, b $ with $ p(a) = b $ and $ p(b) = a $.**p18** Let $ ABC $ be a triangle with orthocenter $ H $ and circumcircle $ \Gamma $ such that $ AB = 13 $, $ BC = 14 $, and $ CA = 15 $. $ BH $ and $ CH $ meet $ \Gamma $ again at points $ D $ and $ E $, respectively, and $ DE $ meets $ AB $ and $ AC $ at $ F $ and $ G $, respectively. The circumcircles of triangles $ ABG $ and $ ACF $ meet BC again at points $ P $ and $ Q $. If $ PQ $ can be expressed as $ \frac{a}{b} $ for positive integers $ a, b $ with $ gcd (a, b) = 1 $, find $ a + b$.
|
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|
**p1.** Suppose that $x$ and $y$ are positive real numbers such that $\log_2 x = \log_x y = \log_y 256$ . Find $xy$ .**p2.** Let the roots of $x^2 + 7x + 11$ be $r$ and $s$ . If f(x) is the monic polynomial with roots $rs + r + s$ and $r^2 + s^2$ , what is $f(3)$ ?**p3.** Call a positive three digit integer $\overline{ABC}$ fancy if $\overline{ABC} = (\overline{AB})^2 - 11 \cdot \overline{C}$ . Find the sum of all fancy integers.**p4.** In triangle $ABC$ , points $D$ and $E$ are on line segments $BC$ and $AC$ , respectively, such that $AD$ and $BE$ intersect at $H$ . Suppose that $AC = 12$ , $BC = 30$ , and $EC = 6$ . Triangle $BEC$ has area $45$ and triangle $ADC$ has area $72$ , and lines $CH$ and $AB$ meet at $F$ . If $BF^2$ can be expressed as $\frac{a-b\sqrt{c}}{d}$ for positive integers $a$ , $b$ , $c$ , $d$ with $c$ squarefree and $gcd(a, b, d) = 1$ , then find $a + b + c + d$ .**p5.** Find the minimum possible integer $y$ such that $y > 100$ and there exists a positive integer $x$ such that $x^2 + 18x + y$ is a perfect fourth power.**p6.** Let $ABCD$ be a quadrilateral such that $AB = 2$ , $CD = 4$ , $BC = AD$ , and $\angle ADC + \angle BCD = 120^o$ . If the sum of the maximum and minimum possible areas of quadrilateral $ABCD$ can be expressed as $a\sqrt{b}$ for positive integers $a, b$ with $b$ squarefree, then find $a + b$ .
PS. You had better use hide for answers. Collected [here](https://artofproblemsolving.com/community/c5h2760506p24143309).
|
3.I will assume $a\not=0$ . Rewriting $100a+10b+c=(10a+b)^2-11c$ gives $12c+25=100(a^2-a)+(b-5)^2+20ab$ .
Then $100(a^2-a)\le 108+25=133\implies a^2-a\le 1\implies a=1\implies 12c+25=(b+5)^2\implies 12c=b(b+10)$ .
Then $b$ is even and RHS is divisible by 3, which gives $100,122,168$ .
5. Note that $x=6,y=112$ gives a solution. Now suppose that $x^2+18x+y=k^4$ with $k>0$ and $y<112$ . Then $t:=y-81=k^4-(x+9)^2=(k^2+x+9)(k^2-x-9)$ and $t<31$ . Write $a,b$ for the factors, then $t=ab, a,b>0$ and $a-b=2(x+9)\ge 20\implies a\ge b+20 \implies 31>t\ge b(b+20)\implies b=1 \implies k^2=x+10\implies t=2k^2-1\implies 2k^2-1<31\implies k\le 3$ . But then $k^2=x+10$ is impossible. Hence 112 is minimal.
|
[
"P2:\n<details><summary>Click to expand</summary>$r + s = -7, rs = 11$ $f(x) = x^{2} - (rs + r + s)x + (r^{2} + s{2})$ $= x^{2} - (11 - 7)x + (r + s)^{2} - 2rs$ $ = x^{2} - 4x + 49 - 23$ $ = x^{2} - 4x + 27$ $f(3) = 9 - 12 + 27 = 24$</details>",
"P1:\n<details><summary>Click to expand</summary>Changing bases to $2$ , we get $(\\log_2 x)^{3} = 8$ $\\log_2 x = 2$ $x = 4$ $\\log_2 y = 4$ $y = 16$ Therefore $xy = 64$ Checking, we find $\\log_2 x = 2, \\log_x y = \\log_4 16 = 2, \\log_16 256 = 2$</details>"
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}
|
**p1.** Find the smallest positive integer $N$ such that $2N -1$ and $2N +1$ are both composite.**p2.** Compute the number of ordered pairs of integers $(a, b)$ with $1 \le a, b \le 5$ such that $ab - a - b$ is prime.**p3.** Given a semicircle $\Omega$ with diameter $AB$ , point $C$ is chosen on $\Omega$ such that $\angle CAB = 60^o$ . Point $D$ lies on ray $BA$ such that $DC$ is tangent to $\Omega$ . Find $\left(\frac{BD}{BC} \right)^2$ .**p4.** Let the roots of $x^2 + 7x + 11$ be $r$ and $s$ . If $f(x)$ is the monic polynomial with roots $rs + r + s$ and $r^2 + s^2$ , what is $f(3)$ ?**p5.** Regular hexagon $ABCDEF$ has side length $3$ . Circle $\omega$ is drawn with $AC$ as its diameter. $BC$ is extended to intersect $\omega$ at point $G$ . If the area of triangle $BEG$ can be expressed as $\frac{a\sqrt{b}}{c}$ for positive integers $a, b, c$ with $b$ squarefree and $gcd(a, c) = 1$ , find $a + b + c$ .**p6.** Suppose that $x$ and $y$ are positive real numbers such that $\log_2 x = \log_x y = \log_y 256$ . Find $xy$ .**p7.** Call a positive three digit integer $\overline{ABC}$ fancy if $\overline{ABC} = (\overline{AB})^2 - 11 \cdot \overline{C}$ . Find the sum of all fancy integers.**p8.** Let $\vartriangle ABC$ be an equilateral triangle. Isosceles triangles $\vartriangle DBC$ , $\vartriangle ECA$ , and $\vartriangle FAB$ , not overlapping $\vartriangle ABC$ , are constructed such that each has area seven times the area of $\vartriangle ABC$ . Compute the ratio of the area of $\vartriangle DEF$ to the area of $\vartriangle ABC$ .**p9.** Consider the sequence of polynomials an(x) with $a_0(x) = 0$ , $a_1(x) = 1$ , and $a_n(x) = a_{n-1}(x) + xa_{n-2}(x)$ for all $n \ge 2$ . Suppose that $p_k = a_k(-1) \cdot a_k(1)$ for all nonnegative integers $k$ . Find the number of positive integers $k$ between $10$ and $50$ , inclusive, such that $p_{k-2} + p_{k-1} = p_{k+1} - p_{k+2}$ .**p10.** In triangle $ABC$ , point $D$ and $E$ are on line segments $BC$ and $AC$ , respectively, such that $AD$ and $BE$ intersect at $H$ . Suppose that $AC = 12$ , $BC = 30$ , and $EC = 6$ . Triangle BEC has area 45 and triangle $ADC$ has area $72$ , and lines CH and AB meet at F. If $BF^2$ can be expressed as $\frac{a-b\sqrt{c}}{d}$ for positive integers $a$ , $b$ , $c$ , $d$ with c squarefree and $gcd(a, b, d) = 1$ , then find $a + b + c + d$ .**p11.** Find the minimum possible integer $y$ such that $y > 100$ and there exists a positive integer x such that $x^2 + 18x + y$ is a perfect fourth power.**p12.** Let $ABCD$ be a quadrilateral such that $AB = 2$ , $CD = 4$ , $BC = AD$ , and $\angle ADC + \angle BCD = 120^o$ . If the sum of the maximum and minimum possible areas of quadrilateral $ABCD$ can be expressed as $a\sqrt{b}$ for positive integers $a$ , $b$ with $b$ squarefree, then find $a + b$ .
PS. You had better use hide for answers. Collected [here](https://artofproblemsolving.com/community/c5h2760506p24143309).
|
<details><summary>Solution for p1</summary>Checking $N=1, 2, 3, \ldots$ going up, we see that the answer is $$ N = \boxed{13}. $$ $\square$</details>
|
[
"<details><summary>p6</summary>real solutions $$ (x=4,y=16) \\Longrightarrow xy=64 $$</details>",
"P4:\n<details><summary>Click to expand</summary>$r + s = -7, rs = 11$ $rs + r + s = 11 - 7 = 4$ , $r^2 + s^2 = (r + s)^2 - 2rs = 49 - 22 = 27$ $f(x) = x^2 - 4x + 27$ $f(3) = 9 - 12 + 27 = 24$</details>",
"P3:\n<details><summary>Click to expand</summary>Angle chasing, we get $\\angle BCD = 120, \\angle BCD = 30$ . $BC = r \\sqrt{3}$ .\nSine rule gives $BD = 3r$ .\nRequired ratio $= 3$ .</details>",
"P6 proof:\n<details><summary>Click to expand</summary>Let $a=log_2 x$ , we have \n\\[ a= \\frac{\\ln x}{\\ln 2} = \\frac{\\ln y}{\\ln x} = \\frac{ 8 \\ln 2}{\\ln y} \\]\nThen $a^3=8$ and $a=2$ (we 're in reals). Then $x=2^2=4$ and $y=x^2=4^2=16$ , and $\\log_{16}{256}=2$ too. Then $(x,y)=(4,16)$ is the only real solution, and $xy=64$ .</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Girls in Math at Yale"
] |
{
"answer_score": 1108,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Girls in Math at Yale/3195043.json"
}
|
(a) Find the value of the real number $k$ , for which the polynomial $P(x)=x^3-kx+2$ has the number $2$ as a root. In addition, for the value of $k$ you will find, write this polynomial as the product of two polynomials with integer coefficients.
(b) If the positive real numbers $a,b$ satisfy the equation $$ 2a+b+\frac{4}{ab}=10, $$ find the maximum possible value of $a$ .
|
<blockquote>(b) If the positive real numbers $a,b$ satisfy the equation $$ 2a+b+\frac{4}{ab}=10, $$ find the maximum possible value of $a$ .</blockquote>
We have $5-a=\frac b2+\frac 2{ab}>0$ and so $a<5$ Squaring, we have $(5-a)^2=\frac{b^2}4+\frac 2a+\frac 4{a^2b^2}$ $=\frac 4a+\left(\frac b2-\frac 2{ab}\right)^2\ge \frac 4a$ And so $a(a-5)^2\ge 4$ , which is $(a-4)(a^2-6a+1)\ge 0$ , which, associated to $a<5$ , implies $\boxed{a\le 4}$ , which indeed fits (with $b=1$ )
|
[
"<blockquote>(a) Find the value of the real number $k$ , for which the polynomial $P(x)=x^3-kx+2$ has the number $2$ as a root. In addition, for the value of $k$ you will find, write this polynomial as the product of two polynomials with integer coefficients.</blockquote> $2^3-2k+2=0$ $\\implies$ $\\boxed{k=5}$ and then, just dividing by $x-2$ : $\\boxed{x^3-5x+2=(x-2)(x^2+2x-1)}$ ",
"<blockquote><blockquote>(b) If the positive real numbers $a,b$ satisfy the equation $$ 2a+b+\\frac{4}{ab}=10, $$ find the maximum possible value of $a$ .</blockquote>\n $(a-4)(a^2-6a+1)\\ge 0$ , which, associated to $a<5$ , implies $\\boxed{a\\le 4}$ , which indeed fits (with $b=1$ )</blockquote>\n\nWhy cannot be $3+\\sqrt 2 \\leqslant a < 5$ ?\n\nEDIT: oops.. I made a mistake when I did the expansion of the $a^2 -6a+1$ .",
"For part b\n<blockquote>b) If the positive real numbers $a,b$ satisfy the equation $$ 2a+b+\\frac{4}{ab}=10 $$ </blockquote> $$ 2a=10-\\left(b+\\frac{4}{ab}\\right)\\leq 10-4a^{-1/2} $$ \nhence for max value $$ 2a-10+4a^{-1/2}=0 $$ , which yields the max value as $4$ .",
"<blockquote> Why cannot be $3+\\sqrt 2 \\leqslant a < 5$ ?</blockquote>\nBecause for $3+\\sqrt 2 \\leqslant a < 5$ , we have $a-4>0$ and $a^2-6a+1<0$ and so $(a-4)(a^2-6a+1)<0$ ; not suitable\n",
"***a)***\nAs 2 is root of $P(x)$ , $2^3 - 2k + 2 = 0$ $<=>$ $10 - 2k = 0$ $<=>$ $-2k = -10$ $<=>$ $2k = 10$ $<=>$ $k = 5$ Now, our polynomial $P(x)$ states $x^3 - 5x + 2$ . \nSince 2 is a root, we shall divide $P(x)$ with $2$ , we will use long division. $(x^3 - 5x + 2) : (x - 2) = x^2 + 2x - 1$ .**b)**\n\nAs $b + \\frac{4}{ab}$ is positive, $10 - 2a$ is also positive. That means $a < 5$ Similar calculations give us $b < 10$ For $b = 1$ , the equation becomes $2a + \\frac{4}{a} = 9$ and it's solutions is $a = 4$ .\nFor $b > 1$ , $a$ gets smaller and smaller.\nFor $b < 1$ , $a$ either gives us $a > 5$ or $a < 4$ .\n\nHence the maxiumum value od $a$ is $4$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 Greece Junior Math Olympiad"
] |
{
"answer_score": 1026,
"boxed": false,
"end_of_proof": false,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790115.json"
}
|
Let $ABC$ be an isosceles triangle, and point $D$ in its interior such that $$ D \hat{B} C=30^\circ, D \hat{B}A=50^\circ, D \hat{C}B=55^\circ $$
(a) Prove that $\hat B=\hat C=80^\circ$ .
(b) Find the measure of the angle $D \hat{A} C$ .
|
For the part (b) we denote by $\alpha$ measure of the $\angle DAC$ . By the Law of Sines in $\triangle ABD$ and $\triangle ACD$ we have that $$ 1 = \frac{BA}{AD}\cdot \frac{DA}{AC} = \frac{\sin (70^\circ - \alpha)}{\sin 50^\circ} \cdot \frac{\sin 25^\circ}{\sin (25^\circ + \alpha)} $$ **Claim.** $\alpha = 5^\circ$ works
*Proof.* We need to show that $\sin 30^\circ \cdot \sin 50^\circ = \sin 25^\circ \cdot \sin 65^\circ$ . Which is clear, since $2\sin 25^\circ \cdot \sin 65^\circ = 2\sin 25^\circ \cos 25^\circ = \sin 50^\circ$ . $\square$ Notice that measure of $\alpha$ is define in unique way by the condition of the problem, so we conclude that answer is $\boxed{5^\circ}$
|
[
"If I am not wrong, Part a is trivial because \\[\\angle ABC=\\angle DBA+\\angle DBC=80\\] and since it's isoceles we're done.",
"Pretty sure this is the intended solution:\n\nLet $E$ be the point on the extension of $BD$ such that $AB=AE$ . Thus, $AE=AC$ .\n\nThen, $\\angle ABE=\\angle AEB=50$ , and $\\angle CAE=\\angle BAE-\\angle ABC=60$ .\nThus, $AEC$ is an equilateral triangle and so $AE=EC$ . \n\nAt the same time, $\\angle AED$ is twice of $\\angle ACD$ , so $E$ is the circumcenter of $ACD$ .\n\nThis means $AE=AD$ , so $\\angle DAE=65$ . Finally, $\\angle DAC=5$ .\n",
"<blockquote>Let $ABC$ be an isosceles triangle, and point $D$ in its interior such that $$ D \\hat{B} C=30^\\circ, D \\hat{B}A=50^\\circ, D \\hat{C}B=55^\\circ $$ \n\n(a) Prove that $\\hat B=\\hat C=80^\\circ$ .\n(b) Find the measure of the angle $D \\hat{A} C$ .</blockquote>\n\nLet E a point on segment BD, such that $\\angle ECB=30$ . Point $E$ is on the bicector of angle $A$ .\n\nIt is very easy to see that rays $CD,ED$ are agnle bisectors in triangle $AED$ , so point $D $ in the incenter of triangle $AEC$ . \n\nThe rest is obvious, since $\\angle EAC=10^0$ .\n\n",
"<blockquote>If I am not wrong, Part a is trivial because \\[\\angle ABC=\\angle DBA+\\angle DBC=80\\] and since it's isoceles we're done.</blockquote>\n\nIt is but you have to check the cases $AC=BC$ and $AB=BC$ ",
"<blockquote><blockquote>If I am not wrong, Part a is trivial because \\[\\angle ABC=\\angle DBA+\\angle DBC=80\\] and since it's isoceles we're done.</blockquote>\n\nIt is but you have to check the cases $AC=BC$ and $AB=BC$ </blockquote>\n\nActually, there are 3 cases.\n* $AB$ $=$ $BC$ (sides where B lies are equal)\n* $AB$ $=$ $AC$ (sides where A lies are equal)\n* $AC$ $=$ $BC$ (sides where C lies are equal)",
"<blockquote><blockquote>If I am not wrong, Part a is trivial because \\[\\angle ABC=\\angle DBA+\\angle DBC=80\\] and since it's isoceles we're done.</blockquote>\n\nIt is but you have to check the cases $AC=BC$ and $AB=BC$ </blockquote>\n\nBut, assuming $AB=BC$ , $D$ will be outside triangle, so there is no question!",
"a) casework\nb) <details><summary>look for incenter</summary>Let M be the midpoint of base BC and O the interesection of AM with BD.\n\nBO=OC =><OBC=<OCB=30^o => <BOM=<MOB=60^o\n=> <AOM=<BOM=60^o , finally <DOC=180^o-60^o-60^o=60^o, so OD bisects <AOC (1)\n\n<OCD=<ACD=25^o => DC bisects <ACO (2)\n\nCombining (1),(2) we get that D is incenter of triangle AOC, therefore AD bisects <OAC, which gives <DAC=5^o[\\hide]</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Greece Junior Math Olympiad"
] |
{
"answer_score": 1124,
"boxed": false,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790118.json"
}
|
On the board we write a series of $n$ numbers, where $n \geq 40$ , and each one of them is equal to either $1$ or $-1$ , such that the following conditions both hold:
(i) The sum of every $40$ consecutive numbers is equal to $0$ .
(ii) The sum of every $42$ consecutive numbers is not equal to $0$ .
We denote by $S_n$ the sum of the $n$ numbers of the board. Find the maximum possible value of $S_n$ for all possible values of $n$ .
|
Let out sequence is $a_1, a_2, ... , a_{40}, ...$ and notice that after $a_{40}$ we must have $a_1$ again, then $a_2$ , etc. So, our sequence is repeated by the first $40$ elements. By the second condition we have that $a_i + a_{i+1} \neq 0$ . Now we have $a_1=a_2=...=a_{20}=x$ and $a_{21}=a_{22}=...=a_{40}=-x$ , where $x\in \{-1,1\}$ . One can see that this type of sequence satisfies to the condition of the problem. Write $n=40m+k$ , where $m\geqslant 1$ and $40>k\geqslant 0$ . We don't care about sum of first $m$ $40$ tuples. So our problem asked us to find the maximum of the sum of first $k$ elements of the $a_1,a_2,...,a_{40}$ . Of course, its happen when $k=20$ and $x=1$ , so the answer is $\boxed{20}$
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Greece Junior Math Olympiad"
] |
{
"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790120.json"
}
|
Find all couples of non-zero integers $(x,y)$ such that, $x^2+y^2$ is a common divisor of $x^5+y$ and $y^5+x$ .
|
Let $\gcd(x,y)=d,x=ad,y=bd$ . Obviously, $\gcd(a,b)=1$ .
Note that $d^2\mid d^2(a^2+b^2)\implies d^2\mid x^2+y^2$ .
But $x^5+y=d(a^5d^4+b)\implies d^2\mid d(a^5d^4+b)\implies d\mid b$ Similarly, $d\mid a$ . Thus, $d\mid \gcd(a,b)=1\implies d=1\implies\gcd(x,y)=1$ . $x^2+y^2\mid (x^5+y)(x^5-y)=(x^2)^5-y^2\implies (-y^2)^5-y^2\equiv (x^2)^5-y^2\equiv 0\pmod{x^2+y^2}$ . $x^2+y^2\mid y^{10}-y^2\implies x^2+y^2\mid y^8-1$ since $\gcd(x^2+y^2,y^2)=1$ .
So, $x^2+y^2\mid y^8-1-y^3(y^5+x)=-(xy^3+1)$ .
Similarly, $x^2+y^2\mid x^8-1\implies x^2+y^2\mid x^8-1-x^3(x^5+y)=-(x^3y+1)$ . $x^2+y^2\mid (x^3y+1)-(xy^3+1)=xy(x^2-y^2)$ .
Since $\gcd(x^2+y^2,x)=\gcd(x^2+y^2,y)=1$ , $x^2+y^2\mid x^2-y^2$ .
If $x^2-y^2\neq 0$ then $|x^2-y^2|\geq |x^2+y^2|\implies \max(x^2,y^2)\geq x^2+y^2>\max(x^2,y^2)$ which is absurd.
Thus, $x^2=y^2\implies x=\pm y$ . But since $\gcd(x,y)=1$ , $x,y\in\{-1,1\}$ .
It turn out that $(x,y)=(-1,1),(1,-1),(-1,-1),(1,1)$ all works.
|
[
"My solution as I solved it in the competition:\nFirst imply $\\gcd(x,y)=1$ as above. Then we have $x^2+y^2 \\mid x^2+y^2+x^2(x^3y-1)\\implies x^2+y^2 \\mid x^3y-1$ since $\\gcd(x^2,x^2+y^2)=1$ Same we get $x^2+y^2 \\mid y^3x-1$ . Combining these two we get $x^2+y^2 \\mid 2$ and we get the solutions $(x,y)=(-1,1),(1,-1),(-1,-1),(1,1)$ ",
"Claim : $\\gcd(x,y) = 1$ .\nProof : $\\gcd(x,y) = d$ and $x = dx', y = dy', \\gcd(x',y') = 1$ . $d^2(x'^2 + y'^2) \\mid d(d^4x' + y') \\implies d \\mid (d^4x' + y') \\implies d \\mid y'$ and $d^2(x'^2 + y'^2) \\mid d(d^4y' + x') \\implies d \\mid (d^4y' + x') \\implies d \\mid x'$ so $d = 1$ . $x^2 + y^2 \\mid x^5 + x^3y^2 , x^2 + y^2 \\mid x^5 + y \\implies x^2 + y^2 \\mid x^3y^2 - y \\implies x^2 + y^2 \\mid x(y^3x-1) \\implies x^2 + y^2 \\mid y^3x-1$ . $x^2 + y^2 \\mid y^3x + x^3y , x^2 + y^2 \\mid y^3x-1 \\implies x^2 + y^2 \\mid x^3y - 1 \\implies x^2 + y^2 \\mid xy(x^2 + y^2) - 2 \\implies x^2 + y^2 \\mid 2$ .\nAnswers : $(x,y)=(1,1),(1,-1),(-1,1),(-1,-1)$ ",
" Good morning. I solve this problem in 2021. \n WLOG x>y . x^2+y^2 | x^5+y then x^2+y^2 | x^6+yx and\n x^2+y^2 | yx+y^6. x^2+y^2 | 2xy we have x,y>0 cause if x<0 then -x>0 we may change x and -x.\n2xy>x^2+y^2 (x-y)^2<0 contraddiction.Then\nx=y 2x^2 |x(x^4+1) \n2x | x^4+1 then sols (1,1);(1,-1)(-1,1);(-1;-1)\n ",
"why post same thing 2 times? \nalso this is a 2022 olympiad problem so u can’t have done it in 2021 lol",
"<blockquote>why post same thing 2 times? \nalso this is a 2022 olympiad problem so u can’t have done it in 2021 lol</blockquote>\n\nHello, some problems featured on olympiads have been tampered with in some form in the past; the problem or a similar version of it can exist well before the inclusion of it on the olympiad or the exam",
"oh ok i didn’t know that "
] |
[
"origin:aops",
"2022 Contests",
"2022 Greece Junior Math Olympiad"
] |
{
"answer_score": 40,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Greece Junior Math Olympiad/2790121.json"
}
|
Let $ABC$ be a triangle such that $AB<AC<BC$ . Let $D,E$ be points on the segment $BC$ such that $BD=BA$ and $CE=CA$ . If $K$ is the circumcenter of triangle $ADE$ , $F$ is the intersection of lines $AD,KC$ and $G$ is the intersection of lines $AE,KB$ , then prove that the circumcircle of triangle $KDE$ (let it be $c_1$ ), the circle with center the point $F$ and radius $FE$ (let it be $c_2$ ) and the circle with center $G$ and radius $GD$ (let it be $c_3$ ) concur on a point which lies on the line $AK$ .
|
Let $I$ be a incenter of $\triangle ABC$ . Denote by $2\alpha, 2\beta, 2\gamma$ angles $\angle A, \angle B, \angle C$ of the $\triangle ABC$ . Let $\odot (ABI)$ intersect side $\overline{BC}$ in $E'$ .**Claim.** $E' \equiv E$ *Proof.* We have that $\angle KE'C = 180^\circ - \angle AE'B = 180^\circ - \angle AIB = 90^\circ - \gamma \Longrightarrow \angle E'AC = 90^\circ - \gamma \Longrightarrow CE'=CA$ . $\square$ Analogously, for point $D$ .
[asy]
size(10cm);
import olympiad;
pair A,B,C,K,F,E,D,T, G, P;
A = dir(110);
B = dir(170);
C = dir(10);
K = incenter(A,B,C);
E = intersectionpoints(B--C, circle(C, 1.53))[0];
D = intersectionpoints(B--C, circle(B, 1))[0];
G = intersectionpoint(B--K, A--E);
F = intersectionpoint(C--K, A--D);
P = 11*K-10*A;
T = intersectionpoints(A--P, circumcircle(K,D,E))[1];
draw(A--B--C--A);
draw(A--D);
draw(A--E);
draw(circumcircle(K,D,E), dashed);
draw(circumcircle(A,B,K), dotted);
draw(circumcircle(A,C,K), dotted);
draw(A--T);
draw(F--E);
draw(F--T);
draw(G--D);
draw(G--T);
draw(T--E);
draw(T--D);
draw(F--G);
draw(B--K);
draw(C--K);
draw(K--E);
draw(K--D);
label(scale(0.8)*" $A$ ", A, dir(110), filltype = Fill(white));
label(scale(0.8)*" $B$ ", B, dir(200), filltype = Fill(white));
label(scale(0.8)*" $C$ ", C, dir(-20), filltype = Fill(white));
label(scale(0.8)*" $K$ ", K, dir(60), filltype = Fill(white));
label(scale(0.8)*" $T$ ", T, dir(-70), filltype = Fill(white));
label(scale(0.8)*" $D$ ", D, dir(-40), filltype = Fill(white));
label(scale(0.8)*" $E$ ", E, dir(200), filltype = Fill(white));
label(scale(0.8)*" $G$ ", G, dir(160), filltype = Fill(white));
label(scale(0.8)*" $F$ ", F, dir(50), filltype = Fill(white));
dot(A);
dot(B);
dot(C);
dot(E);
dot(D);
dot(K);
dot(G);
dot(F);
dot(T);
[/asy]
Observe that $\angle AGK = \angle BGA + \angle BAG = \beta + (2\alpha + \gamma - 90^\circ) = \alpha = \angle KAC = \angle KDC$ , so $G$ lie on the circumcircle of the $\triangle KDE$ . By the same argument we have that $F,G \in \odot (KDE)$ .**Claim.** $I \equiv K$ *Proof.* Just notice that $\angle AIE = 180^\circ - 2\beta$ and $\angle ADE = \angle ACE + \angle CAE = 2\gamma + (2\alpha - (90^\circ - \beta)) = 90^\circ - \beta$ , so $\angle AIE = 2\angle ADE$ . Similarly, $\angle AID =2\angle AED$ and the conclusion is follows. $\square$ Set $T = \odot (KDE) \cap \overline{AK}$ . To finish the proof we just notice that $\angle GTD = 180^\circ - \angle GKD = 180^\circ - \angle GKA = \angle GKT = \angle GDT$ , so $GT=GT$ and similarly $FT=FE$ . $\blacksquare$
|
[
"<blockquote>Let $I$ be a incenter of $\\triangle ABC$ . Denote by $2\\alpha, 2\\beta, 2\\gamma$ angles $\\angle A, \\angle B, \\angle C$ of the $\\triangle ABC$ . Let $\\odot (ABI)$ intersect side $\\overline{BC}$ in $E'$ .**Claim.** $E' \\equiv E$ *Proof.* We have that $\\angle KE'C = 180^\\circ - \\angle AE'B = 180^\\circ - \\angle AIB = 90^\\circ - \\gamma \\Longrightarrow E'AC = 90^\\circ - \\gamma \\Longrightarrow CE'=CA$ . $\\square$ Analogously, for point $D$ .**Claim.** $I \\equiv K$ *Proof.* Just notice that $\\angle AIE = 180^\\circ - 2\\beta$ and $\\angle ADE = \\angle ACE + \\angle CAE = 2\\gamma + (2\\alpha - (90^\\circ - \\beta)) = 90^\\circ - \\beta$ , so $\\angle AIE = 2\\angle ADE$ . Similarly, $\\angle AID =2\\angle AED$ and the conclusion is follows. $\\square$ Set $T = \\odot (KDE) \\cap \\overline{KL}$ . To finish the proof we just notice that $\\angle GTD = 180^\\circ - \\angle GKD = 180^\\circ - \\angle GKA = \\angle GKT = \\angle GDT$ , so $GT=GT$ and similarly $FT=FE$ . $\\blacksquare$ </blockquote>\n\nfantastic!!! amazing dude",
"Note that $ABD$ is isosceles so $BK$ is angle bisector. with same approach for $CK$ we have $K$ is incenter of $ABC$ . $\\angle AEB = \\angle 90 + \\frac{\\angle C}{2} = \\angle AKB \\implies AKEB$ is cyclic. with same approach $AKDC$ is cyclic. $\\angle GEK = \\angle AEK = \\angle EAK = \\angle GDK \\implies GEDK$ is cyclic. with same approach $FDEK$ is cyclic. Let $KGEDF$ meet $AK$ at $S$ . $\\angle KSG = \\angle KDG = \\angle KAG \\implies AG = SG \\implies S$ lies on circle with center $G$ . with same approach $G$ lies on circle with center $F$ .\nwe're Done.",
"Here is my solution.\nlemma 1: K is the incenter of ABC.\nThis is simple .We draw the perpendicular lines to AB and BC from K. Then we prove two triangles are congruent. Thus K is on bisector of B. Similarly we prove K lies on bisector of C.\nlemma2:DFKGE is cyclic\nFirst note that $\\angle KDE$ =A/2.Since K is the incenter of ABC it can be easily proved that $\\angle KGA$ =A/2.So KDEG is cyclic. Similarly it can be shown KDEF is cyclic. So our claim is proved.\nWith our lemmas and some angle chasing it is easy to prove that $\\angle AFE$ =A and $\\angle FEA$ = $\\angle FAE$ =90-A/2.Therefore A lies on w2.Similarly A lies on w3. Call the intersection point of w2 and w3 T. We have :FE=FT,AG=GT. So FGA and FGT are congruent triangles. $\\angle FTG$ = $\\angle FAG$ =90-A/2. K Is the incenter of ABC. So $\\angle FKG$ =90+A/2.Therefore FKGT is cyclic. According to lemma2 we have KDET is cyclic.\nIt remains to prove that A,K and T are collinear. In AKF triangle we'd have $\\angle AKF$ =90+B/2. $\\angle FKT$ = $\\angle FGT$ = $\\angle FGA$ = $\\angle FDE$ =90-B/2.SO A,K and T are collinear.\nQED.",
"A very nice and enjoyable problem! Here's the solution. We first deal with $K$ . \n\n<span style=\"color:#f00\">**Claim :** $K$ is the incenter of $\\triangle ABC$ </span>\n*Proof :* Note that, \n\\[\\measuredangle EKA = 2\\measuredangle EDA = 2\\measuredangle BDA = \\measuredangle DBA = \\measuredangle EBA\\]\nwhich implies that $ABEK$ is cyclic. Similarly, we obtain that $AKDC$ is also cyclic. Now, \n\\[\\measuredangle ABK = \\measuredangle AEK = \\measuredangle KAE = \\measuredangle KBE\\]\nas a result of $ABEK$ being cyclic. Similarly, we can also show that $\\measuredangle ACK = \\measuredangle KCD$ which proves that $K$ is indeed the incenter of $\\triangle ABC$ as claimed.\n\nNow, we can also make the following observations.\n\n<span style=\"color:#f00\">**Claim :**Points $G$ and $F$ lie on the circle $(KDE)$ .</span>\n*Proof :* First, note that \n\\[\\measuredangle GBE = \\measuredangle KBE = \\measuredangle ABK\\]\nand\n\\[\\measuredangle BEG = \\measuredangle BEA = \\measuredangle BKA\\]\nwhich implies that $\\triangle BEG \\sim \\triangle BKA$ . This then gives us that $\\measuredangle EGB = \\measuredangle KAB$ from which we conclude that\n\\[\\measuredangle EGB = \\measuredangle KAB = \\measuredangle CAK = \\measuredangle EDK\\]\nwhich implies that $EDKG$ is cyclic. Similarly, we can also show that $EDFK$ is cyclic, which proves the claim.\n\n<span style=\"color:#f00\">**Claim :** $GA=GD$ and $FA=FE$ .</span>\n*Proof :* We have that\n\\begin{align*}\n2\\measuredangle DAE &= 2 \\measuredangle DAB + 2 \\measuredangle CAE + 2\\measuredangle BAC\n&= \\measuredangle DBA + \\measuredangle ACE + 2\\measuredangle BAC\n&= (\\measuredangle CBA + \\measuredangle BAC + \\measuredangle ACB) + \\measuredangle BAC\n&= \\measuredangle BAC\n&= 2\\measuredangle BKC\n\\end{align*}\nwhere the last equality is since $K$ is the incenter of $\\triangle ABC$ . Then, it follows that $\\measuredangle DAE = \\measuredangle BKC$ . Thus, \n\\[\\measuredangle DAG = \\measuredangle DAE = \\measuredangle BKC = \\measuredangle GKF = \\measuredangle GDF\\]\nfrom which it is quite clear that $GA=GD$ . Similarly, we can show that $FA=FE$ as well.\n\nNow comes the key claim of the problem.\n\n<span style=\"color:#f00\">**Claim :**Let $X= \\odot (G,D) \\cap (KED)$ and $Y= \\odot (G,D) \\cap (KED)$ . Then, $A-K-X$ and $A-K-Y$ .</span>\n*Proof:* We prove one - the other is entirely similar. First, observe that \n\\[\\measuredangle AGK = \\measuredangle EGB = \\measuredangle EDK = \\measuredangle KED = \\measuredangle KGD\\]\nNow, \n\\[2\\measuredangle AXD = \\measuredangle AGD = 2\\measuredangle KGD = 2\\measuredangle KXD\\]\nfrom which it follows that $\\measuredangle AXD = \\measuredangle KXD$ , which implies that $A-X-D$ as desired.\n\nThus, the intersections of the circles $c_1$ and $c_2$ and $c_1$ and $c_3$ both lie on $\\overline{AK}$ which implies that $X=Y$ and thus, the circumcircle of triangle $KDE$ , the circle with center the point $F$ and radius $FE$ and the circle with center $G$ and radius $GD$ concur on a point which lies on the line $AK$ .",
"By symmetry it suffices to show that $c_1, c_2$ concur at a point which lies on $AK$ .**Claim 1:** $KDCA$ are cyclic.\n*Proof:* Since $CA = CE$ then $\\angle KAE = \\angle KEA \\implies \\angle KAC = \\angle KEC = \\angle KDE$ . $\\square$ **Claim 2:** $F \\in c_1$ \n*Proof:* By $SAS$ $\\triangle CAK \\cong \\triangle CEK$ so $\\angle AKC = \\angle CKE \\implies \\angle EKC = \\angle AKC = \\angle ADC$ . $\\square$ Let $AK$ intersect $c_1$ again at $H$ . It suffices to show that $FH = FE = FA$ . Indeed, $\\angle KAD = \\angle KDA = \\angle KEF$ so $FA = FE$ . Then, $\\angle KHF = \\angle KDA = \\angle KAD$ and we're done. $\\blacksquare$ "
] |
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"origin:aops",
"2022 Greece National Olympiad",
"2022 Contests"
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"answer_score": 172,
"boxed": false,
"end_of_proof": true,
"n_reply": 6,
"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790100.json"
}
|
Let $n>4$ be a positive integer, which is divisible by $4$ . We denote by $A_n$ the sum of the odd positive divisors of $n$ . We also denote $B_n$ the sum of the even positive divisors of $n$ , excluding the number $n$ itself. Find the least possible value of the expression $$ f(n)=B_n-2A_n, $$ for all possible values of $n$ , as well as for which positive integers $n$ this minimum value is attained.
|
I claim the answer is $4$ , and is attained iff $n=4p$ , where $p>2$ is a prime or when $n=8$ . Check that indeed when $n=4p$ , then $A_n=1+p$ whereas $B_n=2+4+2p$ , yielding $B_n-2A_n =4$ .
Let $n=2^k\cdot m$ , where $m$ is odd and $k\ge 2$ . Notice that $A_n=\textstyle\sum_{d\mid m}d$ . On the other hand, $B_n = 2A_n+4A_n+\cdots+2^k A_n - n$ . Assume first that $k=2$ . Then, $B_n-2A_n = 4A_n - n =4(A_n-m)$ . Since $A_n\ge m+1$ (as $n>4$ ), it follows $B_n-2A_n\ge 4$ . Equality holds iff $A_n = m+1$ , that is when $m$ itself is an (odd) prime.
Next, assume $k=3$ . Then, $B_n=2A_n+4A_n+8A_n-n$ , yielding $B_n-2A_n = 4A_n + 8(A_n-m)$ . As $A_n\ge 1$ and $A_n-m\ge 0$ , we find $B_n-2A_n\ge 4$ with equality iff $k=3$ and $m=1$ , that is $n=8$ .
Finally, let $k\ge 4$ . Then, $B_n = 2A_n + 4A_n + 8A_n +2^k A_n - n$ , yielding $B_n-2A_n\ge 4A_n+8A_n\ge 12$ , yielding a strictly worse value.
|
[] |
[
"origin:aops",
"2022 Greece National Olympiad",
"2022 Contests"
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{
"answer_score": 64,
"boxed": false,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790101.json"
}
|
The positive real numbers $a,b,c,d$ satisfy the equality $$ a+bc+cd+db+\frac{1}{ab^2c^2d^2}=18. $$ Find the maximum possible value of $a$ .
|
Note that we can write $a$ as $a=18-\left(bc+cd+db+\frac{1}{ab^2c^2d^2}\right)\leq18-\left(\frac{4}{\sqrt[4]{a}}\right)$
since we need to find $a_{\text{max}}$ , minimizing this part $\left(bc+cd+db+\frac{1}{ab^2c^2d^2}\right)$ will work, hence the above part which is true by AM-GM, we need to solve this $$ a+4a^{-\frac{1}{4}}-18=0 $$ which yields $a=16$ as the value.
|
[
"Using AM-GM on the last four terms, we obtain\n\\[\n18\\ge a+ 4a^{-\\frac14}\\iff t^4 +\\frac4t-18\\le 0 \\qquad\\text{where}\\qquad a=t^4.\n\\]\nFrom here, $t\\le 2$ and therefore $a\\le 16$ .\n\nWe now give an example attaining the equality. Note that if $b=c=d$ and $b^2 = (ab^6)^{-1}\\iff b^8 = \\frac1a$ , we enjoy equality. Namely, $a=16$ is indeed the maximal value, concluding the case.",
"If the positive real numbers $a,b$ satisfy the equality $2a+b+\\frac{4}{ab}=10. $ [url=https://artofproblemsolving.com/community/c6h2790115p24533216]Find the maximum value of $a$ .\nIf the positive real numbers $a,b,c,d$ satisfy the equality $a+bc+bd+cd+\\frac{1}{ab^2c^2d^2}=18.$ Find the maximum value of $a$ .**Funny twins.**",
"Let $a,b>0$ and $2a+b+\\frac{4}{ab }=6. $ Prove that $$ 2a^2+b^2 \\leq 6 $$ $$ 2a^2+\\frac{1}{b^2}\\leq \\frac{9}{4} $$ ",
"Let $a,b\\in (0,1]$ and $2a+b+\\frac{4}{ab}=10. $ Prove that $$ a+2b\\leq \\frac{5}{2} $$ $$ a+ \\frac{1}{b}\\leq 2+\\frac{\\sqrt 3}{2} $$ ",
"Similar\nhttps://artofproblemsolving.com/community/c6h2543153p21690156",
" $$ \\dfrac{31a}{32}+\\dfrac{a}{32}+bc+cd+db+\\dfrac{1}{ab^2c^2d^2}\\ge \\dfrac{31a}{32}+\\dfrac{5}{2}\\le 18\\Longrightarrow \\dfrac{31a}{16}\\le 31 \\longrightarrow a\\le 16 $$ "
] |
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"origin:aops",
"2022 Greece National Olympiad",
"2022 Contests"
] |
{
"answer_score": 14,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790106.json"
}
|
Let $Q_n$ be the set of all $n$ -tuples $x=(x_1,\ldots,x_n)$ with $x_i \in \{0,1,2 \}$ , $i=1,2,\ldots,n$ . A triple $(x,y,z)$ (where $x=(x_1,x_2,\ldots,x_n)$ , $y=(y_1,y_2,\ldots,y_n)$ , $z=(z_1,z_2,\ldots,z_n)$ ) of distinct elements of $Q_n$ is called a *good* triple, if there exists at least one $i \in \{1,2, \ldots, n \}$ , for which $\{x_i,y_i,z_i \}=\{0,1,2 \}$ . A subset $A$ of $Q_n$ will be called a *good* subset, if any three elements of $A$ form a *good* triple. Prove that every *good* subset of $Q_n$ contains at most $2 \cdot \left(\frac{3}{2}\right)^n$ elements.
|
<details><summary>BRUH</summary>We proceed by induction on $n$ , with the base case, $n=1$ being clear.
For $t=0,1,2$ , let $J_t$ be the set of sequences with $x_n=t$ . We know that any three sequences in $J_t\cup J_u$ must have $x_j,y_j,z_j$ distinct, but $j\ne n$ . First observe $(x_1,\cdots,x_{n-1})$ cannot appear in $J_t$ and $J_u$ simultaneously. By inductive hypothesis, there are at most $2\cdot (\frac 32)^{n-1}$ possibilities for the first $n-1$ elements in $J_t\cup J_u$ . Since each of the possibilities can only appear in at most one of $J_t$ , $J_u$ , $|J_t|+|J_u|\le 2\cdot (\frac 32)^{n-1}$ sequences by inductive hypothesis. Thus, $2(|J_0|+|J_1|+|J_2|) \le 6\cdot (\frac 32)^{n-1}$ , completing the induction</details>
|
[
"<details><summary>Can we just?</summary>Note that total number of **unordered** triplets $(x,y,z) \\in Q^3 _{n} = \\frac{6! ^n}{6!} = 2^{n-1}3^{n-1}$ , suppose for the sake of contradiction assume that there exist a good subset with $ > 2 \\cdot (\\frac{3}{2})^n$ , then note that $\\binom{|A|}{3} \\le 2^{n-1}3^{n-1}$ , which yields $$ (3^n -2^{n-1}) (3^n -2^n) \\le 2^{4n-3} $$ which is not true for $n \\ge 2$ . Hence we are done.</details>",
"The final inequality doesn't hold for large $n$ because LHS tends to $9^n$ while RHS tends to $16^n$ . I don't understand your argument. Can you write it in more detail"
] |
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"origin:aops",
"2022 Greece National Olympiad",
"2022 Contests"
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"answer_score": 36,
"boxed": false,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Greece National Olympiad/2790111.json"
}
|
Positive integers $a$ , $b$ , and $c$ are all powers of $k$ for some positive integer $k$ . It is known that the equation $ax^2-bx+c=0$ has exactly one real solution $r$ , and this value $r$ is less than $100$ . Compute the maximum possible value of $r$ .
|
Since we must have $b^2=4ac$ , this implies $a$ , $b$ , and $c$ are powers of $2$ , so substitute $a=2^m$ , $b=-2^n$ , $c=2^\ell$ . Then $r=-\tfrac{b}{2a}$ , hence we want $r$ to be the largest power of $2$ less than $100$ , which just so happens to be $64$ .
Now we find a construction for this. $b^2=4ac$ translates to $2n=m+\ell + 2$ , and we want to make $n-m-1=6$ and $\ell-n +1=6$ . Taking $(m, n, \ell )=(1, 8, 13)$ satisfies this and indeed, the solution to $2x^2-256 x + 8192=0$ is $x=\boxed{64}$ .
|
[
"solution\n\n**Attachments:**\n\n[Hmmt 2022 P1.docx](https://cdn.artofproblemsolving.com/attachments/e/2/be004fd69ac8fcb133fa636153ecc729ddb017.docx)",
" $\\text{Discriminant }D=b^2-4ac=0 \\implies b \\text{ is even } \\implies k=2.$ It follows that $r$ is a power of $2$ . So max the power of $2$ smaller than $100$ is $64,$ which is achievable. $\\square$ ",
"<blockquote> $\\text{Discriminant }D=b^2-4ac=0 \\implies b \\text{ is even } \\implies k=2.$ It follows that $r$ is a power of $2$ . So max the power of $2$ smaller than $100$ is $64,$ which is achievable. $\\square$ </blockquote>\n\nI want to elaborate on what ZETA_in_olympiad meant by $k = 2$ : We can replace $a$ , $b$ , and $c$ with $k^A$ , $k^B$ , $k^C$ . So far, we have noticed that the discriminant is 0 - therefore, $b$ and $k$ are even. Since $k$ is even, we can replace $k$ with the following: $$ k = 2^m \\cdot p $$ where $2^m$ is the even component of $k$ and $p$ is the odd component. Substituting $k$ into the discriminant will give you $$ 2^{2mB} \\cdot p^{2B} - 4 \\cdot (2^{mA} \\cdot p^A)(2^{mC} \\cdot p^C) = 0 $$ which will give us $$ \\begin{cases} 2mB = 2+mA+mC 2B = A + C \\end{cases} $$ However, this system is not true, as $0 \\neq 2$ . Therefore, we can deduce that $p = 1$ and $k = 2$ .",
"Let $a=k^{a'}$ , $b=k^{b'}$ , and $c=k^{c'}$ . The condition that $ax^2-bx+c$ has a double root is equivalent to its discriminant $b^2-4ac$ being equal to $0$ , and substituting yields $(k^{b'})^2-4k^{a'}k^{c'}=0$ . Equivalently, $k^{2b'-a'-c'}=4$ , so $k=2$ or $k=4$ . WLOG let $k=2$ , since $\\{4^n|n\\in\\mathbb{Z}^+\\}\\subseteq\\{2^n|n\\in\\mathbb{Z}^+\\}$ . Then, $r=\\tfrac{b}{2a}=2^{2b'-a'-1}$ , which is at most $2^6=\\boxed{64}$ , given that it is less than $100$ . It can be easily verified that this value of $r$ is attainable.",
"This is a fairly stupid problem even for problem 1.\n\nThe second condition implies $b^2=4ac$ , so $2 \\mid k$ . If any other prime $p \\mid k$ , then combining $\\nu_p$ and $\\nu_2$ yields a contradiction. Thus, $k$ is some power of 2, and thus $r = \\frac b{2a}$ must be some power of 2 as well, which is at most $r = \\boxed{64}$ .",
"silly ahh problem\n\n-----\n\nLet $a = k^x$ , $b=k^y$ , and $c = k^z$ . We have $2r = \\tfrac{b}{a} = k^{y-x}$ and $r^2 = \\tfrac{c}{a} = k^{z-x}$ . Obviously, $k=2$ or $k=4$ ; we WLOG set $k=2$ since $4$ is a power of $2$ anyways. Then, we just want the largest power of $2$ less than $100$ , or $\\boxed{64}$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1040,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798886.json"
}
|
Compute the number of positive integers that divide at least two of the integers in the set $\{1^1,2^2,3^3,4^4,5^5,6^6,7^7,8^8,9^9,10^{10}\}$ .
|
First, notice that each of the integers from $1$ to $10$ (excluding $1$ , because it’s hard to define?) have at most two distinct prime factors. Therefore, the integers that divide *at least two* of these power things have only one distinct prime factor, since if it had two prime factors, there’s at most one integer with those factors and thus cannot exist. Then they are powers of $2$ , $3$ , or $5$ .
For powers of $2$ , we can have the exponent range from $0$ to $10$ ; powers of $3$ go from $0$ to $6$ ; powers of $5$ from $0$ to $5$ . Adding gives $(10+1)+(6+1)+(5+1)=24$ , but we must subtract off $2$ since we overcounted $1$ twice, ergo $\boxed{22}$ .
|
[
"22 is the answer ",
"Factorising the set will give you $$ {1^{1}, 2^2, 3^{3}, 2^{8}, 5^{5}, 2^{6} \\cdot 3^{6}, 7^7, 2^{24}, 3^{18}, 2^{10} \\cdot 5^{10}} $$ Let $x = 2^{a} \\cdot 3^{b} \\cdot 5^{c}$ , where $a$ , $b$ , and $c$ are nonnegative integers. (The case when $x = 1$ is obvious, so we will add 1 to our final answer later.)\nWe can then set some upper bounds for $a$ , $b$ , $c$ , and $d$ by looking at the second biggest powers: $a \\in {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}$ $b \\in {1, 2, 3, 4, 5, 6}$ $c \\in {1, 2, 3, 4, 5}$ Therefore, the number of positive integers that divide at least two of the integers is $10 + 6 + 5 + 1 = 22$ . ",
"Uhh I undercounted at first and got $18$ . Pretty much like the one above. Basically we get the prime factorisation of the numbers given. Notice the highest and second highest power of $2$ available is $24$ and $10$ . So $ 2^{1},2^{2}, \\dots , 2^{10}$ would divide at least two of those. Similar way we get $3^{1}, \\dots , 3^{6}$ and $5^{1}, \\dots , 5^{5}$ would divide at least $2$ numbers in the given set. And a trivial case of $1$ which divides every of them, so the answer would be $1+10+6+5=22$ .",
"The main idea is that the only integers that could possibly work are powers of primes. Otherwise, for any two primes $p$ and $q$ , there is at most one integer in $\\{1, 2, 3, \\cdots, 10\\}$ that is a multiple of $pq$ . The particular integers that work are divisors of $2^{10}$ by considering $8^8$ and $10^{10}$ , divisors of $3^6$ by considering $6^6$ and $9^9$ , and divisors of $5^5$ by considering $5^5$ and $10^{10}$ .\n\nFinally, because 1 is overcounted twice, the answer is $(10+1) + (6+1) + (5+1) - 2 = \\boxed{22}$ integers.",
"Here shouldn't we also consider $2^k$ where $k>10$ and $k$ is an integer?",
"Ok, I now understand that we must take the second highest power of any prime in two different terms.",
"Rewrite the set in terms of its prime factorizations:\n\n\\[\\{1, 2^2, 3^3, 2^8, 5^5, 2^6 \\cdot 3^6, 7^7, 2^{24}, 3^{27}, 2^{10} \\cdot 5^{10} \\}.\\]\n\nNo two terms share two common prime factors, so our common divisor must be either $1$ or $p^k$ . When $p=2$ , we can see that the exponent is at most $10$ ; when $p=3$ , the exponent is at most $6$ ; when $p=5$ , the exponent is at most $5$ . Thus, we have $1+10+6+5 = \\boxed{22}$ values. "
] |
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"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 8,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798887.json"
}
|
Let $x_1, x_2, . . . , x_{2022}$ be nonzero real numbers. Suppose that $x_k + \frac{1}{x_{k+1}} < 0$ for each $1 \leq k \leq 2022$ , where $x_{2023}=x_1$ . Compute the maximum possible number of integers $1 \leq n \leq 2022$ such that $x_n > 0$ .
|
We note that if $x_{k}$ is positive, then $x_{k+1}$ must be negative. Therefore, $n$ must be less than 1012, as if we group our numbers by pairs of $(x_{k}, x_{k+1})$ then we would have 1011 groups (or at most 1011 positive integers).
Now, we will prove that $n = 1011$ is impossible through contradiction: First, let's suppose that the maximum possible number of integers that satisfy the conditions is 1011. We will then prove that the sequence $x_{1}, x_{2}, ..., x_{2021}, x_{2022}$ alternates: With $x_{2023} = x_{1}$ , the only possible sequences are $+, -, +, -, ..., +, -, +$ or $-, +, -, +, ..., -, +, -$ .
We can look at the case where the odd elements are positive, and apply the same arguments to the latter case. There are two subcases to the ordering of the elements:
[list=1]
[*] **Odd/even:** $x_{2n} +\frac{1}{x_{2n+1}} < 0$ and $x_{2n+1} \cdot x_{2n} < -1$ [*] **Even/odd:** $x_{2n+1} +\frac{1}{x_{2n}} < 0$ and $x_{2n+1} \cdot x_{2n} > -1$ [/list]
As the two subcases contradict each other, we have proved that $n = 1011$ is not possible.
Finally, we will prove that $n = 1010$ works with a simple example that ranges from $x_{1}$ to $x_{2023}$ : $1, -\frac{1}{2}, 3, -\frac{1}{4}, ..., 2019, -\frac{1}{2020}, \textrm{[any negative real number a]}, \textrm{[any negative fraction b such that the absolute value of b is less than 1]}, 1$ .
Therefore, $\boxed{n = 1010}$ .
|
[
"My answer is 1010.",
"This is fairly easy to do though prickly to actually prove; I'll only give an outline of how to prove $n \\geq 1011$ is impossible.\n\nFirst, observe as we cannot have consecutive $+$ signs, so $n$ is at most 1011. Consider the case when $n=1011$ . Now, assume without loss of generality that $|x_1| \\geq 1$ . We can check that $$ |x_1| < |x_3| < |x_5| < \\cdots < |x_{2023}| = |x_1| $$ by repeatedly applying the condition, which is obviously a contradiction.\n\nThus, $n=1010$ is attainable and easily constructable as we can simply make the final two terms negative.",
"kinda scuffed notation, but should be good enough :|\n\n-----\n\nNote that we cannot have consecutive terms be positive, so $n$ is at most $1011$ . Suppose that $n=1011$ . Obviously, the signs alternative from positive and negative. Suppose that $x_i = y_i$ for odd indices and $x_i = -y_i$ for even indices, where $\\{y_i\\}$ are positive real numbers. We end up getting $y_{2k-1}y_{2k} < 1$ and $y_{2k}y_{2k+1} > 1$ . Multiplying cyclicly for both sets of equations yields $1<y_1y_2\\dots y_{2022}<1$ , a contradiction. \n\nHence, $n \\le 1010$ , and $n=\\boxed{1010}$ is probably achievable since we can have two consecutive negatives at the end. "
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798905.json"
}
|
Compute the sum of all 2-digit prime numbers $p$ such that there exists a prime number $q$ for which $100q + p$ is a perfect square.
|
Note: $p\equiv11,31,41,61,71,19,29,59,79,89\pmod{100}
\Rightarrow 100q+p\equiv k^2\equiv 11,31,41,61,71,19,29,59,79,89\pmod{100}$ Now observe that the values $k^2$ are congruent to are nothing but the last $2$ digits of the perfect squares. So, we can reduce our calculations to just checking the last $2$ digits of the squares of 2 digit natural numbers as for three and for digits, we'll get the same value as we are working with $\pmod{100}$ Also, notice that $k^2\equiv 1,9\pmod{10}\Rightarrow k\equiv 1,3,7,9\pmod{10}$ So, our calculations are reduced to the numbers $11,21,31,41,51,61,71,81,91, $ $13,23,43,53,63,73,83,93,$ $ 17,27,37,47,57,67,87,97,$ $19,29,39,49,59,69,79,89,99$ Well, here (though I shouldn't have) I used a bit of java (coz I don't know python :P ) to find out what the last $2$ digits might be and the possibilites that came up were $41,61,29,89$ .
Now note that $41\equiv 1\pmod{8}$ and as $k^2$ is an odd perfect square, $100q+41\equiv 1\pmod {8}
\Rightarrow 100q+1\equiv 1\pmod {8}\Rightarrow 100q\equiv 0\pmod 8\Rightarrow q\equiv 0\pmod 8,
\Rightarrow\Leftarrow$ (a contradiction)
Hence the required answer is $29+61+89=\boxed{179}$ .
P.S1: Here's the code I used:
```
class congruence
{
public static void main()
{
for(int i=10;i<=99;i++)
{
if(i%10==1||i%10==3||i%10==7||i%10==9)
System.out.print((i*i)%100+" ,");
}//end for
}//end main()
}//end class
```
P.S2: My first solution to a problem in LATEX
|
[
"With mod 4 and mod 10 we obtain that p must be in the form 4k+1 and the last digit must be 1 or 9. Moreover, since a perfect square must have even number as its tens digit we got the only possibilities are 29, 41, 61, and 89. 89 obviously works when q=2 (17²). 61 when q=3 (19²) . 29 when q= 5 (23²). 41 is rejected since perfect square ending with 41 will have even number as its third hundreds digit. So the sum is 29+61+89= 179. ",
"Taking mod 5 and mod 4, we can narrow the possibilities for $p$ down to $29, 41, 61, 89$ . Notice that $529, 361, 289$ validate three of these numbers: for $n=41$ , notice that if $$ 100 \\mid x^2-y^2 = (x+y)(x-y) $$ and $x$ and $y$ are both odd, then one of $x+y$ and $x-y$ is a multiple of four, and the other is a multiple of two; this implies that $200 \\mid x^2-y^2$ , and as $21^2 = 441$ has an even hundreds digit, all perfect squares ending in 41 have an even hundreds digit as well. Furthermore, $241$ is not a perfect square, so $p = 41$ does not work. The answer is thus $29+61+89 = \\boxed{179}$ .",
"Note that squares end in $0,1,4,5,6,9$ , so the only options for $p$ to end in is $1,9$ . Furthermore, odd squares are $1 \\pmod{4}$ , so we can reduce the possible values of $p$ to $29, 41, 61, 89$ . It is easy to see that $529, 361, 289$ verify all but $p =41$ . When $p=41$ , we get\n\n\\[100q+41 \\equiv 4q+1 \\pmod{8}.\\]\n\nHowever, odd squares are always $1 \\pmod{8}$ , so $q$ is even. The only even prime is $2$ , but $241$ is not a perfect square. Summing up the valid values of $p$ gives $\\boxed{179}$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1032,
"boxed": true,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798908.json"
}
|
Given a positive integer $k$ , let $||k||$ denote the absolute difference between $k$ and the nearest perfect square. For example, $||13||=3$ since the nearest perfect square to $13$ is $16$ . Compute the smallest positive integer $n$ such that $\frac{||1|| + ||2|| + ...+ ||n||}{n}=100$ .
|
Let $k = \lfloor \sqrt n \rfloor$ and let $n = k^2+r$ for some positive integers $r, k$ . Notice that $$ ||a^2|| + ||a^2+1|| + \cdots + ||(a+1)^2|| = a(a+1), $$ so we can write $$ 100n = 100k^2+100r = ||1||+||2||+ \cdots + ||n|| \leq \sum_{i=2}^k i(i-1) + \frac 12 r(r+1) \leq \frac 13 (k^3-k) + \frac 12 r(r+1). $$ Notice that $\frac 12 r(r+1) > 100r$ if and only if $r > 199$ , so we need the $\frac 13(k^3-k)$ term to be significantly larger than the $100k^2$ term. The inequality $$ \frac 13(k^3-k) \geq 100k^2 \iff k \geq 300 $$ under the integers. Thus, the smallest solutions should be in the neighborhood of 300; a solution with $k = 299$ should conspicuously exist as the margin for $k = 299$ is quite small. Indeed, we can rewrite this expression as $n = (k+1)^2 - r$ and plug in $k=299$ to obtain that $r = 200$ works. As a result, the smallest possible value is $300^2 - 200 = \boxed{89800}$ .
|
[
"I'm too lazy to make a mathematical solution xD.\nThe answer is 89800, checked by C++: \n```\n#include <bits/stdc++.h>\nusing namespace std;\n\nint f(int n) {\n int lo_sqrt_n = floor(sqrt(n));\n int up_sqrt_n = lo_sqrt_n + 1;\n int dif1 = n - lo_sqrt_n * lo_sqrt_n;\n int dif2 = up_sqrt_n * up_sqrt_n - n;\n if (dif1 <= dif2)\n return dif1;\n return dif2;\n}\n\nint main() {\n int s = 0;\n for (int i = 1; ; i++) {\n s += f(i);\n if (s == 100 * i) {\n cout << i ;\n return 0;\n }\n }\n}\n```\n",
"thats what I got too ",
"I find it cleaner to take cases on the nearest perfect square? Specifically, there are $2i$ integers closest to $i^2$ , and the sum of their distances is $i^2$ . Hence, we approximately need\n\\[\\frac{\\textstyle\\sum i^2}{\\textstyle\\sum 2i}=\\frac{\\frac{m(m+1)(2m+1)}{6}}{m(m+1)}=\\frac{2m+1}{6}\\approx100,\\]\nor $m\\approx299,300$ . Refining, the ratio is $\\tfrac{599}{6}=100-\\tfrac{1}{6}$ at $n=299\\cdot300$ , a deficiency of $50\\cdot299$ from the desired $100$ . Tacking on the $100$ values $100+199,\\dots,100+101$ balances the check, ergo $n=299\\cdot300+100=\\boxed{89800}$ .",
"here is a solution with all of the major computation taken out :) fsr, i felt like taking cases on $n = k^2+k$ rather than perfect squares.\n----- \n\nWith some basic computation, one can realize that when $\\sum ||n||$ is taken up to $n = k^2+k$ , we simply get $1^2+2^2+\\dots + k^2$ , which has closed form $\\tfrac{k(k+1)(2k+1)}{6}$ . Plugging this into the numerator, one can realize that the desired value occurs when $\\tfrac{2k+1}{6} = 100$ , which yields $k \\approx 299$ . We can set $n = 299 \\cdot 300+k$ . Notice that $||299 \\cdot 300+k|| = 300-k$ , so plugging this in to find an exact value, we get $k = 100, 299$ ; the smaller value is $299 \\cdot 300+100 = \\boxed{89800}$ .\n\n"
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1038,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798911.json"
}
|
Let f be a function from $\{1, 2, . . . , 22\}$ to the positive integers such that $mn | f(m) + f(n)$ for all $m, n \in \{1, 2, . . . , 22\}$ . If $d$ is the number of positive divisors of $f(20)$ , compute the minimum possible value of $d$ .
|
The answer is $\boxed{2016}$ .
Claim: The function $f(n) = n\cdot \mathrm{lcm}(1,2,\ldots, 22)$ works.
Proof: Let $a = \mathrm{lcm}(1,2,\ldots, 22)$ . We must check that \[mn |a(m+n)\] for all $1\le m,n\le 22$ . Suppose for some fixed $m,n$ , and prime $p$ , we had $\nu_p(mn) >\nu_p(a(m+n))$ . Then, \[\nu_p(m) + \nu_p(n) > \nu_p(a) + \nu_p(m+n)\]WLOG that $\nu_p(m) \ge \nu_p(n)$ . Then we have \begin{align*}
\nu_p(a) + \nu_p(m+n)
\ge \nu_p(m) + \nu_p(n),
\end{align*}contradiction to our assumption. $\square$ Claim: $20a\mid f(20)$
Proof. Let $P(m,n)$ denote the given assertion. $P(m,m): m^2 \mid 2f(m)$ . If $m$ is odd, then $m^2 \mid f(m)$ , and if $m$ is even, $\frac{m^2}{2}\mid f(m)$ . Both imply $m\mid f(m)$ for all $m$ within range.
So $m\mid f(m) + f(n)\implies m\mid f(n)$ for all $m,n\in \{1,2,\ldots, 22\}$ . Thus $a$ divides $f(n)$ for each $n$ .
Now suppose there existed a prime $p$ such that $\nu_p(20a) > \nu_p(f(20))$ . Since $a\mid f(20)$ , we have $p=2$ or $p=5$ . Notice that $\nu_5(20a) = 2$ and $\nu_2(20a) = 6$ . $P(20,20): 200\mid f(20)$ , so $\nu_5(20a) \le \nu_p(f(20))$ .
This implies $\nu_2(f(20))<6$ . $P(16,16): \nu_2(f(16)) \ge 7$ . $P(16,20): \nu_2(f(16) + f(20)) \ge 6$ , which implies $\nu_2(f(20)) \ge 6$ , contradiction. $\square$ This implies the minimum number of divisors of $f(20)$ is when $f(20) = 20a$ , which is equal to \[2^6 \cdot 3^2 \cdot 5^2\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19,\]which has \[7\cdot 3\cdot 3\cdot 2^5 = 2016\] divisors.
|
[
"Does $f(n)=n\\cdot \\mathrm{lcm}(1,2,\\ldots,22)$ work?",
"2016 is my answer .",
"For every prime $p$ , we can look at the minimum power of $p$ that must divide $f(20)$ . In general, we can use the following strategy: if $k$ is the number less than or equal to 22 with the maximum $\\nu_p$ , then we can consider the expressions \n\\begin{align*}\n20k \\mid f(20) + f(k) \nk^2 \\mid 2f(k)\n\\end{align*}\nto obtain $20k \\mid f(20)$ . (Notice that the coefficient of 2 doesn't matter, as $20$ does not maximize $\\nu_2$ and for other primes dividing by 2 does not change the divisibility.) Thus, we can conclude $f(20) = 20\\cdot \\text{lcm}(1, 2, 3, \\cdots, 22)$ through this logic, which yields $d = \\boxed{2016}$ .\n\nA construction exists simply by letting $f(n) = n\\cdot \\text{lcm}(1, 2, 3, \\cdots, 22)$ for each $n$ .",
"Obviously, $f(m)+f(n)\\equiv 0\\pmod{m}$ for all $m$ . \n\nWe attempt to prove that $f(m)\\equiv 0\\pmod{m}$ :\n\nAssume that $m\\nmid{f(m)}$ but $m^2\\mid{2f(m)}$ . \n\nLet $\\nu_2(m)=s \\Longrightarrow \\nu_2{f(m)}=r$ . This implies that $s>r$ and $2s\\leq{r+1}$ which is very clearly a contradiction.\n\nTherefore, $m\\mid{f(m)} \\Longrightarrow m\\mid{f(n)}$ for all $m$ .\n\nFocusing on $f(20)$ , we get that it must divide $\\{1, 2, 3\\cdots{20}\\}$ but we cannot just take the LCM of $\\{1,2,3\\cdots{20}\\}$ because of the following:\n\nConsider $\\gcd\\{20, n\\}\\neq{1}$ :\n\nIf we assume $f(20)=$ LCM $\\{1,2,3\\cdots{20}\\}$ , $20, n\\mid{f(20)}$ but $20n\\nmid{f(20)}$ because the LCM negates \"overlapping\" factors between 2 numbers.\n\nIf $20n\\nmid{f(20)} \\Longrightarrow 20n\\nmid{f(20)+f(n)}$ . \n\nIf we let $\\nu_2{(n)}=a$ and $\\nu_2{f(n)}=b$ , then $n^2\\mid{2f(n)} \\Longrightarrow 2a\\leq{b+1}$ . If $p\\neq2$ and we let $\\nu_p(n)=a$ and $\\nu_pf(n)=b$ , then $2a\\leq{b}$ .\n\n If we substitute $n=2^4=16$ , $\\nu_2f(16)\\geq{8-1=7}$ . Then, $20(16)\\mid{f(20)+f(16)} \\Longrightarrow f(20)\\equiv 0\\pmod{2^6}$ . Now similarly if we substitute $n=5$ , $f(20)\\equiv 0\\pmod{5^2}$ . Every other prime factor beside 2 and 5 will stay the same as in the LCM since they are relatively prime to 20. Therefore, the minimum of $f(20)=$ LCM $\\{1,2,3\\cdots{20}\\}\\cdot2^2\\cdot5$ .\n\nSimply calculating, we get that $f(20)=2^6\\cdot3^2\\cdot5^2\\cdot7\\cdot11\\cdot13\\cdot17\\cdot19$ . The number of factors of $f(20)$ would consequently be $7\\cdot3\\cdot3\\cdot2\\cdot2\\cdot2\\cdot2\\cdot2=\\boxed{2016}$ .\n\n\n",
"It appears there is a typo on the official answer key. In the last line, $2^6 \\mid f(20)$ and $6 = \\nu_2(20L)$ I believe.",
"For all pairs $x$ where $x$ is the power of a prime $p$ with optimal exponent, we have that $x^2|f(x)$ . Therefore, $20x|f(20)+f(x)$ . Thus, we have $20x|f(20)$ . Therefore, $f(20)$ is the least common multiple of the first $22$ positive integers times $20$ , which leads to $f(20)=2^6 \\cdot 3^2 \\cdot 5^2 \\cdot 7 \\cdot 11 \\cdot 13 \\cdot 17 \\cdot 19 \\implies \\boxed{2016}$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1178,
"boxed": true,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798917.json"
}
|
Let $(x_1, y_1)$ , $(x_2, y_2)$ , $(x_3, y_3)$ , $(x_4, y_4)$ , and $(x_5, y_5)$ be the vertices of a regular pentagon centered at $(0, 0)$ . Compute the product of all positive integers k such that the equality $x_1^k+x_2^k+x_3^k+x_4^k+x_5^k=y_1^k+y_2^k+y_3^k+y_4^k+y_5^k$ must hold for all possible choices of the pentagon.
|
Much more difficult than #6, but a super interesting problem.
Throughout this solution, we drop all constants. By scaling, we may assume that the vertices of the pentagon lie on the unit circle. Now, let $(x_1, y_1) = \exp(\theta)$ for some angle $\theta$ . Then, $$ x_1^k + x_2^k + x_3^k + x_4^k + x_5^k = (\exp(\theta) + \exp(-\theta))^k + (\omega \cdot \exp(\theta) + \omega^{-1} \cdot \exp(-\theta))^k + \cdots + (\omega^4 \cdot \exp(\theta) + \omega^{-4} \cdot \exp(-\theta))^k. $$ Here $\omega$ is a fifth root of unity. Treating this as a homogeneous polynomial in $\exp(\theta)$ and $\exp(-\theta)$ , it expands as
\begin{align*}
&\phantom{+} \exp(\theta)^k (1+\omega^k + \omega^{2k} + \omega^{3k} + \omega^{4k})
&+ \exp(\theta)^{k-1} \exp(-\theta) \left({k \choose 1} + {k \choose 1}\omega^{k-2} + \cdots + {k \choose 1}\omega^{4k-4}\right)
&+ \cdots
&+ \exp(-\theta)^k \left(1+\omega^{-k} + \omega^{-2k} + \omega^{-3k} + \omega^{-4k}\right).
\end{align*}
Index these terms with $i=0, i=1, i=2, \cdots$ . Now, notice that because $\cos \theta = \cos(-\theta) = \sin(90^\circ + \theta)$ , the expression for the $y_i$ is simply the same polynomial with the value $\exp\left(\theta + \frac{\pi}2\right)$ . As this has to be true for all values of $\theta$ , this implies that all terms of $\exp(\theta)\exp(-\theta)$ with coefficients not subtracting to a multiple of 4 are each zero. Furthermore, observe that each of the terms is not equal to zero if and only if the exponent of the first term $k-2i$ is a multiple of five. Thus, if $k$ is odd, we need all terms to be zero, implying $k < 5$ . If $k$ is even, then the term with $i=5$ will fail, so we have $k < 10$ . Thus, the possible values multiply to $$ 1 \cdot 2 \cdot 3 \cdot 4 \cdot 6 \cdot 8 =\boxed{1152}. $$
|
[
"Nice problem, my answer is 1×2×3×4×6×8=1152.",
"Here's a fast unrigorous solution: let $\\arg(x_i,y_i)=\\theta_i$ . Observe $\\textstyle\\sum\\cos k\\theta_i=\\sum\\sin k\\theta_i$ for $5\\nmid k$ ; expanding by Chevyshev gives the sum of degrees $k,k-2,\\dots$ , and so we can inductively build $1\\mapsto3\\not\\mapsto5$ and $2\\mapsto4\\mapsto6\\mapsto8\\not\\mapsto10$ , stopping when $5\\mid k$ . Extracting, $\\textstyle\\prod k=3!!\\cdot8!!=\\boxed{1152}$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1042,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798920.json"
}
|
Positive integers $a_1, a_2, ... , a_7, b_1, b_2, ... , b_7$ satisfy $2 \leq a_i \leq 166$ and $a_i^{b_i} \cong a_{i+1}^2$ (mod 167) for each $1 \leq i \leq 7$ (where $a_8=a_1$ ). Compute the minimum possible value of $b_1b_2 ... b_7(b_1 + b_2 + ...+ b_7)$ .
|
Note that $167$ is prime.
We have \[a_i^{2^{i-1}} \equiv a_1^{b_1 b_2 \cdots b_{i-1}}\pmod{167}\]for all $1<i\le 8$ .
This implies $a_1^{128}\equiv a_1^{b_1 b_2 \cdots b_7}\pmod{167}$ , so $a_1^{128 - b_1 b_2 \cdots b_7}\equiv 1\pmod{167}$ .
Case 1: $a_1 = 166$ .
If $a_i = 166$ , then since $-1$ isn't a QR mod $167$ , we get $b_i$ is even, then $a_{i+1} = 166$ . So $a_n = 166$ for all $1\le n\le 7 $ and $b_n$ is even. This gives a minimum possible value of $2^7 \cdot 7\cdot 2 = 1792$ .
Case 2: $a_1\ne 166$ .
Then the order of $a_1$ mod $167$ is either $83$ or $166$ . This implies $b_1 b_2 \cdots b_7\equiv 45\pmod{83}$ . Since we want to minimize $b_1 b_2 \cdots b_7(b_1 + b_2 + \cdots + b_7)$ , we can assume $b_1 b_2 \cdots b_7= 45$ since all other possible values are too big.
Now we want to minimize the value of $b_1 + b_2 + \cdots + b_7$ . We can replace a $(1,9)$ with $(3,3)$ , a $(1,15)$ with $(3,5)$ , and a $(1,45)$ with $(3,15)$ . If we want our construction to be optimal, then we must not have any 9's, 15's 45's. This leaves us with $(1,1,1,1,3,3,5)$ , so the ans is \[45\cdot (1+1+1+1+3+3+5) = \boxed{675}\]
|
[
"The official solutions are here:\n[https://hmmt-archive.s3.amazonaws.com/tournaments/2022/feb/algnt/solutions.pdf](https://hmmt-archive.s3.amazonaws.com/tournaments/2022/feb/algnt/solutions.pdf).",
"Let $p = b_1b_2\\dots b_7$ . Note that\n\n\\[a_1^p = a_1^{b_1b_2\\dots b_7} \\equiv a_1^{2b_2b_3\\dots} \\dots \\equiv a_1^{128} \\pmod{167}.\\]\n\nA similar process can be used on the other $a_i$ to get $a_i^{p-128} \\equiv 1 \\pmod{167}$ . Since $\\operatorname{ord}_{167} (a_i) \\mid 166 = 2 \\cdot 83$ , we only have a few cases. \n\nIf $83 \\mid p-128$ , we either have $p = 45$ or $p \\ge 128$ . When $p =45$ , we can pick $(b_1, b_2, b_3, b_4, b_5, b_6, b_7) = (5,3,3,1,1,1,1)$ , yielding a value of $675$ . Otherwise, AM-GM gives\n\n\\[b_1b_2\\dots b_7 (b_1+b_2 + \\dots b_7) \\ge 7 p^{8/7} \\ge 7 \\cdot 2^8 > 675.\\]\n\nIf $83 \\nmid p-128$ , then we must have $\\operatorname{ord}_167(a_i) = 2$ , meaning $a_i^2 \\equiv 1 \\pmod{167}$ . Since $a_i \\not\\equiv 1 \\pmod{167}$ , we have $a_i \\equiv -1 \\pmod{167}$ for all $i$ . Consequently, all of the $b_i$ are even, making the minimum desired value $2^7 \\cdot 14 > 675$ . \n\nThus, the answer is $\\boxed{675}$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1060,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2798921.json"
}
|
Suppose $P(x)$ is a monic polynomial of degree $2023$ such that $P(k) = k^{2023}P(1-\frac{1}{k})$ for every positive integer $1 \leq k \leq 2023$ . Then $P(-1) = \frac{a}{b}$ where $a$ and $b$ are relatively prime integers. Compute the unique integer $0 \leq n < 2027$ such that $bn-a$ is divisible by the prime $2027$ .
|
bruh i'm not gonna lie this is probably the hardest algebra problem i've ever done :skull:
Firstly, we note that $P(k)-k^{2023}P\left(1-\frac{1}{k}\right)$ evidently has roots $1,2,3, \cdots 2023$ . Therefore, we can write $P(k)-k^{2023}P\left(1-\frac{1}{k}\right)=k(x-1)(x-2) \cdots (x-2023)$ for some $k$ . Let us write $P(x)=x^{2023}+a_{2022}x^{2022}+a_{2021}x^{2021}+\cdots +a_0$ , thus $x^{2023}P\left(1-\frac{1}{x}\right)=(x-1)^{2023}+a_{2022}(x-1)^{2022}x+a_{n-2}(x-1)^{2021}x^2+ \cdots a_0x^n$ . We now compare coefficients: the absolute difference of $x^{2023}$ in our polynomials should be $k$ . Therefore, $1-P(1)=k$ . Now we compare constant terms and receive $P(0)+1=-(1-P(1))2023!$ . Also, since $P(1)=P(0)$ from the given problem condition, we solve for $P(1)=\frac{2023!+1}{2023!-1}$ and $k=-\frac{2}{2023!-1}$ . Let us now plug in $P(2), P\left(\frac{1}{2}\right),$ and $P(-1)$ to solve for $P(-1)$ . We receive $P(2)=2^{2023}P\left(\frac{1}{2}\right)$ , $P\left(\frac{1}{2}\right)-\frac{P(-1)}{2^{2023}}=\frac{2 \cdot 4045!!}{(2023!-1) \cdot 2^{2023}} \implies 2^{2023}P\left(\frac{1}{2}\right)=P(-1)+\frac{2 \cdot 4045!!}{2023!-1}$ , and $P(-1)+P(2)=2024! \cdot \frac{2}{2023!-1}$ . We solve for $P(-1)=\frac{2024!-4045!!}{2023!-1}$ . Therefore, we want $n(2023!-1) \equiv 2024!-4045!! \pmod{2027}$ . We proceed with the extremely inelegant modular arithmetic bash. Note that $4045!!\equiv 0 \pmod{2027}$ so $n(2023!-1) \equiv 2024! \pmod{2027}$ . By Wilson, we have $2024! \equiv \frac{-1}{-1 \cdot -2} \equiv (-2)^{-1} \equiv 1013 \pmod{2027}$ and $2023! \equiv \frac{-1}{-1 \cdot -2 \cdot -3} \equiv 6^{-1} \implies337 \pmod{2027}$ . We wish to find $n$ such that $337n \equiv 1013 \pmod{2027}$ . We notice that $337 \cdot 6 \equiv -5 \pmod{2027}$ . Thus, we can express $n=6a+b$ . and therefore we want to find $-5a+337b \equiv 1013 \equiv -1014 \pmod{2027}$ . We find that $a=68$ and $b=-2$ works, therefore $n=6 \cdot 68 -2 = \boxed{406}$ .
|
[
"Here's a version of the official solution that I think is a bit easier (or at the very least more time-efficient) for a non-olympiad like the HMMT:\n\nWe begin by constructing the following equation: $P(x) - x^{2023}P(1-\\frac{1}{x}) = c(x-1)(x-2)\\ldots(x-2023)$ ( $c$ is some constant)\n\nWe first evaluate the $x^{2023}$ coefficient of this equation. $1 - P(1) = c$ Thus, $P(x) - x^{2023}P(1-\\frac{1}{x}) = (1-P(1))(x-1)(x-2)\\ldots(x-2023)$ We then evaluate the $x^0$ term $P(0) + 1 = (P(1)-1)2023!$ Thus gives us $P(1) = \\frac{2023!+1}{2023!-1}$ We now have enough information to finish the problem:\nPlugging in $x=\\frac{1}{2}$ and $x=2$ gives us $P(-1) = P(2)$ Substituting $x=-1$ gives us $P(-1) + P(2) = 2P(-1) = 2024!(P(1)-1)$ Since we know the value of $P(1)$ simple modulo calculation gives us the answer of 406\n\nOverall quite a difficult problem that I think is pretty much impossible to solve in a high-pressure test environment.\nNot only is the problem-killing idea hard to find, but it is incredibly easy to make small mistakes in the intermediate processes.",
"Too many places to silly :mad:. Solution by Plugging In Stuff:\n\nLet $P(k) = \\sum a_ik^i$ . From the roots, $$ \\sum a_i(k-1)^ik^{2023 - i} - \\sum a_ik^i = a \\prod (k - i). $$ First, comparing leading coefficients yields, $\\sum a_i - a_{2023} = \\sum a_i -1 = a.$ Plugging in $0$ , $-a_{2023}-a_0 = -a(2023!).$ Plugging in $1$ gives, $P(1) = P(0) = a_0.$ Therefore,\n\\begin{align*}\na &= \\frac{a_0 + 1}{2023!} = \\frac{P(1) + 1}{2023!} \n&= \\frac{\\sum{a_i} + 1}{2023!} = \\frac{a + 2}{2023!} \na &= \\frac{2}{2023! - 1}.\n\\end{align*}\n\nNow plugging in $-1$ yields, $-P(2) - P(-1) = a(-2024!)$ .\nPlugging in $2$ yields, $P(2) = 2^{2023}P(\\frac{1}{2}).$ Plugging in $\\frac{1}{2}$ yields $\\frac{1}{2^{2023}} P(-1) - P(\\frac{1}{2}) = a \\prod_{i = 1}^{2023} - \\frac{2i - 1}{2}.$ Synthesizing,\n\\begin{align*}\nP(-1) + P(2) &= 2024!a \n\\frac{1}{2^{2023}} P(-1) - P(\\frac{1}{2}) &= \\frac{1}{2^{2023}} P(-1) - \\frac{1}{2^{2023}} P(2) \n&= -a \\frac{4045!!}{2^{2023}} \nP(-1) &= a\\frac{2024! - 4045!!}{2} = \\frac{2024! - 4045!!}{2023! - 1}.\n\\end{align*}\n\nBy inspection, the numerator and denominator are relatively prime. By Wilsons, the numerator is $\\equiv \\frac{-1}{-1 \\cdot -2}$ and the denominator is $\\equiv \\frac{-1}{-1 \\cdot -2 \\cdot -3} - 1$ . Therefore,\n\\begin{align*}\nn \\equiv \\frac{-1/2}{1/6 - 1} \\equiv \\frac{3}{5} = \\boxed{406}.\n\\end{align*}",
"Misplaced? Setting $k=1$ gives $P(0)=P(1)$ , which we will use later on.\n\nExecute the classic trick of constructing a polynomial of roots $[2023]$ :\n\\[k^{2023}P(\\tfrac{k-1}{k})-P(k)=c(k-1)\\dots(k-2023).\\]\nEquating degrees gives $P(0)-1=c$ , and setting $k=0$ gives $-1-P(0)=-\\tfrac{1}{2023!}(P(0)-1)$ . Solving, $c=\\tfrac{2}{2023!-1}$ .\n\nNow the idea is to substitute a \"cyclic set\" $k\\in(a_0~a_1~\\dots~a_{n-1})$ where $a_{i+1}=\\tfrac{a_i-1}{a_i}$ , such that the $k$ equations may be solved for $P(a_0),\\dots,P(a_{n-1})$ . Here, $a_0=-1$ gives $(-1~\\tfrac{1}{2}~2)$ , whereby $P(-1)=\\mathbf{\\tfrac{2024!-4045!!}{2023!-1}}$ . The rest is extraction."
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Compute the smallest positive integer $n$ for which there are at least two odd primes $p$ such that $\sum_{k=1}^{n} (-1)^{v_p(k!)} < 0$ . Note: for a prime $p$ and a positive integer $m$ , $v_p(m)$ is the exponent of the largest power of $p$ that divides $m$ ; for example, $v_3(18) = 2$ .
|
[
"It was very tough one in the individual round, though I have solved and found the answer 229"
] |
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|
Let $(a_1, a_2, ..., a_8)$ be a permutation of $(1, 2, ... , 8)$ . Find, with proof, the maximum possible number of elements of the set $$ \{a_1, a_1 + a_2, ... , a_1 + a_2 + ... + a_8\} $$ that can be perfect squares.
|
The answer is $\boxed{5}$ , achievable with $(1,3,5,7,2,4,6,8)$ .
For the bound, begin by noticing that $1+2+\dots+8 = 36$ , so we can have at most $6$ squares among the set. Thus, it suffices to show we cannot have all $6$ of them. Obviously, the squares must appear in increasing order, and also must be consecutive in order to keep the $a_i$ small enough. However, the difference between consecutive squares is odd, so we require $5$ odd numbers to have a construction with $6$ squares, a contradiction because we only have $4$ odd numbers.
|
[
"<blockquote>I submitted this and got full credit</blockquote>\n\ni like ur handwriting"
] |
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Find, with proof, the maximum positive integer $k$ for which it is possible to color $6k$ cells of $6 \times 6$ grid such that, for any choice of three distinct rows $R_1$ , $R_2$ , $R_3$ and three distinct columns $C_1$ , $C_2$ , $C_3$ , there exists an uncolored cell $c$ and integers $1 \le i, j \le 3$ so that $c$ lies in $R_i$ and $C_j$
|
The answer is $\boxed{4}$ , and a construction can easily be found (too lazy to use asy).
Obviously, $k=6$ isn't possible. Consider $k=5$ . There are $6$ uncolored squares; pick three of them and report their respective rows as $r_1, r_2, r_3$ . Let the columns of the other three uncolored squares be $c_1, c_2, c_3$ . Note that picking $R_i$ out of the set $\{1,2,3,4,5,6\} \setminus \{r_1,r_2,r_3\}$ and picking $C_i$ out of the set $\{1,2,3,4,5,6\} \setminus \{c_1, c_2, c_3\}$ suffices as a contradiction.
|
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"origin:aops",
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Let triangle $ABC$ be an acute triangle with circumcircle $\Gamma$ . Let $X$ and $Y$ be the midpoints of minor arcs $AB$ and $AC$ of $\Gamma$ , respectively. If line $XY$ is tangent to the incircle of triangle $ABC$ and the radius of $\Gamma$ is $R$ , find, with proof, the value of $XY$ in terms of $R$ .
|
Let the incenter be $I = \overline{CX} \cap \overline{BY}$ . Since $\overline{XY}$ is tangent to the incircle, we have $\triangle IBC \cong \triangle IXY$ . Thus, $BI = XI = XB$ , meaning that $\angle BXI = \angle BXC = \angle BAC = 60^\circ$ . Hence, $\angle OBC = \angle OXY = 120^\circ$ , giving $XY = \boxed{R\sqrt{3}}$ .
|
[
"Rsqrt3=XY"
] |
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"2022 Contests",
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"answer_score": 1014,
"boxed": true,
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"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801547.json"
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Let $ABC$ be a triangle with centroid $G$ , and let $E$ and $F$ be points on side $BC$ such that $BE = EF = F C$ . Points $X$ and $Y$ lie on lines $AB$ and $AC$ , respectively, so that $X$ , $Y$ , and $G$ are not collinear. If the line through $E$ parallel to $XG$ and the line through $F$ parallel to $Y G$ intersect at $P\ne G$ , prove that $GP$ passes through the midpoint of $XY$ .
|
Trivial. Equivalent to $S_{GPX} = S_{GPY}$ and the parallel lines rewrite the latter into $S_{GEX} = S_{GFY}$ . But $GE \parallel AB$ (consider $M = CG \cap AB$ and note $BE/EM = CG/GM$ ) and similarly $GF \parallel AC$ , so we reduce to $S_{GEB} = S_{GFC}$ , which is clear.
|
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"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801549.json"
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Let $P(x) = x^4 + ax^3 + bx^2 + x$ be a polynomial with four distinct roots that lie on a circle in the complex plane. Prove that $ab\ne 9$ .
|
This is an incredibly deep problem. BTW, I don't think any teams solved this problem at the contest.
One of the roots is $0$ . Let the other roots be $z_1, z_2, z_3$ . By Vieta's, $z_1+z_2+z_3=-a\implies \frac{z_1+z_2+z_3}{3}=-\frac{a}{3}$ . Also by Vieta's, $\frac{\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}}{3}=-\frac{b}{3}$ . Now they key is to **invert said circle about the unit circle**. Because the circle passes through $0$ , it's image under this inversion is a line! Now the key is that we want to show that $-\frac{3}{a}\ne -\frac{b}{3}$ . It is enough to show that **the image of the centroid of $z_1, z_2, z_3$ is not on the line** from the Vieta's equations above and the fact that the centroid of $\frac{1}{z_1}, \frac{1}{z_2}, \frac{1}{z_3}$ is on the line (which is obvious). Therefore it is enough to show **the centroid of three points on a circle does not also lie on this circle.** This is true.
|
[
"Suppose by way of contradiction that some $a,b$ work. Let $0,p,q,r$ be the roots, which lie on the circle in that order. From Vieta, we have $pqr = -1$ and\n\\[ab = -(p + q + r)(pq + qr + rp) = (p + q + r)\\left(\\frac1p + \\frac1q + \\frac1r\\right) = 9\\text{.}\\]\nNow we deal with the circle condition. By Ptolemy, $0,p,q,r$ are concyclic iff\n\\[|q(r - p)| = |r(q - p)| + |p(r - q)| \\iff |1/p - 1/r| = |1/p - 1/q| + |1/q - 1/r|\\text{.}\\]\nBut this is simply the equality case of the triangle inequality, so this implies $\\frac1p,\\frac1q,\\frac1r$ are collinear. We see\n\\[\\frac{\\frac1p + \\frac1q + \\frac1r}{3}\\]\nis also on the line. But since the map $z \\to \\frac1z$ is a Mobius transformation, it sends clines to clines, so\n\\[\\frac{3}{\\frac1p + \\frac1q + \\frac1r} = \\frac{p + q + r}{3}\\]\nis on the same circle as $0,p,q,r$ . But this is impossible as this point is the centroid of triangle $PQR$ ."
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Find, with proof, all functions $f : R - \{0\} \to R$ such that $$ f(x)^2 - f(y)f(z) = x(x + y + z)(f(x) + f(y) + f(z)) $$ for all real $x, y, z$ such that $xyz = 1$ .
|
<details><summary>good game!</summary>Let $P(x,y,z)$ be the assertion $P(1,1,1)\implies f(1)^2-f(1)^2=9f(1)$ , i.e $\boxed{f(1)=0}$ $P(x,\frac{1}{x},1) \implies f(x)^2=x(x+\frac{1}{x}+1)(f(x)+f(\frac{1}{x}))$ $P(\frac{1}{x},x,1)\implies f(\frac{1}{x})^2=\frac{1}{x}(\frac{1}{x}+x+1)f(\frac{1}{x})+f(x))$ which gives $(f(x)-f(\frac{1}{x}))(f(x)+f(\frac{1}{x}))=(f(x)+f(\frac{1}{x}))(x^2-\frac{1}{x^2}+x-\frac{1}{x})$ **<u>case1:-</u>** $f(x)+f(\frac{1}{x})\neq 0$ then $f(x)-f(\frac{1}{x})=(x-\frac{1}{x})(x+\frac{1}{x}+1)$ $P(1,x,\frac{1}{x})\implies f(x)f(\frac{1}{x})=-(1+x+\frac{1}{x})(f(x)+f(\frac{1}{x}))$ set $a=f(x) , b=f(\frac{1}{x})$ gives $a-b=(x-\frac{1}{x})(x+\frac{1}{x}+1)$ and $ab=-(1+x+\frac{1}{x})(a+b)$ solving for $b$ we get $b=\pm \frac{1}{2}\cdot (x^2+\frac{1}{x^2}+x+\frac{1}{x}+2)$ **<u>subcase 1.1:-</u>** solving for "+" case we get $b=\frac{1}{x^2}-x$ , which gives $a=x^2-\frac{1}{x}$ **<u>subcase 1.2:-</u>** solving for "-" case we get $b=-x^2-2x-\frac{1}{x}-2$ which actually don't gives $f(1)=0$ , hence subcase 1.2 don't holds**<u>Case2:-</u>** $f(x)+f(\frac{1}{x})=0$ we get then $f(x)\cdot f(\frac{1}{x})=0$ which gives $f(x)=0$ on concluding we get $f(x)\equiv \boxed{x^2-\frac{1}{x},0}$ as our functions $\blacksquare$</details>
|
[
"Bump this",
"plug 1,1,1 to get f(1)=0 and then x,1/x,1 and 1/x,x,1 and solve for f(x), f(1/x)",
"Find, with proof, all functions $f : R - \\{0\\} \\to R$ such that $$ f(x)^2 - f(y)f(z) = x(x + y + z)(f(x) + f(y) + f(z)) $$ for all real $x, y, z$ such that $xyz = 1$ The only solutions are $\\boxed{f\\equiv 0}$ and $\\boxed{f(x) = x^2 - \\frac{1}{x}}$ , which work.\n\nLet $P(x,y,z)$ denote the given assertion. $P(1,1,1): f(1) = 0$ . $P\\left(x,\\frac{1}{x},1\\right): f(x)^2 = x\\left(x+\\frac{1}{x} + 1\\right)(f(x) +f(1/x))$ . $P\\left(1,x,\\frac{1}{x}\\right): -f(x)f\\left(\\frac{1}{x}\\right) = \\left(x+\\frac{1}{x} + 1\\right)(f(x) +f(1/x))$ Thus we have $f(x)^2 = -xf(x)f\\left(\\frac{1}{x}\\right)$ . If $f(x) = 0$ , then setting $x\\to \\frac{1}{x}$ here gives $f\\left(\\frac{1}{x}\\right) = 0$ .\n\nIf $f(x) \\ne 0$ , then\\[f\\left(\\frac{1}{x}\\right) = - \\frac{f(x)}{x}\\]\nThus,\\[f(x)^2 = (x^2 + x + 1) \\left(f(x) - \\frac{f(x)}{x}\\right)\\]\nIf $f(x) \\ne 0$ , dividing both sides by $f(x)$ gives\\[f(x) = (x^2 + x + 1)\\left(1-\\frac{1}{x}\\right) = x^2 + x + 1 - x - 1 - \\frac{1}{x} = x^2 - \\frac{1}{x}\\]\nThis implies for each $x$ , we have\\[f(x)\\in \\left\\{0,x^2 - \\frac{1}{x}\\right\\}\\]\nSuppose now that there existed $a,b\\ne 1$ , so that $f(a) = 0$ and $f(b) = b^2 - \\frac{1}{b}$ .\n\nCase 1: $f\\left(\\frac{1}{ab}\\right) = 0$ . $P\\left(a,b, \\frac{1}{ab}\\right): a\\left(a+b+\\frac{1}{ab}\\right)\\left(b^2 -\\frac{1}{b} \\right) =0$ .\n\nSince $a$ and $b^2 - \\frac{1}{b}$ are nonzero, we have\\[a+b+\\frac{1}{ab} = 0\\]\nHowever, this implies the RHS of the equation is also zero when we do $P\\left(b, a, \\frac{1}{ab}\\right)$ , which implies $f(b)^2 = 0$ , contradiction.\n\nCase 2: $f\\left(\\frac{1}{ab}\\right) = \\frac{1}{a^2 b^2} - ab$ .\n\\[P\\left(a,b,\\frac{1}{ab}\\right): - \\left(b^2 - \\frac{1}{b}\\right)\\cdot \\left(\\frac{1}{a^2 b^2} - ab\\right) = a\\left(a+b+\\frac{1}{ab}\\right)\\left(b^2 -\\frac{1}{b} + \\frac{1}{a^2 b^2} - ab \\right)\\] $P\\left(b,a,\\frac{1}{ab}\\right): f(b)^2 =- \\frac{b}{a} \\cdot f(b)f\\left(\\frac{1}{ab}\\right),$ so\\[f(b) = -\\frac{b}{a}\\cdot \\left(\\frac{1}{a^2 b^2} - ab\\right) = - \\frac{1}{a^3 b} + b^2\\]Then $-\\frac{1}{a^3b} = -\\frac{1}{b}$ , so $a=1$ , which we know is not true.\n\n\n\nThus, either $f(x) = x^2 - \\frac{1}{x}$ for all $x$ , or $f\\equiv 0$ .",
"It is obvious that $f(x)=0$ is a solution.\n\nLet $x=y=z=1, 0=9f(1), f(1)=0$ Now, consider assertion $P: (x,1,\\frac{1}{x}), f(x)^2=x(x+1+\\frac{1}{x})=(f(x)+f(\\frac{1}{x}))$ ; Moreover, consider assertion $P(1,x,\\frac{1}{x}): -f(x)f(\\frac{1}{x})=(x+1+\\frac{1}{x})(f(x)+f(\\frac{1}{x}))$ Thus, we have $f(x)^2=-xf(x)f(\\frac{1}{x});$ yielding $f(x)=0; f(x)=-xf(\\frac{1}{x})$ .\n\nNow, plug this back, we have $f(x)^2=(x^2+x+1)f(x)(1-\\frac{1}{x}), f(x)=x^2+x+1-x-1-\\frac{1}{x}=x^2-\\frac{1}{x}$ "
] |
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"boxed": false,
"end_of_proof": false,
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"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801553.json"
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Let $P_1P_2...P_n$ be a regular $n$ -gon in the plane and $a_1, . . . , a_n$ be nonnegative integers. It is possible to draw $m$ circles so that for each $1 \le i \le n$ , there are exactly $a_i$ circles that contain $P_i$ on their interior. Find, with proof, the minimum possible value of $m$ in terms of the $a_i$ .
.
|
[] |
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"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801554.json"
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|
Let $\Gamma_1$ and $\Gamma_2$ be two circles externally tangent to each other at $N$ that are both internally tangent to $\Gamma$ at points $U$ and $V$ , respectively. A common external tangent of $\Gamma_1$ and $\Gamma_2$ is tangent to $\Gamma_1$ and $\Gamma_2$ at $P$ and $Q$ , respectively, and intersects $\Gamma$ at points $X$ and $Y$ . Let $M$ be the midpoint of the arc $XY$ that does not contain $U$ and $V$ . Let $Z$ be on $\Gamma$ such $MZ \perp NZ$ , and suppose the circumcircles of $QVZ$ and $PUZ$ intersect at $T\ne Z$ . Find, with proof, the value of $T U + T V$ , in terms of $R$ , $r_1$ , and $r_2$ , the radii of $\Gamma$ , $\Gamma_1$ , and $\Gamma_2$ , respectively.
|
[] |
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"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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"answer_score": 0,
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"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801557.json"
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On a board the following six vectors are written: $$ (1, 0, 0), (-1, 0, 0), (0, 1, 0), (0, -1, 0), (0, 0, 1), (0, 0, -1). $$ Given two vectors $v$ and $w$ on the board, a move consists of erasing $v$ and $w$ and replacing them with $\frac{1}{\sqrt2} (v + w)$ and $\frac{1}{\sqrt2} (v - w)$ . After some number of moves, the sum of the six vectors on the board is $u$ . Find, with proof, the maximum possible length of $u$ .
|
[] |
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"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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"answer_score": 0,
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"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801562.json"
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|
Suppose $n \ge 3$ is a positive integer. Let $a_1 < a_2 < ... < a_n$ be an increasing sequence of positive real numbers, and let $a_{n+1} = a_1$ . Prove that $$ \sum_{k=1}^{n}\frac{a_k}{a_{k+1}}>\sum_{k=1}^{n}\frac{a_{k+1}}{a_k} $$
|
The LaTeX is supposed to be
\[ \sum^{n}_{k=1} \frac{a_k}{a_{k+1}} > \sum^{n}_{k=1} \frac{a_{k+1}}{a_k}. \]
Anyway here's my solution. It's quite different to the official solutions.
<details><summary>Alternate Solution</summary>We'll show it for $n=3$ first. Notice that we want to show
\[ \frac{a_1}{a_2} + \frac{a_2}{a_3} + \frac{a_3}{a_1} > \frac{a_2}{a_1} + \frac{a_3}{a_2} + \frac{a_1}{a_3}. \]
Multiplying by $-1$ and adding $\frac{a_3}{a_1} + \frac{a_3}{a_2} + \frac{a_3}{a_3}$ , we obtain:
\begin{align*}
\frac{a_3 - a_1}{a_2} + \frac{a_3 - a_2}{a_3} + \frac{a_3 - a_3}{a_1} < \frac{a_3 - a_2}{a_1} + \frac{a_3 - a_3}{a_2} + \frac{a_3 - a_1}{a_3}
\end{align*}
which is true, since $\frac{a_3 - a_2}{a_3} < \frac{a_3 - a_1}{a_3}$ . Hence we need to show that $\frac{a_3 - a_1}{a_2} < \frac{a_3 - a_2}{a_1} \iff \frac{a_3 - a_2 - a_1}{a_2} < \frac{a_3 - a_2 - a_1}{a_1}$ which is equivalent to $a_1 < a_2$ , so it's true.
Now we'll induct. Suppose $\sum^{n}_{k=1} \frac{a_k}{a_{k+1}} > \sum^{n}_{k=1} \frac{a_{k+1}}{a_k}$ for some natural number $n \geq 3$ . Then,
\[ \sum^{n+1}_{k=1} \frac{a_k}{a_{k+1}} = \sum^{n}_{k=1} \frac{a_k}{a_{k+1}} - \frac{a_n}{a_1} + \frac{a_n}{a_{n+1}} + \frac{a_{n+1}}{a_1} \]
and \[ \sum^{n+1}_{k=1} \frac{a_{k+1}}{a_k} = \sum^{n}_{k=1} \frac{a_{k+1}}{a_k} - \frac{a_1}{a_n} + \frac{a_{n+1}}{a_n} + \frac{a_1}{a_{n+1}}. \]
By our induction hypothesis it suffices to show that:
\[ - \frac{a_n}{a_1} + \frac{a_n}{a_{n+1}} + \frac{a_{n+1}}{a_1} > - \frac{a_1}{a_n} + \frac{a_{n+1}}{a_n} + \frac{a_1}{a_{n+1}} \]
which, similar to the base case proof, is equivalent to
\[ \frac{a_{n+1} - a_n}{a_1} + \frac{a_{n+1} - a_{n+1}}{a_n} + \frac{a_{n+1} - a_1}{a_{n+1}} > \frac{a_{n+1} - a_1}{a_n} + \frac{a_{n+1} - a_n}{a_{n+1}} + \frac{a_{n+1} - a_{n+1}}{a_1}. \]
Since $\frac{a_{n+1} - a_1}{a_{n+1}} > \frac{a_{n+1} - a_n}{a_{n+1}}$ , we need to show that
\begin{align*}
\frac{a_{n+1} - a_n}{a_1} &> \frac{a_{n+1} - a_1}{a_n}
\iff \frac{a_{n+1} - a_n - a_1}{a_1} &> \frac{a_{n+1} - a_n - a_1}{a_n},
\end{align*}
which is true.</details>
|
[
"On the other hand, you can also solve this by smoothing $a_i$ between $a_{i-1}$ and $a_{i+1}$ , and using calculus to note that the inequality is closer when $a_i$ is very near $a_{i-1}$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 18,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801569.json"
}
|
Let $ABC$ be a triangle with $\angle A = 60^o$ . Line $\ell$ intersects segments $AB$ and $AC$ and splits triangle $ABC$ into an equilateral triangle and a quadrilateral. Let $X$ and $Y$ be on $\ell$ such that lines $BX$ and $CY$ are perpendicular to ℓ. Given that $AB = 20$ and $AC = 22$ , compute $XY$ .
|
<details><summary>Solution</summary>Let the intersection of $\ell$ with $\overline{AB}$ and $\overline{AC}$ be $M$ and $N,$ respectively. Then, let the side length of $\triangle{AMN}$ be $x.$ Then, it follows that $AM=AN=x,$ so we get $BM=20-x$ and $CN=22-x.$ Since $\triangle{AMN}$ is equilateral, we know that $\angle{AMN}=\angle{ANM}=60^\circ,$ so by vertical angles, it follows that $$ \angle{BMX}=\angle{CNY}=60^\circ. $$ Therefore, $\triangle{BMX}$ and $\triangle{CNY}$ are $30-60-90$ triangles, implying that $$ MX = \frac{20-x}{2}=10-\frac{x}{2}, NY=\frac{22-x}{2} = 11-\frac{x}{2}. $$ Hence, the answer is
\begin{align*}
XY &= XM + MN + NY
&= 10-\frac{x}{2}+x+11-\frac{x}{2}
&= \boxed{21}.
\end{align*} $\square$</details>
|
[
"Let line l intersect AB and AC at P and Q. From 30-60-90 triangle properties, XP = (20-x)/2 and QY = (22-x)/2, where x is the length of the equilateral triangle with angle A of 60 degrees. XY = XP + PQ + QY = (20-x)/2 + (22-x)/2 + x = **21**.",
"Perfect for problem one: clean and simple.\n\nLet the side length of the equilateral triangle be $x$ . Then using 30-60-90 triangles, $$ XY = x+\\frac 12(20-x) + \\frac 12(22-x) = \\frac 12(20+22) = \\boxed{21}. $$ ",
"...bad solution featuring coordbashing and half cheesing\n[asy]\npair A = (11, 11*sqrt(3));\npair B = (1, sqrt(3));\npair C = (22, 0);\npair O = (0, 0);\ndraw(A--C--O--cycle);\ndraw(B--C);\ndraw(B--(1, 0));\ndraw((-2, 6*sqrt(3))--(24, 6*sqrt(3)), grey);\ndraw(B--(1, 6*sqrt(3)), grey + dashed);\ndraw(C--(22, 6*sqrt(3)), grey + dashed);\ndot(\" $A$ \", A, dir(A));\ndot(\" \", B, dir(B));\nlabel(\" $B$ \", (-0.5, sqrt(3)+0.3));\ndot(\" $C$ \", (22, 0));\ndot(\" \", (1, 0));\nlabel(\" $D$ \", (2, -1.2));\ndot(\" \", (1, 6*sqrt(3)));\nlabel(\" $X$ \", (1, 6*sqrt(3)+1.7));\ndot(\" \", (22, 6*sqrt(3)));\nlabel(\" $Y$ \", (22, 6*sqrt(3)+1.7));\n[/asy]\nFirst, note we can shift $X$ and $Y$ to $D$ and $C$ , where $D$ is on the line parallel to $\\ell$ passing through $C$ . (Hence $D$ is the foot of the altitude from $B$ to this new line)\n\nNote that the side length of this new equilateral triangle is $22$ . Then, we coordbash. Set the unlabeled vertex to be the origin, $A=(11, 11\\sqrt3)$ , $C=(22, 0)$ , and since $AB=20$ , $B=(1, \\sqrt3)$ , so $DC=\\boxed{21}$ . Bruh.",
"My compulsion to trig everything:\n\nBy LoC, $BC = 2 \\sqrt{111}.$ Then,\n\\begin{align*}\nXY &= BC \\sin{(C + 30^\\circ)} \n&= (2 \\sqrt{111})(\\frac{\\sqrt{3}}{2}\\sin{C} + \\frac{1}{2}\\cos{C}) \n&= (2 \\sqrt{111})(\\frac{\\sqrt{3}}{2} \\frac{5\\sqrt{3}}{\\sqrt{111}} + \\frac{1}{2} \\frac{6}{\\sqrt{111}}) = \\boxed{21}\n\\end{align*}\nwhere the angles were found with LoC again."
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1140,
"boxed": true,
"end_of_proof": true,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801591.json"
}
|
Let $ABCD$ and $AEF G$ be unit squares such that the area of their intersection is $\frac{20}{21}$ . Given that $\angle BAE < 45^o$ , $\tan \angle BAE$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$ . Compute $100a + b$ .
|
<details><summary>Solution</summary>Without loss of generality, let $G$ lie inside square $ABCD.$ Also, let the second intersection of the squares be $X \neq A.$ Furthermore, let $\angle{BAX}=\angle{GAX}=\theta.$ We wish to find $\tan(90^\circ-2\theta).$ Clearly, the area of the overlapping region between the two squares is $\tan(\theta),$ so it follows that $\tan(\theta)=\tfrac{20}{21},$ from which it follows that
\begin{align*}
\tan(2\theta) &= \frac{2\tan(\theta)}{1-\tan^2(\theta)}
&= \frac{2\cdot\tfrac{20}{21}}{1-\left(\tfrac{20}{21}\right)^2}
&= \frac{\tfrac{40}{21}}{1 - \tfrac{400}{441}}
&= \frac{\tfrac{40}{21}}{\tfrac{41}{441}}
&= \frac{840}{41}.
\end{align*}
Therefore, we get
\begin{align*}
\tan(90^\circ-2\theta) &= \frac{1}{\tan(2\theta)}
&= \frac{1}{\tfrac{840}{41}}
&= \frac{41}{840},
\end{align*}
yielding an answer of
\begin{align*}
100\cdot41+840 &= 4100+840
&= \boxed{4940}
\end{align*} $\square$</details>
|
[
"Again, trivial problem. Note the area of the intersection of the two unit squares is the sum of two triangles, each of which has a base of 1 and another side x, x < 1. Hence the intersection area = (1*x/2)*2 = x = 20/21. Let the two angles of these triangles, centered at point A, be alpha and beta. The tangent of each of alpha and beta is equal to 20/21, so tangent of (alpha + beta) = (40/21)/(1-(20/21)^2 = 21*40/41 = 840/41. The tangent of BAE is thus cot(90-BAE) = cot(alpha+beta) = 1/(tan (alpha+beta) = 1/(840/41) = 41/840, which is already in reduced fraction form. Hence the answer is 100(41)+840 = **4940**.",
"Let $\\theta = \\angle BAE$ . Then, equating the area of overlap, we have $$ \\tan\\left(45^\\circ - \\frac{\\theta}2\\right) = \\frac{20}{21} \\iff \\tan(90^\\circ - \\theta) = \\frac{840}{41}, $$ so $\\tan \\theta = \\frac{41}{840}$ . We can extract $\\boxed{4940}$ .",
"We let $\\angle BAE=2\\theta$ , and it’s easy to get that the overlapping area is $\\tan (45-\\theta)$ . We would like to have the value of $\\tan (2\\theta)$ , so note that \n\\[\\tan(90-2\\theta)=\\frac{2\\cdot \\tfrac{20}{21}}{1-(\\tfrac{20}{21})^2}=\\frac{840}{41}\\implies \\tan 2\\theta=\\tfrac{41}{840}\\implies \\boxed{4940}.\\]"
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1116,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801593.json"
}
|
Parallel lines $\ell_1$ , $\ell_2$ , $\ell_3$ , $\ell_4$ are evenly spaced in the plane, in that order. Square $ABCD$ has the property that $A$ lies on $\ell_1$ and $C$ lies on $\ell_4$ . Let $P$ be a uniformly random point in the interior of $ABCD$ and let $Q$ be a uniformly random point on the perimeter of $ABCD$ . Given that the probability that $P$ lies between $\ell_2$ and $\ell_3$ is $\frac{53}{100}$ , the probability that $Q$ lies between $\ell_2$ and $\ell_3$ can be expressed as $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$ .
|
Here's a surprisingly clean trigonometric solution.
First, check that $B$ and $D$ must lie between $\ell_2$ and $\ell_3$ ; to prove this is the case, simply notice that the border case yields a probability of $\frac 12$ and the more "tilted" square $ABCD$ is, the greater portion of it will be contained between $\ell_2$ and $\ell_3$ .
Now, let $\ell$ be a fixed line perpendicular to the $\ell_i$ , and define $\theta = \measuredangle(\ell, \overline{AD})$ . Furthermore, assume that the distances between consecutive $\ell_i$ 's is 1. Then, notice that $$ AB^2 = \frac 32 \cdot \frac 1{\cos^2(45^\circ - \theta)} $$ and the area bounded by the square between $\ell_1$ and $\ell_2$ is given by $\tan \theta + \cot \theta$ . Thus,
\begin{align*}
\frac 32 \cdot \frac 1{\cos^2(45^\circ - \theta)} \cdot \frac{47}{100} &= \tan \theta + \cot \theta
\frac 32 \cdot \frac 1{(\frac{\sqrt 2}2 \sin \theta + \frac{\sqrt 2}2 \cos \theta)^2} \cdot \frac{47}{100} &= \tan \theta + \cot \theta
\frac 9{(\sin \theta + \cos \theta)^2} \cdot \frac{47}{100} &= \frac 1{\sin \theta \cos \theta}
\frac 9{1+\sin(2\theta)} \cdot \frac{47}{100} &= \frac 2{\sin(2\theta)}.
\end{align*}
This is linear in $\sin(2\theta)$ and yields $\sin(2\theta) = \frac{200}{223}$ . Then, the portion of perimeter not contained between $\ell_2$ and $\ell_3$ is $$ \frac{2\left(\frac 1{\cos \theta} + \frac 1{\sin \theta}\right)}{4 \cdot \frac 3{\sin \theta + \cos \theta}} = \frac{\frac{\sin \theta + \cos \theta}{\sin(2\theta)}}{\frac 3{\sin \theta + \cos \theta}} = \frac{1+\sin(2\theta)}{3 \sin(2\theta)} = \frac{141}{200}, $$ so the answer is $\frac{59}{200}$ which yields an answer of $\boxed{6100}$ .
|
[
"Different clean trig solution.\n\nFix a unit square $ABCD$ and let $l_1$ be incident from $A$ with angle $\\theta.$ By inspection, $\\theta < 45^\\circ$ <details><summary>equivalent to</summary><blockquote>First, check that $B$ and $D$ must lie between $\\ell_2$ and $\\ell_3$ ; to prove this is the case, simply notice that the border case yields a probability of $\\frac 12$ and the more \"tilted\" square $ABCD$ is, the greater portion of it will be contained between $\\ell_2$ and $\\ell_3$ .\n</blockquote></details>\n\nThen the distance between parallel lines, $d$ , satisfies $3d = \\sqrt{2} \\sin{(\\theta + 45^\\circ)} = (\\sin{\\theta} + \\cos{\\theta}).$ The area not in between $l_2, l_3$ is $\\frac{47}{100}$ made of two triangles. So, $$ 2(\\frac{1}{2} (\\frac{d}{\\sin{\\theta}} \\frac{d}{\\cos{\\theta}})) = \\frac{1}{9} \\frac{(\\sin{\\theta} + \\cos{\\theta})^2}{(\\sin{\\theta}\\cos{\\theta})} $$ found by noting $d$ is the altitudes of the two triangles.\n\nThe desired probability is,\n\\begin{align*}\n\\frac{4 - \\frac{2d}{\\sin{\\theta}} - \\frac{2d}{\\cos{\\theta}}}{4} &= 1 - \\frac{1}{2}(\\frac{d}{\\sin{\\theta}} + \\frac{d}{\\cos{\\theta}}) \n&= 1 - \\frac{d}{2} (\\frac{\\sin{\\theta} + \\cos{\\theta}}{\\sin{\\theta}\\cos{\\theta}}) = 1 - \\frac{1}{6} \\frac{(\\sin{\\theta} + \\cos{\\theta})^2}{(\\sin{\\theta}\\cos{\\theta})} \n&= 1 - \\frac{3}{2} \\frac{47}{100} = \\frac{59}{200}.\n\\end{align*}",
"Let $AB$ , $BC$ , $CD$ , and $DA$ intersect $\\ell_2$ and $\\ell_3$ at $W$ , $X$ , $Y$ , and $Z$ , respectively. As $[WAZ] + [XCY] < [BWZDYX]$ , then $B$ and $D$ both must lie between $\\ell_2$ and $\\ell_3$ .\n\nSuppose $[ABCD] = 100$ . Then $[AWZ] = 23.5$ and $[BWX] = 3$ . Define $a = AZ$ and $b = AW$ . So:\n\\begin{align*}\n[AWZ] &= \\tfrac{1}{2} \\cdot AW \\cdot AZ = \\tfrac{ab}{2} = 23.5 \\quad \\longrightarrow \\quad ab = 47 \n[BWX] &= \\tfrac{1}{2} \\cdot BW \\cdot BX = \\tfrac{(10 - a)(10 - b)}{2} = \\tfrac{100 + ab - 10(a + b)}{2} = 3 \\quad \\longrightarrow \\quad a + b = \\tfrac{141}{10}.\n\\end{align*}\nWe want to find the value of:\n\\begin{align*}\n\\tfrac{40 - 2(a + b)}{40} = 1 - \\tfrac{a+b}{20} = 1 - \\tfrac{141}{200} = \\tfrac{59}{200}.\n\\end{align*}\nTherefore, the desired answer is $\\boxed{6100}$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1050,
"boxed": true,
"end_of_proof": false,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801597.json"
}
|
Let triangle $ABC$ be such that $AB = AC = 22$ and $BC = 11$ . Point $D$ is chosen in the interior of the triangle such that $AD = 19$ and $\angle ABD + \angle ACD = 90^o$ . The value of $BD^2 + CD^2$ can be expressed as $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$ .
|
Let $C' \neq B$ be a point such that $\triangle ABC \cong \triangle ACC'$ . In addition, let $D'$ be a point in $\triangle ACC'$ such that $\angle DAD' \cong \angle BAC$ .
Notice, $BD^2 + CD^2 = CD^2 + CD'^2 = DD'^2$ , but $\angle DCD' = 90^{\circ}$ . As $\triangle DAD' \sim \triangle BAC$ , then
\[ DD' = AD \cdot \tfrac{BC}{BA} = 19 \cdot \tfrac{11}{22} = \tfrac{19}{2} \quad \longrightarrow \quad DD'^2 = \tfrac{361}{4} \]
Hence, the answer is $\boxed{36104}$ .
|
[
"Rotate $\\triangle{ADC}$ and let $AC$ meet $AB$ , after rotation, we call the triangle be $\\triangle{ABD'}$ with $\\angle{ABD'}=\\angle{ACD}, \\angle AD'B=90^{\\circ}$ We can see that $\\angle{D'AB}=\\angle{DAC}, \\triangle{AD'D}\\sim \\triangle{ABC}, D'D=\\frac{19}{2}$ $BD^2+CD^2=BD^2+D'B^2=DD'^2=\\frac{361}{4}$ leads to $\\boxed{36104}$ ",
"Found this to be much easier than problem four.\n\nLet $D'$ be the point such that $\\triangle AD'B \\sim \\triangle ADC$ . Then, by the angle condition, $$ \\angle D'BD = \\angle D'BA + \\angle ABD = \\angle ABD + \\angle ACD = 90^\\circ. $$ Then, $$ BD^2+CD^2 = DD'^2 = \\left(11 \\cdot \\frac{19}{22}\\right)^2 = \\frac{361}4, $$ so we can extract $\\boxed{36104}$ .",
"**<span style=\"color:#f00\">Remark:</span>** Before you do anything, think about the simplest way you can formulate that idea. Dang man that really sucks.\n\nRotate $\\triangle DAC$ so that $AC$ coincides with $AB$ . Then $\\angle D'BD = 90^{\\circ}$ , so $BD^2+CD^2=D'D^2$ . Since $D'AD \\sim BAC$ , we have $D'D=\\tfrac{19}{22} \\cdot 11 = \\tfrac{19}{2}$ , so the answer is $\\boxed{\\tfrac{361}{4}}$ . \n\n\n\n",
"Alternatively, consider an inversion about $A$ with radius $22$ , and let $P^*$ denote the image of point $P$ under this inversion. The condition then becomes $\\angle B^*D^*C^* = 90$ . By the inversion distance formula, $AD^* = \\frac{22^2}{19}$ , and\n\\[BD^2 + CD^2 = \\left(\\frac{22^2}{AB^*AD^*}B^*D^*\\right)^2 + \\left(\\frac{22^2}{AC^*AD^*}C^*D^*\\right)^2 = \\frac{19^2}{22^2} (B^*C^*)^2 = \\frac{361}{4}\\]\ngiving an answer of $\\boxed{36104}$ and we are done."
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1018,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801601.json"
}
|
Let $ABCD$ be a rectangle inscribed in circle $\Gamma$ , and let $P$ be a point on minor arc $AB$ of $\Gamma$ . Suppose that $P A \cdot P B = 2$ , $P C \cdot P D = 18$ , and $P B \cdot P C = 9$ . The area of rectangle $ABCD$ can be expressed as $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers and $b$ is a squarefree positive integer. Compute $100a + 10b + c$ .
|
Note that we have $PA \cdot PB=2, PC \cdot PD = 18, PB \cdot PC = 9$ , which implies that $PA \cdot PD=4$ . Then, by applying the Law of Cosines on $\triangle PAD, \triangle PDC, \triangle PBC,$ and $\triangle PAB$ , and setting the minor arcs subtended by lines $AB$ and $AD$ to be $2 \beta$ and $2\theta$ , we get that $$ y^2=PA^2+PB^2+4\cos \beta = PC^2+PD^2-36 \cos \beta $$ and $$ x^2=PA^2+PD^2-8 \cos \theta=PB^2+PC^2-18 \cos \theta $$ Let $PA^2+PB^2+PC^2+PD^2=S$ . Since $PA^2+PC^2=PB^2+PD^2=x^2+y^2$ as $AC$ and $BD$ are diameters of the circle, we note that $S=2x^2+2y^2$ . Also, $2x^2=S-26\cos\theta$ and $2y^2=S-32\cos\beta$ . Looking at triangle $ADC$ , we see that $\cos\beta=\frac{x}{\sqrt{x^2+y^2}}$ and $\cos\theta=\frac{y}{\sqrt{x^2+y^2}}$ . Therefore, $\cos\beta=\frac{x\cos\theta}{y}$ and $\cos\theta=\frac{y\cos\beta}{x}$ . We plug in our earlier value of $S$ into the equations with terms $2x^2$ and $2y^2$ to receive that $y^2=13\cos\theta$ and $x^2=16\cos\beta$ . Substituting our earlier values for $\cos\beta$ and $\cos\theta$ in terms of the other cosine, we see that $x=\frac{16\cos\theta}{y} \implies xy=16\cos\theta$ and $y=\frac{13\cos\beta}{x} \implies xy=13\cos\beta$ . Thus $16\cos\theta=13\cos\beta$ . We also know that $\cos(\beta+\theta)=\cos\beta\cos\theta-\sin\beta\sin\theta=0$ . Now it is simply a matter of solving for $\cos\beta$ and $\cos\theta$ . We get $\cos\theta=\frac{13}{5\sqrt{17}}$ . But $xy$ is the area, and we derived earlier that $xy=16\cos\theta$ . Then the answer is $16\cdot\frac{13}{5\sqrt{17}}=\frac{208\sqrt{17}}{85}$ , and the requested sum is $\boxed{21055}$ .
|
[
"Firstly we can observe that $\\frac{PD}{PB}=2; \\frac{PC}{PA}=\\frac{9}{2}$ Then, since rectangle $ABCD$ is inscribed in the circle, so $\\angle{APC}=\\angle{PBD}=90^{\\circ}; AC=BD$ We denote that $PB=a,PD=2a; PA=2m, PC=9m$ , now with the application of PT, $5a^2=85m^2, a=\\sqrt{17}m$ , the diameter of the circle can be expressed as $\\sqrt{85}m$ So we put this result to $PA\\cdot PB=2; a\\cdot 2m=2; \\sqrt{17}m^2=1; m^2=\\frac{\\sqrt{17}}{17}$ Then we can apply Ptolemy theorem in cyclic quadrilateral $ADCP, PBCA$ separately. We consider that $BC=x,AB=y$ In quadrilateral $ADCP$ , $x\\cdot 9m+y\\cdot 2m=2\\sqrt{17}m \\cdot \\sqrt{85}m$ , after eliminate $m$ , we have $9x+2y=34\\sqrt{5}m$ Now, in quadrilateral $APBC$ , $2m\\cdot x+\\sqrt{17}m\\cdot \\sqrt{85}m=9m\\cdot y$ , we have $17\\sqrt{5}m=9y-2x$ Now, we can find the relationship between $x,y; 18y-4x=9x+2y; 13x=16y$ Finally, solve $x,y$ in term of $m$ , write $y=\\frac{13x}{16}, 9x+\\frac{13x}{8}=\\frac{85x}{8}=34\\sqrt{5}m$ , getting that $x=\\frac{16\\sqrt{5}m}{5}; y=\\frac{13\\sqrt{5}m}{5}$ So $xy=\\frac{16\\cdot 13}{5}\\cdot m^2=\\frac{208\\sqrt{17}}{85}$ as the answer\n\nSo, the desired answer should be $20800+170+85=\\boxed{21055}$ ",
"Not bad for problem 6; here's a quick solution.\n\nLet $\\theta = \\angle DPC$ . Then, we may angle chase to obtain $\\angle APD = \\angle BPC = 90^\\circ - \\theta$ and $\\angle APB = 180^\\circ - \\theta$ . Notice that the three conditions imply $PD \\cdot PA = 4$ . Then, \n\\begin{align*}\n[PAD]+[PBC ] &=[ABCD] = [PDC] - [PAB] \n\\sin \\theta (PC \\cdot PD - PA \\cdot PB) &= \\cos \\theta (PA \\cdot PD + PB \\cdot PC) \n16 \\sin \\theta &= 13 \\cos \\theta.\n\\end{align*}\nThus, $\\sin \\theta = \\frac{13\\sqrt{17}}{85}$ , and the desired area is $16 \\sin \\theta = \\frac{208\\sqrt{17}}{85}$ . The requested answer is $\\boxed{21055}$ .",
"Ptolemy & LoC:\n\nLet $PA = a, PB = b.$ Note $PC = \\frac{9}{2}a, PD = 2b.$ Let $AB = x, BC = y, AC = z = \\sqrt{x^2 + y^2}.$ By Ptolemy on $PBCD$ and $PADB$ we have,\n\\begin{align*}\nby + 2bx &= \\frac{9}{2}az\naz + bx &= 2by.\n\\end{align*}\nThe equations result in $y = \\frac{13}{16}x$ or, $x = \\frac{16\\sqrt{17}}{85}z, y = \\frac{13 \\sqrt{17}}{85}.$ By Ptolemy on $PABC$ , \n\\begin{align*}\nay + \\frac{9}{2}ax &= 2bz \na &= \\frac{2}{\\sqrt{17}} b \\implies a = \\frac{2}{17^{1/4}}, b =17^{1/4}.\n\\end{align*}\nThen $\\cos{\\angle ABC} = \\frac{y}{z} = \\frac{13 \\sqrt{17}}{85}.$ So by LoC on $\\triangle APD$ ,\n\\begin{align*}\nx^2 &= a^2 + (2b)^2 - 2(2ab) \\cos{\\angle APD} \n&= a^2 + 4b^2 - 8 \\frac{13\\sqrt{17}}{85} \n[ABCD] &= \\frac{13}{16}x^2= \\boxed{\\frac{208\\sqrt{17}}{85}}.\n\\end{align*}",
"The conditions imply $PA \\cdot PD = 4$ . Note that \\[ \\tfrac{1}{2} \\cdot [ABCD] = [PCD] - [PAB] = [PAD] + [PBC]. \\]\nLet $\\alpha = \\angle CPD$ , then $\\angle APB = 180 - \\alpha$ and $\\angle APD = \\angle BPC = 90 - \\alpha$ . Thus:\n\\begin{align*}\n[ABCD] = 2 \\cdot \\left(\\tfrac{\\sin \\alpha}{2} \\cdot \\left(PC \\cdot PD - PA \\cdot PB\\right)\\right) = 2 \\cdot \\left(\\tfrac{\\cos \\alpha}{2} \\cdot \\left(PA \\cdot PD + PB \\cdot PC\\right)\\right).\n\\end{align*}\nSo, $16 \\sin \\alpha = 13 \\cos \\alpha$ . Combining with Pythagorean Identity yields $\\sin \\alpha = \\tfrac{13}{5\\sqrt{17}}$ . So, $[ABCD] = \\tfrac{208\\sqrt{17}}{85}$ . Hence, the answer is $\\boxed{21055}$ ."
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1084,
"boxed": true,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801604.json"
}
|
Point $P$ is located inside a square $ABCD$ of side length $10$ . Let $O_1$ , $O_2$ , $O_3$ , $O_4$ be the circumcenters of $P AB$ , $P BC$ , $P CD$ , and $P DA$ , respectively. Given that $P A+P B +P C +P D = 23\sqrt2$ and the area of $O_1O_2O_3O_4$ is $50$ , the second largest of the lengths $O_1O_2$ , $O_2O_3$ , $O_3O_4$ , $O_4O_1$ can be written as $\sqrt{\frac{a}{b}}$ , where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$ .
|
Just a quick note:
A shortcut to the official sol is scaling $\square{O_1O_2O_3O_4}$ by a factor of $2$ about $P$ to obtain a cyclic quadrilateral with circumcenter $P$ (This reduces the solution length to 4-5 lines)
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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{
"answer_score": 108,
"boxed": false,
"end_of_proof": true,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801605.json"
}
|
Let $E$ be an ellipse with foci $A$ and $B$ . Suppose there exists a parabola $P$ such that $\bullet$ $P$ passes through $A$ and $B$ , $\bullet$ the focus $F$ of $P$ lies on $E$ , $\bullet$ the orthocenter $H$ of $\vartriangle F AB$ lies on the directrix of $P$ .
If the major and minor axes of $E$ have lengths $50$ and $14$ , respectively, compute $AH^2 + BH^2$ .
|
$AH=BH$ and the distance from $A$ to the directrix of parabola is equivalent to $AF$ which is half of the major axis, $AF=25$ .
Let $D$ be the projection of $A$ onto the directrix, $AD=25$ . Due to symmetry, $DH=\frac{AB}{2}=24$ , $AH^2=25^2+24^2=1201$ . As $AH=BH, AH^2+BH^2=2402$
|
[
" $AH^{2} + BH^{2}=2402$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 20,
"boxed": false,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801606.json"
}
|
Let $A_1B_1C_1$ , $A_2B_2C_2$ , and $A_3B_3C_3$ be three triangles in the plane. For $1 \le i \le3$ , let $D_i $ , $E_i$ , and $F_i$ be the midpoints of $B_iC_i$ , $A_iC_i$ , and $A_iB_i$ , respectively. Furthermore, for $1 \le i \le 3$ let $G_i$ be the centroid of $A_iB_iC_i$ .
Suppose that the areas of the triangles $A_1A_2A_3$ , $B_1B_2B_3$ , $C_1C_2C_3$ , $D_1D_2D_3$ , $E_1E_2E_3$ , and $F_1F_2F_3$ are $2$ , $3$ , $4$ , $20$ , $21$ , and $2020$ , respectively. Compute the largest possible area of $G_1G_2G_3$ .
|
Let $a_i, b_i, c_i, d_i, e_i, f_i, g_i$ be the points' vectors. It suffices to maximize $\frac{1}{2} |(g_2 - g_1) \times (g_3 - g_1)|.$ Observe that,
\begin{align*}
\frac{1}{2} |(g_2 - g_1) \times (g_3 - g_1)| &= \frac{1}{2} |(\frac{(a_2 - a_1) + (b_2 -b_1) + (c_2 - c_1)}{3}) \times (\frac{(a_3 - a_1) + (b_3 -b_1) + (c_3 - c_1)}{3})|
&= \frac{1}{18} |\sum_{cyc} (a_2 - a_1) \times (a_3 - a_1) + \sum_{cyc} (b_2 - b_1)(c_3 - c_1) + (b_3 - b_1)(c_2 - c_1)|
&= \frac{1}{18} |\sum_{cyc} (b_2 - b_1 + c_2 - c_1) \times (b_3 - b_1 + c_3 - c_1) - \sum_{cyc} (a_2 - a_1) \times (a_3 - a_1)|
\end{align*}
We are given $\frac{1}{2} |(a_2 - a_1) \times (a_3 - a_1)|, \dots$ and $\frac{1}{2}|(\frac{b_2 - b_1 + c_2 - c_1}{2}) \times (\frac{b_3 - b_1 + c_3 - c_1}{2})|, \dots$ from the areas of $A_1A_2A_3, \dots$ and $D_1D_2D_3, \dots$ respectively. To maximize $[G_1G_2G_3]$ let the signed areas of the vertex triangles be negative and the midpoint ones positive. Then we have, $[G_1G_2G_3] = \frac{1}{18}(8 \cdot (20 + 21 + 2020) + 2 \cdot (2 + 3 + 4)) = \boxed{917}.$
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1016,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801608.json"
}
|
Suppose $\omega$ is a circle centered at $O$ with radius $8$ . Let $AC$ and $BD$ be perpendicular chords of $\omega$ . Let $P$ be a point inside quadrilateral $ABCD$ such that the circumcircles of triangles $ABP$ and $CDP$ are tangent, and the circumcircles of triangles $ADP$ and $BCP$ are tangent. If $AC = 2\sqrt{61}$ and $BD = 6\sqrt7$ ,then $OP$ can be expressed as $\sqrt{a}-\sqrt{b}$ for positive integers $a$ and $b$ . Compute $100a + b$ .
|
Let $X=AB \cap CD$ and $Y=AD \cap BC$ . By radical axis theorem on $\omega$ , $(ABP)$ , and $(CDP)$ followed by PoP, $XP^2=XO^2-8^2$ . By similar logic, $YP^2=YO^2-8^2$ . By the perpendicularity lemma, $\overline{OP}$ is perpendicular to $\overline{XY}$ . Let $P' = OP \cap XY$ , and note that $P'$ is the projection of $O$ onto $\overline{XY}$ . By the Pythagorean theorem we have $OP \cdot (2OP'-OP)=OP'^2-(OP'-OP)^2=OP'^2-PP'^2=8^2$ . Now one can coordinate bash and compute $OP'=32$ which returns $OP=32-8\sqrt{15}$ , for a final answer of $1024 \cdot 100+960=\boxed{103360}$ .
<details><summary>Idea for potential synthetic solution</summary>Brokard's theorem on $ABCD$ implies that the triangle whose vertices are $Z=AC \cap BD$ , $X$ , and $Y$ has orthocenter $O$ . Then the problem reduces by orthocenter calculations to finding $ZX^2+ZY^2-XY^2$ . Bash using trig and similar triangles induced by the cyclic quadrilateral $ABCD$ .</details>
|
[] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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{
"answer_score": 1048,
"boxed": true,
"end_of_proof": false,
"n_reply": 1,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801610.json"
}
|
Rectangle $R_0$ has sides of lengths $3$ and $4$ . Rectangles $R_1$ , $R_2$ , and $R_3$ are formed such that: $\bullet$ all four rectangles share a common vertex $P$ , $\bullet$ for each $n = 1, 2, 3$ , one side of $R_n$ is a diagonal of $R_{n-1}$ , $\bullet$ for each $n = 1, 2, 3$ , the opposite side of $R_n$ passes through a vertex of $R_{n-1}$ such that the center of $R_n$ is located counterclockwise of the center of $R_{n-1}$ with respect to $P$ .
Compute the total area covered by the union of the four rectangles.
|
<details><summary>Solution</summary>Note that the overlapping region of any two consecutive triangles is a triangle which makes up half of the area of either triangle, so all triangles have the same area. Therefore, the answer is
\begin{align*}
12 + \frac{12}{2} + \frac{12}{2} + \frac{12}{2} &= 12+6+6+6
&= \boxed{30},
\end{align*}
by simply looking at the amount of area added by each new rectangle. $\square$</details>
|
[
"Trivial, can eyeball this problem. Note that R0, R1, R2, and R3 have equal areas, all which equal to 3*4 = 12. The incremental area added by each R_n+1 is simple 12/2 = 6. Hence the total area of the entire figure is 12+6*3 = **30**",
"Notice that the three rectangles have equal areas, which we will call $S$ . Then the total area is simply $$ S+S-\\frac 12 S + S - \\frac 12 S + S - \\frac 12 S = \\frac 52 S = \\boxed{30} $$ by considering the overlap after appending each rectangle.",
"extremely easy, headsolved in 30 seconds :cool: my solution is the same as #3",
"every time we add a new rectangle we add half the area\n\n12+3*6=30"
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1102,
"boxed": false,
"end_of_proof": false,
"n_reply": 5,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2801613.json"
}
|
Sets $A, B$ , and $C$ satisfy $|A| = 92$ , $|B| = 35$ , $|C| = 63$ , $|A\cap B| = 16$ , $|A\cap C| = 51$ , $|B\cap C| = 19$ . Compute the number of possible values of $ |A \cap B \cap C|$ .
|
@above that's incorrect
<details><summary>Solution</summary>Let $|A\cap B\cap C|=x$ . Then $|A\cap B|=16-x$ , so we have $x\leq 16$ . Also, $|C\backslash \{A\cup B\}|=x-7$ , so we have $x\geq 7$ . We can verify that these are the only constraints on $x$ , so the answer is $\boxed{10}$ .</details>
|
[
"<details><summary>solution</summary>$|A \\cup B \\cup C| = |A| + |B| + |C| - |A \\cap B| - |A \\cap C| - |B \\cap C| + |A \\cap B \\cap C|$ . Plugging in the above constants, $|A \\cap B \\cap C| = |A \\cup B \\cup C| - 104$ . The intersection of the three sets is at least 0 and at most equal to the cardinality of the greatest intersection of two of the sets, which is 51. Therefore there are 52 possible values of $|A \\cap B \\cap C|$</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1014,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804643.json"
}
|
Compute the number of ways to color $3$ cells in a $3\times 3$ grid so that no two colored cells share an edge.
|
<details><summary>Solution</summary>If one of the cells is the center cell, then the number of possible colorings is $\tbinom{4}{2}=6,$ and if no cells are the center cell, then it is not difficult to see that there are $4$ possible configuration, each of which yields $4$ possible colorings. Hence, the answer is
\begin{align*}
6+4\cdot4 &= 6+16
&= \boxed{22}.
\end{align*} $\square$</details>
|
[
"There are three cases we need to consider for a configuration that violates the rule (in other words, a grid where two colored cells share an edge)\n1. We have three consecutive colored cells in the same row or column. \n2. We have two consecutive colored squares in the same row or column, and one colored cell that is not in the same row or column but shares an edge with one of the two colored squares. This looks like an \"L\" on the grid.\n3. We have two consecutive colored squares that share an edge and one that does not share an edge with either of the other two colored squares.\n\nThere are six ways to achieve (1), since there are three rows and three columns we could choose for our three cells. For (2), this is the same as counting the number of $2\\times2$ squares and then counting the number of ways we could select one of the cells in that $2\\times2$ subsquare to not be colored, which gives us $4 \\cdot 4 = 16$ ways. For (3), we further break down into two cases. If we pick the two consecutive colored cells to share one of the middle four edges that touches the center square, there are only two ways to pick the other square as not to share an edge. If we pick any of the other 8 edges that does not touch the center square, we have four squares to place the last colored square.\n\nIn total, this gives us $6 + 16 + 4\\cdot 2 + 8\\cdot 4 = 62.$ There are $\\binom{9}{3} = 84$ total choices of three colored cells, so our answer is $84 - 62 = \\fbox{22}$ possible ways.",
"The number of columns in the grid that contain at least one colored cell must be either two or three. If three columns have at least one colored cell, then the number of ways to color them onto the grid is $3\\cdot 2\\cdot 2$ . If two columns have at least one colored cell, then one of these two columns, which we will name $C$ , must have two colored cell, meaning that there is one way to color $C$ . Then if the other column $D$ with a colored cell is adjacent to $C$ , there are $2\\cdot 2$ ways to choose which columns are $C$ and $D$ , and one way to color them. If $D$ is not adjacent to $C$ , then there are $2$ ways to choose which columns are $C$ and $D$ , and $3$ ways to color these columns. This totals to $3\\cdot 2\\cdot 2 + 2\\cdot 2 + 2\\cdot 3 = \\boxed{22}$ ways.",
"<details><summary>Solution</summary>Notice that there is either one colored box in each column or two colored boxes in a column and the other colored box in another column.**Case 1:** Two colored boxes in a column.**Subcase 1.1:** The first column has two colored boxes.\n\nNotice that this means one colored box is in the top left corner and the other colored box is in the bottom left corner. There are $4$ ways for the final box.**Subcase 1.2:** The second column has two colored boxes.\n\nThis means one colored box is in the middle top box and the other is in the middle bottom box. There are $2$ ways for the final box.**Subcase 1.3:** The third column has two colored boxes.\n\nNote this is basically the same as *Subcase 1.1*, so there are $4$ ways for this subcase.**Case 2:** Each column has a colored box.\n\nThere are $\\binom{3}{1}$ ways to pick the first box, $\\binom{2}{1}$ way to pick the second box, and $\\binom{2}{1}$ for the final box. So, $\\binom{3}{1} \\cdot \\binom{2}{1} \\cdot \\binom{2}{1}=12.$ Thus, we have a total of $4+2+4+12=\\boxed{22}.$</details>"
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1108,
"boxed": true,
"end_of_proof": true,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804645.json"
}
|
Michel starts with the string HMMT. An operation consists of either replacing an occurrence of H with HM, replacing an occurrence of MM with MOM, or replacing an occurrence of T with MT. For example, the two strings that can be reached after one operation are HMMMT and HMOMT. Compute the number of distinct strings Michel can obtain after exactly $10$ operations.
|
The final string must be $14$ characters long, and of the form $HM----------MT$ , where each $-$ is either an $O$ or an $M$ such that no two $O$ 's are adjacent. The number of $O$ 's is an integer from $0$ to $5$ inclusive, so we simply do casework on the number of $O$ 's to get $\tbinom{6}{5}+\tbinom{7}{4}+\tbinom{8}{3}+\tbinom{9}{2}+\tbinom{10}{1}+1=\boxed{144}$ .
|
[
"HMMT is the first string\n\n2nd string: HMMMT, HMOMT\n\n3rd string: HMMMMT, HMOMMT, HMMOMT\n\n4th string: HMMMMMT, HMOMMMT, HMMOMMT, HMMMOMT, HMOMOMT\n\nengineering induction tells us its just a Fibonacci sequence, 2-3-5-8-13-21-34-55-89-144, $\\boxed{144}$ is the answer "
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1022,
"boxed": true,
"end_of_proof": false,
"n_reply": 2,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804646.json"
}
|
Compute the number of nonempty subsets $S \subseteq\{-10,-9,-8, . . . , 8, 9, 10\}$ that satisfy $$ |S| +\ min(S) \cdot \max (S) = 0. $$
|
<details><summary>Solution</summary>Note that either $\min(S)$ or $\max(S)$ has an absolute value of $1$ or $(\min(S), \max(S))$ $\in$ $\{(-3,2),$ $(-2,2),$ $(-2,3)\}.$ The former case yields
\begin{align*}
2\binom{2}{2}+2\binom{3}{2}+\ldots+2\binom{10}{2} &=2 \binom{11}{3}
&= 2\cdot 165
&= 330,
\end{align*}
and the latter case yields $5,$ for a total of $330+5=\boxed{335}.$ $\square$</details>
|
[
"<details><summary>Solution</summary>Note that we must have $\\min S <0, \\max S>0.$ Let $-a= \\min S$ and $b=\\max S$ , then we get $|S|\\leq a+b+1$ and so $ab\\leq a+b+1.$ This shows that one of $a,b$ must be equal to $1$ or $(a,b)$ must be one of the pairs $(2,2), (2,3)$ or $(3,2).$ We can now count the number of possible sets for each case.\nIf $a=1, b=k$ (where $k=1$ is impossible for obvious reasons), we can choose $k-2$ elements from $\\{0,\\ldots, k-1\\}$ , giving $\\binom{k}{2}$ total sets. Thus, we have $\\sum_{k=2}^{10} \\binom{k}{2}=165$ possible sets $S$ for which $a=1$ and analogously $165$ sets such that $b=1.$ Finally, there are $3$ possible sets for $(a,b)=(2,2)$ and a unique set for $(a,b)=(2,3)$ and $(a,b)=(3,2)$ , resulting in the final answer $\\boxed{335}.$</details>",
"Note that the maximum $|S|=10$ . \n\nFor every $2\\leq |S|\\leq 10$ , there are $2\\cdot \\binom{|S|}{2}$ ways. So there are in total $2\\cdot \\sum_{i=2}^{10}\\binom{i}{2}=330$ ways.\n\nHowever, there are cases when $|S|=4=-2\\cdot 2$ , there are $\\binom{5-2}{2}=3$ ways, and $|S|=6=-3\\cdot 2=-2\\cdot 3$ , there are $2\\cdot \\binom{6-2}{6-2}=2$ ways. Sum up to attain $\\boxed{335}$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
] |
{
"answer_score": 1122,
"boxed": true,
"end_of_proof": true,
"n_reply": 3,
"path": "Contest Collections/2022 Contests/2022 Harvard-MIT Mathematics Tournament/2804647.json"
}
|
Five cards labeled $1, 3, 5, 7, 9$ are laid in a row in that order, forming the five-digit number $13579$ when read from left to right. A swap consists of picking two distinct cards, and then swapping them. After three swaps, the cards form a new five-digit number n when read from left to right. Compute the expected value of $n$ .
|
By linearity of expectation, we can compute the expected value of each digit.
In general, the probability that the digit remains the same after three swaps is $$ \left(\frac 35\right)^3 + {3 \choose 2} \cdot \frac 35 \cdot \frac 25 \cdot \frac 1{10} + \frac 25 \cdot \left(\frac 25 - \frac 1{10}\right) \cdot \frac 1{10} = \frac 3{10} $$ using casework on how many times that position engages in a swap. Thus, we can compute the expected value of the first digit as $$ \frac 3{10} \cdot 1 + \frac 7{10} \cdot \frac{3+5+7+9}4 = \frac 92 $$ as each of the other four digits are equally likely to be swapped in. Similarly, we may compute the expected values of the other digits to be $\frac{19}4$ , 5, $\frac{21}4$ , $\frac {11}2$ in that order. Thus, the answer is $$ 10000 \cdot \frac 92 + 1000 \cdot \frac {19}4 + 100 \cdot 5 + 10 \cdot \frac{21}4 + \frac{11}2 = \boxed{50308}. $$
|
[
"Find the sum of $1+3+5+7+9=25$ . Find the probability that it will be the same digit. Then, we can have $\\frac{6^3+3\\cdot 6\\cdot 4+4\\cdot 3}{10^3}=\\frac{3}{10}$ . Thus, for each one its essentially the same. By LOE, we have $10000\\cdot \\left(\\frac{3}{10}+\\frac{7}{10}\\cdot \\frac{25-1}{4}\\right)+1000\\left(\\frac{3\\cdot 3}{10}+\\frac{7}{10}\\cdot \\frac{25-3}{4}\\right)+100\\left(\\frac{3\\cdot 5}{10}+\\frac{7}{10}\\cdot \\frac{25-5}{4}\\right)+10\\left(\\frac{3\\cdot 7}{10}+\\frac{7}{10}\\cdot \\frac{25-7}{4}\\right)+1\\left(\\frac{3\\cdot 9}{10}+\\frac{7}{10}\\cdot \\frac{25-9}{4}\\right)$ Which gives $10000\\cdot \\frac{9}{2}+1000\\cdot \\frac{9}{10}+25\\cdot 7\\cdot 22+150+7\\cdot 5\\cdot 10+21+\\frac{63}{2}+\\frac{27}{10}+\\frac{14}{5}$ $=45000+900+3850+500+21+\\frac{63}{2}+\\frac{11}{2}=45900+500+3850+21+37=46400+3850+58=50250+58=\\boxed{50308}$ . "
] |
[
"origin:aops",
"2022 Contests",
"2022 Harvard-MIT Mathematics Tournament"
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{
"answer_score": 1018,
"boxed": true,
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The numbers $1, 2, . . . , 10$ are randomly arranged in a circle. Let $p$ be the probability that for every positive integer $k < 10$ , there exists an integer $k' > k$ such that there is at most one number between $k$ and $k'$ in the circle. If $p$ can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$ , compute $100a + b$ .
|
The idea is to use recursion. Let $a_n$ be the number of *ways* to permute $1, 2, 3, \cdots, n$ cyclically such that the condition holds. Call a number $a$ *close enough* to $b$ if there is at most one number between $a$ and $b$ .
To develop a recursive formula for $a_n$ , we consider two cases. In the first case, say that $n$ and $n-1$ are adjacent. Then, by removing $n$ from the circle of numbers, any number that satisfied the condition because $n$ was close enough to it now has $n-1$ the same distance from it, so every one of these numbers still works. Thus, the remaining circular permutation is a valid permutation under $a_{n-1}$ . Thus, every permutation with $n$ numbers and the two numbers adjacent can be obtained by inserting $n$ into one of the positions consecutive to $n-1$ , which yields $2a_{n-1}$ cases.
In the second case, say that $n$ and $n-1$ have one number between them, say $k$ . Then, consider removing both $n$ and $k$ . Any number that was close enough to $n$ is close enough to $n-1$ . Furthermore, $k$ is still adjacent to $n-1$ , so it satisfies the condition as well. Thus, we are left with a permutation of $n-2$ numbers that satisfies the given condition, for $a_{n-2}$ ways. Finally, every permutation with $n$ and $n-1$ not adjacent can be obtained by taking one of the $a_{n-2}$ smaller circular permutations, inserting $n$ consecutive to $n-1$ (the largest number in this permutation) and choosing one of $n-2$ numbers to omit from the outside permutation and instead insert between $n$ and $n-1$ .
From our previous discussion, we then have the recursive relation $$ a_n = 2a_{n-1} + 2(n-2) a_{n-2}. $$ Letting $p_n = \frac{a_n}{(n-1)!}$ , $$ p_n(n-1)! = 2p_{n-1}(n-2)! + 2p_{n-2} (n-2)! \iff p_n = \frac 2{n-1}(p_{n-1} + p_{n-2}). $$ Because $p_4 = p_5 = 1$ , we have $p_6 = \frac 45$ , $p_7 = \frac 35$ , $p_8 = \frac 25$ , $p_9 = \frac 14$ , and $p_{10} = \frac{13}{90},$ so the answer is $\boxed{1390}$ .
|
[
"<details><summary>solution</summary>There are at least 3 numbers less than or equal to 10 that are greater than numbers 1-7, so they will never be directly adjacent to all the numbers that are greater than them. The condition is only not satisfied when 9 and 10 are adjacent or 8 is between 9 and 10. Since reflections and rotations are irrelevant in this count, $\\frac{a}{b} = \\frac{8! + 7!}{9!} = \\frac{8+1}{9*8} = \\frac{1}{8}$ , $100a + b = 108$ .</details>",
"<details><summary>Think-About-It</summary>Let $A_n$ be the successful probability for numbers $1, 2, ..., n$ . We want to find $A_{10}$ .\n\nFor a placement to succeed, $n-1$ and $n$ must either be directly adjacent or have exactly one number between them. If they're directly adjacent, delete $n$ from the circle and we're left with $A_{n-1}$ . If they're semi-adjacent, delete $n$ and the number that was between $n-1$ and $n$ , and we're left with $A_{n-2}$ .\n\nThe probability that each case happens is $\\frac{2}{n-1}$ so we come up with the recursion $$ A_n = \\frac{2}{n-1} \\left (A_{n-1} + A_{n-2} \\right) $$ Given $A_4 = A_5 = 1$ (as it's impossible to come up with a \"bad\" placement for those cases), we can calculate $A_6 = \\frac{4}{5}, A_7 = \\frac{3}{5}, A_8 = \\frac{2}{5}, A_9 = \\frac{1}{4},$ and $A_{10} = \\frac{13}{90}$ .\n\nSo our desired answer is $\\boxed{1390}$</details>",
"I claim the answer is $\\boxed{1390}$ .\nLet $p_n$ be the probability for a circle with numbers $1, 2, \\dots, n$ . Because the condition is true for $n$ all of the time, we are inspired to write a recursion based on how $n$ is inserted into the $n-1$ case. Due to the condition, then $n$ and $n-1$ must have a distance of at most $2$ .\n-------------------\n<span style=\"color:blue\">**Case 1: the distance between $n$ and $n-1$ is 1.**</span> In this scenario, there is first a $\\frac{2}{n-1}$ probability that $n-1$ is next to $n$ .\n[asy]\nunitsize(5cm); size(4cm); draw(unitcircle);\nint n = 10; real tickLength = 0.1;\nfor (int i = 0; i < n; ++i) { real theta = pi/2 - 2*pi*i/n; pair p1 = dir(degrees(theta)); pair p2 = (1 - tickLength) * p1; draw(p1--p2); } \nlabel(\" $n$ \", 1.2 * dir(90), fontsize(10pt)); label(\" $n-1$ \", 1.2 * dir(54), fontsize(10pt)); [/asy]\nThen treating the $n$ and $n-1$ like just $n-1$ , we are in the $n-1$ case. Thus the probability here is $\\frac{2}{n-1}\\cdot p_{n-1}$ .\n-------------------\n<span style=\"color:blue\">**Case 2: the distance is 2.**</span> Again, there is a $\\frac{2}{n-1}$ probability of this case.\n[asy]\nunitsize(5cm); size(4cm); draw(unitcircle);\nint n = 10; real tickLength = 0.1;\nfor (int i = 0; i < n; ++i) { real theta = pi/2 - 2*pi*i/n; pair p1 = dir(degrees(theta)); pair p2 = (1 - tickLength) * p1; draw(p1--p2); }\nlabel(\" $n$ \", 1.2 * dir(90), fontsize(10pt)); label(\" $a$ \", 1.2 * dir(54), fontsize(10pt)); label(\" $n-1$ \", 1.2 * dir(18), fontsize(10pt)); [/asy]\nNotice that $a$ could be anything, and then the condition for $k = a$ is satisfied. Treating the arc $n-a-(n-1)$ like one number again, we are in the $n-2$ case. Thus the probability here is $\\frac{2}{n-1}\\cdot p_{n-2}$ .\nThus, \\[ p_n = \\frac{2}{n-1}(p_{n-1} + p_{n-2}) \\] with $p_5 = 1$ and $p_4 = 1$ , we solve and find $p_{10} = \\frac{13}{90}$ which gives the requested answer."
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Let $S = \{(x, y) \in Z^2 | 0 \le x \le 11, 0\le y \le 9\}$ . Compute the number of sequences $(s_0, s_1, . . . , s_n)$ of elements in $S$ (for any positive integer $n \ge 2$ ) that satisfy the following conditions: $\bullet$ $s_0 = (0, 0)$ and $s_1 = (1, 0)$ , $\bullet$ $s_0, s_1, . . . , s_n$ are distinct, $\bullet$ for all integers $2 \le i \le n$ , $s_i$ is obtained by rotating $s_{i-2}$ about $s_{i-1}$ by either $90^o$ or $180^o$ in the
clockwise direction.
|
Observe that consecutive moves are either one forward move or one left move. Define a *run* as a maximal subsequence of $s$ in a line. Then $s$ can be segmented into alternating runs horizontally and vertically.
Given a set of numbers, we can determine the direction of the runs in the sequence due to the non-intersecting property. E.g for $1, 2, 4, 6$ the is a $x$ -runs span $1 \to 6, 6 \to 2, 2 \to 4$ in $x$ -coordinates respectively. With this property, it suffices to choose subsets $A \in \{0, 1, \dots, 11\}$ (starting $0$ doesn't count) and $B\in \{0, 1, \dots, 9\}$ ( $0 \in B$ is guaranteed) to determine the path $s.$ However, since the runs alternate, either $|A| = |B|, |B + 1|.$
We can enumerate the sequence as (note $n \ge 2$ ),
\begin{align*}
& \left ( \left (\binom{12}{1} - 2 \right ) \binom{9}{0}+ \left (\binom{12}{1} - 1 \right) \cdot \binom{9}{1} \right ) + \left (\binom{12}{2}\binom{9}{1} + \binom{12}{2} \cdot \binom{9}{2} \right ) + \cdots
&= \sum_{i = 0}^{9} \binom{12}{i + 1}\binom{9}{i} + \sum_{i = 0}^{9} \binom{12}{i}\binom{9}{i} - 12
&= \sum_{i = 0}^{9} \binom{12}{11-i}\binom{9}{i} + \sum_{i = 0}^{9} \binom{12}{12-i}\binom{9}{i} - 12
&= \binom{21}{11} + \binom{21}{12} - 12 = \boxed{646634}
\end{align*}
|
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Random sequences $a_1, a_2, . . .$ and $b_1, b_2, . . .$ are chosen so that every element in each sequence is chosen independently and uniformly from the set $\{0, 1, 2, 3, . . . , 100\}$ . Compute the expected value of the smallest nonnegative integer $s$ such that there exist positive integers $m$ and $n$ with $$ s =\sum^m_{i=1} a_i =\sum^n_{j=1}b_j . $$
|
[] |
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|
Consider permutations $(a_0, a_1, . . . , a_{2022})$ of $(0, 1, . . . , 2022)$ such that $\bullet$ $a_{2022} = 625$ , $\bullet$ for each $0 \le i \le 2022$ , $a_i \ge \frac{625i}{2022}$ , $\bullet$ for each $0 \le i \le 2022$ , $\{a_i, . . . , a_{2022}\}$ is a set of consecutive integers (in some order).
The number of such permutations can be written as $\frac{a!}{b!c!}$ for positive integers $a, b, c$ , where $b > c$ and $a$ is minimal. Compute $100a + 10b + c$ .
|
Typo: In the second condition, $a_i \le \frac{625i}{2022}$ should be $a_i \ge \frac{625i}{2022}$ .
|
[
"<details><summary>ans idk</summary>216695????????????????????????????????????????</details>"
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"answer_score": 4,
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Let $S$ be a set of size $11$ . A random $12$ -tuple $(s_1, s_2, . . . , s_{12})$ of elements of $S$ is chosen uniformly at random. Moreover, let $\pi : S \to S$ be a permutation of $S$ chosen uniformly at random. The probability that $s_{i+1}\ne \pi (s_i)$ for all $1 \le i \le 12$ (where $s_{13} = s_1$ ) can be written as $\frac{a}{b}$ where $a$ and $b$ are relatively prime positive integers. Compute $a$ .
|
We claim that the probability is $\frac{10^{12}+4}{11^{12}}$ . We can count this as follows.
First, note that if it was not in a circle (without the condition at $i=12$ ), then we can fix the first element ( $a_i$ ), and then each element thereafter has a $\frac{10}{11}$ chance of not being the mapping of the previous element through $\pi$ . With this, we would get a "raw" probability of $\frac{10^{11}}{11^{11}}$ . However, we do have the condition holding at $i=12$ , so we have overcounted, since some of the cases we counted could have the sequence "loop around" so that the condition does not hold at $i=12$ .
To account for this, we can subtract such cases where $\pi(a_{12})=a_1$ . Theoretically, if the condition did not have to hold at $i=11$ , we could again similarly fix $a_{12}$ (which then fixes $a_1$ in turn, since $\pi(a_{12})=a_1$ , and the probability $a_1$ is what we want it to be is $\frac{1}{11}$ ), and each element thereafter would have a $\frac{10}{11}$ chance of not being the previous element's mapping through $\pi$ . This gives us the "raw" probability here of $\frac{10^{10}}{11^{10}}*\frac{1}{11}=\frac{10^{10}}{11^{11}}$ . Now, notice that we have again overcounted---we have a possibility of the sequence again "looping around" so that the condition does not hold at $i=11$ .
Again, to account for this, we can add back the cases where $\pi(a_{11})=a_{12}$ and $\pi(a_{12})=a_1$ . Again, here we can also "ignore" the $i=10$ condition, and we would fix $a_{11}$ (in turn fixing $a_{12}$ and $a_1$ , occuring as we want with probability $\frac{1}{11^2}$ ) and the rest follow as we want each with probability $\frac{10}{11}$ . Multiplying these together as we want, we end up with a "raw probability" here of $\frac{10^{9}}{11^{9}}*\frac{1}{11^2}=\frac{10^{9}}{11^{11}}$ .
We can continue this pattern of overcounting and adding/subtracting back, until we end up with an answer of
\[\frac{10^{11}-10^{10}+10^9-10^8+\dots-10^2+10-1}{11^{11}},\]
before we add back our final necessary case, which is if the permutation is such that $\pi(a_i)=a_{i+1}$ holds for all $i$ . We want to know the probability of which this will happen.
First, notice that if this holds, then if we fix what $a_1$ is, the entirety of the $12$ -tuple must be fixed. Additionally, in order for the permutation mappings to loop around in a cycle, we must also have that the "cycle length" of the cycle $a_1$ belongs to in the permutation $\pi$ must be a divisor of $6$ . Essentially, we want to figure out the probability of $a_1$ being in a "cycle" of length $1$ , $2$ , $3$ , $4$ , or $6$ in the permutation $\pi$ , times $\frac{1}{11^{11}}$ to account for the probability of the other $11$ elements of the $12$ -tuple aligning with the values we want. Now, to solve this, I make the following claim.
<span style="color:#960000">***</span>**<span style="color:#960000">Claim.</span>** The probability of $a_1$ being in a cycle of length $k$ (for any $1\leq k\leq 11$ ) in the permutation $\pi$ is exactly $\frac{1}{11}$ .
*Proof.*
If we want $a_1$ to be in a cycle of length $k$ , then we need $k-1$ more elements to be in the cycle, and we need to choose which maps to which in the situation. First, to choose which elements map in which order in the cycle, we have a $10*9*\dots*(12-k)$ ways to choose them. Then, notice that $a_1$ maps to the first element with probability $\frac{1}{11}$ , the first element maps to the second element with probability $\frac{1}{10}$ , and so on (in particular, note that the very last probability accounts for the mapping from the $k$ -th element back to $a_1$ to complete the cycle, which has a probability of $\frac{1}{12-k}$ ). Multiplying all of these out, we get a total probability of $a_1$ being in a cycle of length exactly $k$ in $\pi$ is
\[\frac{10*9*\dots*(12-k)}{11*10*9*\dots*(12-k)}=\frac{1}{11},\]
as we wished to prove.
<span style="color:#960000">***</span>
This means that the probability of $a_1$ being in a "cycle" of length $1$ , $2$ , $3$ , $4$ , or $6$ in the permutation $\pi$ , times $\frac{1}{11^{11}}$ is equal to exactly
\[5*\frac{1}{11}*\frac{1}{11^{11}}=\frac{5}{11^{12}},\]
which we can add back to our original summation. Using geometric series summation, we end up with
\[\frac{10^{11}-10^{10}+10^9-10^8+\dots-10^2+10-1}{11^{11}}+\frac{5}{11^{12}}=\frac{10^{12}-1}{11^{12}}+\frac{5}{11^{12}},\]
or a final probability of $\frac{10^{12}+4}{11^{12}}$ , which we claimed in the first place. This means that $a=10^{12}+4$ (I'm too lazy to type out the whole number and count the number of $0$ 's), finishing the problem.
<span style="color:#960000">***</span>
In general, the **<span style="color:#960000">Claim</span>**'s value of $11$ can be extended to any integer $n$ with a similar argument. Additionally, for this problem, using a similar approach, we can find that if we instead have a $n$ -tuple and a permutation of $k$ -elements (for example, in our problem, $n$ was $12$ and $k$ was $11$ ), if $m$ is the number of factors of $n$ less than or equal to $k$ , then the probability that the condition described in the original problem (extended to match the numbers here) is satisfied is then exactly
\[p=\frac{(k-1)^n+(m-1)}{k^n}.\]
In particular, one should notice that $m$ is always at least $1$ , since $1\leq k$ and $1\mid n$ always. This means that the probability is always at least $\left(\frac{k-1}{k}\right)^n$ , which I thought was cool.
|
[
"<details><summary>Answer</summary>1000000000004</details>\n<details><summary>Solution</summary>Let's count the number $N$ of ordered pairs $(T, \\pi)$ , where $T$ is the 12-tuple, satisfying the given condition.\n\nWe will use inclusion-exclusion on the 12 constraints of the form $s_{i+1} \\neq \\pi(s_i)$ . We have \\[N = \\sum_{A \\subseteq \\{1, 2, \\dots, 12\\}}(-1)^{|A|} \\text{(number of ways to choose}\\ (T, \\pi)\\ \\text{such that for all}\\ i \\in A, s_{i+1} = \\pi(s_i)\\text{)}.\\] For a moment, assume $A \\neq \\{1, 2, \\dots, 12\\}$ . Then we claim that there are $11! \\cdot 11^{12-|A|}$ ways to choose $(T, \\pi)$ for this particular $A$ , since once one of the $11!$ possible $\\pi$ 's is chosen, every $s_i$ such that $i - 1 \\not \\in A$ ( $12 \\not \\in A$ in the case of $i = 0$ ) can be independently assigned to an arbitrary element of $S$ , allowing all other $s_i$ to be uniquely determined afterwards. Therefore we can rewrite this sum as \\[N = \\text{(number of ways to choose}\\ (T, \\pi)\\ \\text{such that for all}\\ i, s_{i+1} = \\pi(s_i)\\text{)}\\ + \\sum_{A \\subsetneq \\{1, 2, \\dots, 12\\}}(-1)^{|A|}11!\\cdot 11^{12-|A|},\\] where the first term comes from the case $A = \\{1, 2, \\dots, 12\\}$ ; call it $M$ . The sum can be rewritten by summing over $x = |A|$ : \\[N = M + 11! \\sum_{x=0}^{11} \\binom{12}{x}(-1)^x11^{12-x}.\\] By the binomial theorem, we have \\[N = M + 11!((11 - 1)^{12} - 1) = M + 11! \\cdot 10^{12} - 11!.\\] Now it remains to evaluate $M$ . Suppose that the permutation $\\pi$ has been fixed. Then, all the elements of $T$ are uniquely determined by $s_1$ . Also, in order for it to be the case that $s_1 = \\pi(s_{12})$ , the size of the cycle of $\\pi$ that $s_1$ is located in must be a factor of 12 (that is, either 1, 2, 3, 4, 6; note that 12 is not possible since $\\pi$ has only 11 elements). Therefore, the number of valid pairs for the fixed $\\pi$ is the sum of the sizes of all cycles of $\\pi$ whose size satisfies this condition.\n\nNow, we evaluate the sum of this value over all $\\pi$ by looking at one element of $S$ at a time and counting the number of permutations that place this element in a cycle of appropriate size, and summing over all elements (in effect, multiplying by 11).**Lemma:** The number of permutations that place a fixed element $x$ of $S$ in a cycle of length $k$ is $10!$ .**Proof:** We need to choose $k-1$ elements from the 10 elements of $S \\setminus \\{x\\}$ to form the cycle with $S$ ; then there are $(k-1)!$ ways to arrange the cycle and $(11-k)!$ ways to arrange the permutation's structure outside the cycle. This gives a count of $\\binom{10}{k-1}(k-1)!(11-k)! = 10!$ , as desired.\n\nBy the lemma, we have \\[M = 11 \\cdot (10! \\cdot 5) = 11! \\cdot 5,\\] since there are 5 possible choices of cycle size. This gives \\[N = 11!(10^{12} + 4),\\] so the desired probability is \\[\\frac{N}{11!\\cdot11^{12}} = \\frac{10^{12} + 4}{11^{12}}.\\] Since $\\gcd(10^{12} + 4, 11^{12}) = 1$ , our answer is $10^{12} + 4$ .</details>"
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|
Find all natural numbers $n$ for which there is a permutation $\sigma$ of $\{1,2,\ldots, n\}$ that satisfies:
\[
\sum_{i=1}^n \sigma(i)(-2)^{i-1}=0
\]
|
Awesome problem, the hardest and nicest problem of the test, very similar to <details><summary>past IMO problem</summary>IMO 2012, P6</details> in terms of the ideas involved. Solved with $\textbf{Aryan-23}$ . $\textbf{Lemma 1:}$ $n \equiv 1 \pmod{3}$ do not work. $\textbf{Proof}$
Assume for the sake of contradiction that some $n \equiv 1 \pmod{3}$ works.
Look at the equation $\pmod{3}$ , then $$ 0 = \sum_{i=1}^{n} \sigma(i) \cdot (-2)^{i-1} = \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \equiv 1 \pmod{3} $$ which is a contradiction. $\blacksquare$
We present the same construction as in #4 with some motivation afterwards. $\textbf{Lemma 2}$ $n \equiv 0,2 \pmod{3}$ work. $\textbf{Proof}$
We induct with base case permutations $(2,1), (2,3,1)$ for $n = 2,3$ , respectively.
Assume that $(\sigma(1), \cdots, \sigma(n))$ work for some $n \in \mathbb{N}$ , $n \equiv 0,2 \pmod{3}$ , then take $\sigma_1(1) = 2, \sigma_1(2) = 3, \sigma(m+3) = 1$ and $\sigma_1(i) = \sigma(i-2)+3$ for $i \in \{3, \cdots, n+2\}$ which can be verified to work for $n+3$ as $(\sigma_1(1), \cdots, \sigma_1(n+3))$ . $\blacksquare$
Summing up our conclusions from $\textbf{Lemma 1}$ and $\textbf{Lemma 2}$ , we can conclude that only $n \equiv 0,2 \pmod{3}$ work.
<details><summary>Terribly Written Motivation</summary>I actually started the problem myself before Aryan joined me and was able to conclude using the same global strategy as IMO 2012, P6 by taking $\pmod{3}$ which is motivated by the fact that $(-2)^{i-1} \equiv 1 \pmod{3}$ giving us that $n \equiv 0,2 \pmod{3}$ , this leaves us to construct, and there should intuitively be some construction; after finding the construction for $n =2,3,5,6$ , we imposed the conditions that $\sigma(1) = 2, \sigma(2) = 3$ and guessed that $i$ gets inserted into the $i-2-$ th place for each new thing and when there is a jump of two we insert them in decreasing order consecutively. This gives the constructions $(2,3,5,6,8,7,4,1)$ and $(2,3,5,6,8,9,7,4,1)$ for $8,9$ , respectively, the rest of the problem is to just formalize the inductive construction which turns out to contain $0,2 \pmod{3}$ in ascending and $1 \pmod{3}$ in descending order.</details>
|
[
"It's easy to get $n \\equiv 0,2 \\pmod 3$ and then you die for the rest of the contest. ",
"NVT's problem!!\n\nThis was easily the best problem on the test according to me. \n\nSoln. Looking mod 3, we get that $3|n^2+n$ and thus, $n\\not\\equiv 1 \\pmod{3}$ . This is kinda a thing that's easy to miss and then struggle for a long while.\n\nNow, we show a construction for $m+3$ assuming one for $m$ . Let $\\sigma$ be the permuation for $m$ .\n\nDefine $\\tau(1)=2, \\tau(2)=3, \\tau(i)=\\sigma(i-2)+3$ for $i\\in \\{3,\\cdots m+2\\}$ and $\\tau(m+3)=1$ . \n\nWe can now verify that this indeed works. This also works out to something quite clean with the first few terms being the 2, 3 mod 3 terms alternating and a bunch of 1 mod 3 terms at the end.\n \nSomething like 2,3,5,6,8,9,...3k+2, 3k+3, 3k+1, 3k-2, ..., 1 I think which is possible to observe with some attempts on small values.",
"<blockquote>Find all natural numbers $n$ for which there is a permutation $\\sigma$ of $\\{1,2,\\ldots, n\\}$ that satisfies:\n\\[\n\\sum_{i=1}^n \\sigma(i)(-2)^{i-1}=0\n\\]</blockquote>\n\nit's 2,3,5,6.... (first group all 2 mod 3 and 0 mod 3) and then 7,4, 1 (all 1 mod 3 in descending)<blockquote>NVT's problem!!! \n\nThis was easily the best problem on the test according to me. \n\nSoln. Looking mod 3, we get that $3|n^2+n$ and thus, $n\\not\\equiv 1 \\pmod{3}$ . This is kinda a thing that's easy to miss and then struggle for a long while.\n\nNow, we show a construction for $m+3$ assuming one for $m$ . Let $\\sigma$ be the permuation for $m$ .\n\nDefine $\\tau(1)=2, \\tau(2)=3, \\tau(i)=\\sigma(i-2)+3$ for $i\\in \\{3,\\cdots m+2\\}$ and $\\tau(m+3)=1$ . \n\nWe can now verify that this indeed works. This also works out to something quite clean with the first few terms being the 2, 3 mod 3 terms alternating and a bunch of 1 mod 3 terms at the end.\n \nSomething like 2,3,5,6,8,9,...3k+2, 3k+3, 3k+1, 3k-2, ..., 1 I think which is possible to observe with some attempts on small values.</blockquote>\n\nhow much for just construct?",
"eek didnt do it cuz i forgot what a bijective function is\n\n<blockquote>NVT's problem!!! \n\nThis was easily the best problem on the test according to me. \n\nSoln. Looking mod 3, we get that $3|n^2+n$ and thus, $n\\not\\equiv 1 \\pmod{3}$ . This is kinda a thing that's easy to miss and then struggle for a long while.\n\nNow, we show a construction for $m+3$ assuming one for $m$ . Let $\\sigma$ be the permuation for $m$ .\n\nDefine $\\tau(1)=2, \\tau(2)=3, \\tau(i)=\\sigma(i-2)+3$ for $i\\in \\{3,\\cdots m+2\\}$ and $\\tau(m+3)=1$ . \n\nWe can now verify that this indeed works. This also works out to something quite clean with the first few terms being the 2, 3 mod 3 terms alternating and a bunch of 1 mod 3 terms at the end.\n \nSomething like 2,3,5,6,8,9,...3k+2, 3k+3, 3k+1, 3k-2, ..., 1 I think which is possible to observe with some attempts on small values.</blockquote>\n\n\n<details><summary>rg</summary>sir <span style=\"font-size:50%\">hehe</span></details> orz",
"The way I motivated the same construction is that I am dealing with mod3, I am sleepy, so it has to be n->n+3 and insertion.\n\n",
"<blockquote>This was easily the best problem on the test according to me. </blockquote>\n\n :( ",
"Any ideas on who was the author of the problem? @below say no more :P",
"<blockquote>This was easily the best problem on the test according to me. \n</blockquote>\n\nIt was proposed by Tejaswi after all :)\n",
"<blockquote><blockquote>This was easily the best problem on the test according to me. \n</blockquote>\n\nIt was proposed by Tejaswi after all :)</blockquote>\nEven though I failed I am pretty sure this is the best problem of the test.The statement is so beautiful itself.",
"Clearly by modulo $3$ , we need to have $\\sum_{i=1}^n i \\equiv 0 \\pmod{3}$ , which means $n \\equiv 0,2 \\pmod{3}$ For the construction, we create a sequence $a_1, \\cdots a_{n-1}$ , and define \n\\[ \\sigma(1) = 2a_1, \\sigma(2) = a_1+2a_2, \\sigma(3) = a_2+2a_3, \\cdots, \\sigma(n-1) = a_{n-2} + 2a_{n-1}, \\sigma(n) = a_n \\]\nFor $n =3k$ , the desired sequence is generated by $1,1,2,2,3,3, ..., k, k, k-1, k-2, ... 1$ For $n =3k-1$ , the desired sequence is generated by $1,1,2,3, ... ,k-1,k, k-1,k-1...,3,3,2,2,1,1$ ",
"i have shown that $n \\equiv 0,2 \\mod 3$ . But i couldn't find the construction.How much can i get?please reply to this because i am very tensed. ",
"Taking the sum $\\mod 3$ , we get that $\\frac{n^2+n}{2}\\equiv 0\\mod 3$ , so $n\\equiv 0,2\\mod 3$ . Realize that we are essentially representing every number from $1$ to $n$ in the form $1\\cdot a+2\\cdot b$ such that the $a$ value of $\\sigma (i+1)$ is the $b$ value of $\\sigma (i)$ . Once we do this, and look at small cases (I got for $6$ that the permutation $2,5,6,4,3,1$ works, it seems everyone else go the second value as $3$ somehow) that we have $|a-b|$ as small as possible. We can now inductively construct $n$ . Base cases are the permutations $2,1$ and $2,3,1$ .\n\nIf $n=3k$ , then our inductive hypothesis is that $3k-1,3k-2$ are consecutive terms in that order, $3k-1$ is represented as $1\\cdot (k-1)+2\\cdot k$ , and $3k-2$ is represented as $1\\cdot k +2\\cdot (k-1)$ . We want to show that $3k$ can fit inside the permutation, with $3k$ being represented as $1\\cdot k+2\\cdot k$ , and $3k-1, 3k$ will be consecutive terms. This just works.\n\nIf $n=3k+2$ , then our inductive hypothesis is that $3k-1,3k$ are consecutive terms in that order, $3k-1$ is represented as $1\\cdot (k-1)+2\\cdot k$ and $3k$ is represented as $1\\cdot k+2\\cdot k$ . We want to show that $3k+1,3k+2$ are represented as $1\\cdot (k+1)+2\\cdot k$ and $1\\cdot k+2\\cdot (k+1)$ respectively, and that they are consecutive in that order. Placing $3k+1,3k+2$ between $3k-1,3k$ works, as desired.\n\nFor $n=14$ this gives the sequence $2,5,8,11,14,13,12,10,9,7,6,4,3,1$ , which I think is the opposite/some sort of inverse of the sequence everyone else has, which starts by alternating $0,2\\mod 3$ and then adds the $1\\mod 3$ numbers at the end.",
"<details><summary>3 hours of dying</summary>First take $\\pmod 3$ of the whole equation to get that $n \\equiv 0,2 \\pmod 3$ .\nSo the satisfying $n$ are of the form $3k+2,3k$ .\nNow replace $\\sigma(i)$ by $a_I$ for now,checking small cases we see that\n\\begin{align*}n=2 &\\implies a_1=2,a_2=1 \nn=3 &\\implies a_1=2,a_2=3,a_3=1\nn=5 &\\implies a_1=2,a_2=3,a_3=5,a_4=4,a_5=1\nn=6 &\\implies a_1=2,a_2=3,a_3=5,a_4=6,a_5=4,a_6=1\n\nn=8 &\\implies a_1=2,a_2=3,a_3=5,a_4=6,a_5=8,a_6=7,a_7=4,a_8=1\nn=9 &\\implies a_1=2,a_2=3,a_3=5,a_4=6,a_5=8,a_6=9,a_7=7,a_8=4,a_9=1\\end{align*}\nThe construction is clearly visible now,\nIf $a_i,b_i$ denote the permutation for $n=3k,3(k+1)$ respectively then take,\n\\begin{align*}a_i &=b_i \\text{ for all } i \\in \\{1,2,....,3k-2\\} \nb_{3k-1} &=3(k+1)-1 \nb_{3k} &=3(k+1) \nb_{3k+1} &=3(k+1)-2 \nb_{3k+2} &=a_{3k-1} \nb_{3(k+1)} &=a_{3k} \\end{align*}\nSimilarly for $n=3k+2,3(k+1)+2$ ,the construction would be,\n\\begin{align*}a_i &=b_i \\text{ for all } i \\in \\{1,2,....,3k\\} \nb_{3k+1} &=3(k+1) \nb_{3k+2} &=3(k+1)+2 \nb_{3(k+1)} &=3(k+1)+1 \nb_{3(k+1)+1} &=a_{3k+1}\nb_{3(k+1)+2} &=a_{3k+2} \\end{align*}\nWhich clearly works and thus we are done $\\blacksquare$</details>\n<details><summary>Emotional damage</summary>First time while testing small values I got for $n=5$ a construction $\\{2,5,4,3,1\\}$ this worked :wallbash_red: and I thought this was the actual construction leading me to die for 3 hours :( :( :( Then I checked the thread and was motivated to concentrate on $\\{2,3, \\cdots ,1\\}$ and after dying for 45 more minutes (idk properly), I finally managed to finish this.</details>\n\n\n\n",
"Special Thanks to **Periwinkle** for motivating me to verify the construction.\nLet $(\\sigma(1),\\sigma(2),....,\\sigma(n))$ be $(a_1,a_2,....,a_n)$ . $a_1+a_2+a_3+...a_n$ = $\\frac{n(n+1)}{2}$ ------ $(1)$ and $a_1+(-2)a_2+...a_n(-2)^n-1$ = $0$ ------ $(2)$ ,\nSubstracting $(2)$ from $(1)$ gives $n=0,2(mod3)$ .\n<details><summary>Construction</summary>$n=2$ $(2,1)$ $n=3$ $(2,3,1)$ $n=5$ $(2,3,5,4,1)$ $n=6$ $(2,3,5,6,4,1)$ $n=8$ $(2,3,5,6,8,7,4,1)$ $n=9$ $(2,3,5,6,8,9,7,4,1)$</details>\nFrom the construction we can observe that for $n=3k$ we have $(2,3,5,...3k-1,3k,3k-2,...1)$ and for $n=3k+2$ we have $(2,3,5,....3k+2,3k+1,...1)$ .\nAlso, note that right sides of $(3k-2)$ and $(3k+1)$ in $n=3k$ and $n=3k+2$ is decreasing by a differnce of $3$ .\nSo, if we verify for $n=3k$ , somehow $n=3k+2$ has the same way but different procedure.\nLet the construction for $n=3k$ be like:- $(2,3,5,....3k-1,3k,3k-2,.....1)$ and let $2+3(-2)+....3k-3(-2)^{2k-2}=x$ , and $3k-5(-2)^{2k+1} +3k-8(-2)^{2k+2} + ...+(-2)^{3k-1} =y$ ,**Claim 1** $y=(-2)^{2k+1}.(k-1)$ <details><summary>Induction proof</summary>Clearly , $k=1$ and $k=2$ works.\nSuppose for $k$ it works, \nSo, for $n=3k$ we have $y=(-2)^{2k+1} .(k-1)$ ,\nNow, for $n=3(k+1)$ ,\nWe have from Construction, $(-2)^{2k+3} . (3k-2+(-2)(k-1)$ which is $(-2)^{2k+3} . (k)$ ------ $(a)$</details>\nFrom, $n=3k$ , we have , $x+y+ (-2)^{2k+1}.(9k-9)=0$ , $x=-y-(-2)^{2k+1} . (9k-9)$ ,\nNow, for $n=3k+3$ we have , $x-63(-2)^{2k-2}.(k-1) -8y$ which should be zero.\nNow, by replacing $x$ we get , $y+8(k-1)(-2)^{2k-2}$ , and from $a$ , clearly it is zero. \nHence verified , similarily, we can do for $n=3k+2$ . So, The problem is completed.\nIf anyone wants to solve with other construction dm me.",
"Nice problem...literally wasted nearly 2 hours....after seeing the pattern for 9 because I am so idiot that I don't know $28 \\times 3 = 84$ and not $54$ .Oops!! I wrote as the format $\\{n,n-1,...,1\\}$ instead of $\\{1,2,...,n\\}$ <details><summary>sketch</summary>So , it is very easy to show that $n \\equiv 0 \\text{or} 2 \\pmod 3 $ .We say a permutation to be good if it satisfies the given property. Note that $\\{1,3,2\\} ; \\{1,3,4,6,5,2\\} , \\{1,3,4,6,7,9,8,5,2\\}$ are all good .( I wrote $384$ $354$ for $9$ and discarded the construction.) Now sort of extend this Now note that $\\sum\\limits_{k=0}^n(3k+2)(-2)^k = -(n+1)(-2)^{n+1}$ by simple AGP sum.\nNow in the inductive step take the good permutation for $n$ . We have $1,3,4,6,7,9...,3n+1, 3n+3 , 3n+2 , 3n-1 , 3n-4,...,2 $ . Now insert $3n+4 , 3n+5$ and $3n+4,3n+6,3n+5$ Between $3n+3$ and $3n+2$ .[\\hide]</details>",
"Let $S=\\sum_{i=1}^n \\sigma(i)(-2)^{i-1}$ , we see that $S\\pmod{2}$ gives $\\sigma(1)$ to be even, let $\\sigma(1)=2a_1$ . Taking mod $4$ we $\\sigma(2)$ and $a_1$ of the same parity. Using this we let $\\sigma(2)=a_1+2a_2$ and in general we get the recurrence as $\\sigma(k)=a_{k-1}+2a_k$ . \nSo \\[S=(-2)^{n-1}a_n\\]\nSince \\[\\frac{n(n+1)}{2}=\\sum_{i=1}^{n}i=\\sum_{i=1}^{n}\\sigma(i)=3\\sum_{i=1}^na_i\\] we see that $3\\mid n(n+1)$ Motivated by this we do casework mod $3$ \n\n\n- For $n=3k$ \nBase case $(2,1)$ works then just induct to get the sequence to be $(2,3,5,6,\\dotsc, 3k-3,3k-1, 3k, 3k-2, 3k-4,\\dotsc, 1)$ .\n- For $n\\equiv 1 \\pmod{3}$ it is easy to see that \\[\\sum_{i=1}^n \\sigma(i)(-2)^{i-1}=\\sum_{i=1}^n\\sigma(i)\\not\\equiv0\\pmod{3}\\] so no permutation in this case\n- For $n=3k+2:$ Base case $(2,3,1)$ works then again induct to get the sequence $(2,3,5,\\dotsc,3k,3k+2,3k+1,3k-2,\\dotsc,1)$ .\n"
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"origin:aops",
"2022 Contests",
"2022 India National Olympiad"
] |
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"answer_score": 176,
"boxed": false,
"end_of_proof": false,
"n_reply": 17,
"path": "Contest Collections/2022 Contests/2022 India National Olympiad/2795409.json"
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Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$ . Let the circumcircle of triangle $ADB$ intersect $AC$ again at $E(\ne A)$ and the circumcircle of triangle $ADC$ intersect $AB$ again at $F(\ne A)$ . Let $AD$ , $BE$ , and $CF$ intersect the circumcircle of triangle $ABC$ again at $D_1(\ne A)$ , $E_1(\ne B)$ and $F_1(\ne C)$ , respectively. Let $I$ and $I_1$ be the incentres of triangles $DEF$ and $D_1E_1F_1$ , respectively. Prove that $E,F, I, I_1$ are concyclic.
|
Guys I found a solution heavily using the properties of miquel point !
<blockquote>Let $D$ be an interior point on the side $BC$ of an acute-angled triangle $ABC$ . Let the circumcircle of triangle $ADB$ intersect $AC$ again at $E(\ne A)$ and the circumcircle of triangle $ADC$ intersect $AB$ again at $F(\ne A)$ . Let $AD$ , $BE$ , and $CF$ intersect the circumcircle of triangle $ABC$ again at $D_1(\ne A)$ , $E_1(\ne B)$ and $F_1(\ne C)$ , respectively. Let $I$ and $I_1$ be the incentres of triangles $DEF$ and $D_1E_1F_1$ , respectively. Prove that $E,F, I, I_1$ are concyclic.</blockquote>
First note that $I_1 \equiv BE \cap CF$ by a simple angle chase.
Now Consider the complete quadrilateral, $BFAECI_1$ clearly $D(\odot(\triangle ABE)\cap \odot(\triangle ACF))$ is its miquel point.
Since $D \in BC$ we get that quadrilateral $AEI_1F$ must be cyclic.Now its well known that if $O$ is the centre of $\odot(AEI_1F)$ , $D \in \odot(\triangle OEF)$ (Check E.G.M.O proposition $10.14$ ) and $OD$ bisects $\angle EDF$ (E.G.M.O proposition $10.15$ ).
So by the well known Fact 5/Incentre-Excentre Lemma we get that $I \equiv OD \cap \odot(AEI_1F)$ .The End $\blacksquare$
|
[
"<details><summary>Thanks INMO :)</summary>\n\n- Angle chase to show $BE_1$ and $CF_1$ are angle bisectors in $D_1E_1F_1$ .\n- Chase more to show $\\angle EDF = \\angle E_1D_1F_1$ (infact the corresponding lines are parallel).\n- Apply $\\angle BIC = 90 + \\angle A/2$ in $DEF$ , $D_1E_1F_1$ to finish.\n</details>\nAlso first irl contest solve but fuzzed up ioqm :D",
"I showed that $A$ lies on $(EFII_1)$ , too :D",
"Clearly $I_1 = BE_1 \\cap BF_1$ because $BE_1$ is the angle bisector of $\\angle F_1E_1D_1$ as $$ \\angle FE_1B = \\angle F_1CB = \\angle FCD = \\angle FAD = \\angle BAD_1 = \\angle BE_1D_1 $$ and similarly $CF_1$ is the angle bisector of $\\angle D_1F_1E_1$ . $\\textbf{Lemma 1}$ \nIn any $\\triangle XYZ$ with incenter $J$ , $\\angle YJZ = 90^{\\circ}+\\frac{\\angle YXZ}{2}$ . $\\textbf{Proof}$ \nOmitted, just angle chase. $\\blacksquare$ \n\nNow, notice that $$ \\angle EDF = \\angle EDA + \\angle ADF = \\angle FCA + \\angle EBA = \\angle F_1CA + \\angle ABE_1 = \\angle E_1D_1F_1 $$ using the two given cyclites, then using that $I_1 = FF_1 \\cap EE_1$ and $\\textbf{Lemma 1}$ , $$ \\angle EI_1F = \\angle E_1I_1F_1 = 90^{\\circ} + \\frac{\\angle E_1D_1F_1}{2} = 90^{\\circ} + \\frac{\\angle EDF}{2} = \\angle EIF $$ giving the desired cyclicity. $\\blacksquare$ Edit (regarding configuration issues pointed out below):\nYou can observe that both $I_1, I$ lie inside $\\triangle DEF$ as $D$ lies in the interior of $BC$ , I did not include it previously because it's an AoPS post.",
"<blockquote>Clearly $I_1 = BE_1 \\cap BF_1$ because $BE_1$ is the angle bisector of $\\angle F_1E_1D_1$ as $$ \\angle FE_1B = \\angle F_1CB = \\angle FCD = \\angle FAD = \\angle BAD_1 = \\angle BE_1D_1 $$ and similarly $CF_1$ is the angle bisector of $\\angle D_1F_1E_1$ . $\\textbf{Lemma 1}$ \nIn any $\\triangle XYZ$ with incenter $J$ , $\\angle YJZ = 90^{\\circ}+\\frac{\\angle YXZ}{2}$ . $\\textbf{Proof}$ \nOmitted, just angle chase. $\\blacksquare$ \n\nNow, notice that $$ \\angle EDF = \\angle EDA + \\angle ADF = \\angle FCA + \\angle EBA = \\angle F_1CA + \\angle ABE_1 = \\angle E_1D_1F_1 $$ using the two given cyclites, then using that $I_1 = FF_1 \\cap EE_1$ and $\\textbf{Lemma 1}$ , $$ \\angle EI_1F = \\angle E_1I_1F_1 = 90^{\\circ} + \\frac{\\angle E_1D_1F_1}{2} = 90^{\\circ} + \\frac{\\angle EDF}{2} = \\angle EIF $$ giving the desired cyclicity. $\\blacksquare$ </blockquote>\n\nig there might pop up some configuration issue, so its better to show that $A$ lies on $\\odot EFII_1$ , which is just another bit of trivial angle chase. my proof was exactly identical to bora's proof, but i just added the last fact, anyways my solution is not even going to be checked and will be thrown like any other random piece of paper, ig ill stop doin MO and concentrate IOQM level bashy sums first",
"Anyone proved the similarity part using homothety?",
"*Walkthrough.*\n\n\n- $DE || D_1 E_1$ and $DF || D_1 F_1$\n- $E_1E$ and $F_1F$ are bisectors of $\\triangle D_1E_1F_1$\n- Complete using $\\angle{QIR} = \\frac{\\pi}{2} + \\frac{\\angle{QPR}}{2}$ when $I$ is incenter of $\\triangle PQR$\n",
"<blockquote>Anyone proved the similarity part using homothety?</blockquote>\n\nDEF is just made larger to match up with D1E1F1. Use angle chasing and parallel lines to prove it.",
"Easy problem it can be solving with only using angel change and prove $A,E,I_1,I,F$ ",
"This was just simple angle chase ",
"<blockquote>Anyone proved the similarity part using homothety?</blockquote>\n\nThe triangles are not similar. Two sides are parallel but the third side may or may not be.\n",
"2nd IRL Contest solve after STEMS P5 :\")\n\nHere's a proof without parallelism.**Claim :** $BE \\cap CF \\equiv I_1$ **Proof**Note that, $$ \\angle BE_1F_1=\\angle BCF_1=\\angle DCF=\\angle DAF=\\angle D_1AB=\\angle BE_1D_1 \\implies \\text{BE bisects } \\angle D_1E_1F_1 $$ Similarly $CF$ is another angle bisector ,so the claim is proved $\\blacksquare$ We observe that $\\angle I_1EA+\\angle I_1FA=\\angle BEA+\\angle CFA=\\angle BDA+\\angle CDA=180 \\implies I_1 \\in \\odot(\\triangle AEF)$ [This part was motivated by ELMOSL 2013 G3]\nNow,observe \n\\begin{align*}\\angle EDF &=\\angle EDA+\\angle FDA \n&=\\angle EBA+\\angle FCA \n&=180-(\\angle A+\\angle AEB)+180-(\\angle A +\\angle CFA)\n&=360-(2\\angle A+\\angle CFA+\\angle BEA)\n&=360-(180+2\\angle A)\n&=180-2\\angle A \\end{align*}\nFinally $\\angle EIF=90+\\frac{\\angle EDF}{2}=90+\\frac{180-2\\angle A}{2}=180- \\angle A \\implies I \\in \\odot(\\triangle AEF)$ Thus as $I,I_1 \\in \\odot(\\triangle AEF)$ we are done $\\blacksquare$ [This was my solution in the contest where I did find out the parallel lines but ended up not using them.]",
"nice problem it was main thing to see was to see that ff1 is the bisector which can be proved easily by angle chase and e is the miquel point.\n\n ",
"Honestly learning so much geo didn’t pay off, this was trivial, probably the easiest inmo geo in the recent years\nanyways my sol is the same as all above and i left the fact the A also lies on this circle as a remark which might help :D also directed angles which might help with configuration issues\n@below No because this might still be a nice problem but it isn't hard but ok if u say so\nALSO WHAT MAKAR IS VIEWING THIS THREAD ORZ ORZ ORZ SIR",
"Sorry, but can we stop calling it trivial? I spent all my time on this and I think it was a nice problem.",
"<blockquote><blockquote>Anyone proved the similarity part using homothety?</blockquote>\n\nThe triangles are not similar. Two sides are parallel but the third side may or may not be.</blockquote>\n\nOh... I fudged up",
"Is there another contest or does it just have 3 pbs this year? ",
"<blockquote>Is there another contest or does it just have 3 pbs this year?</blockquote>\n\nThis year INMO had 3 problems and we had 2.5 hours to attempt them, mainly because of covid",
"I did the parallel proof and showed angle D1=D. However, I did not have time to prove angle F1I1E1= angle FI1E1. I assumed that. How much may I lose for not showing this proof?",
"<blockquote>Sorry, but can we stop calling it trivial? I spent all my time on this and I think it was a nice problem.</blockquote>\n\nEven I also spent 1.5 hours to get everything in place and write it completely and I must say it was a very nice problem (at least for me). If a problem has a short solution or solution using simple stuffs doesn't mean it is easy... observing stuffs which makes a problem trivial are sometimes tough. Btw congratulations for solving it!!",
"<blockquote><blockquote>Sorry, but can we stop calling it trivial? I spent all my time on this and I think it was a nice problem.</blockquote>\n\nEven I also spent 1.5 hours to get everything in place and write it completely and I must say it was a very nice problem (at least for me). If a problem has a short solution or solution using simple stuffs doesn't mean it is easy... observing stuffs which makes a problem trivial are sometimes tough. Btw congratulations for solving it!!</blockquote>\n\nAbsolutely !",
"All angles are directed modulo $180^\\circ$ when we use $\\measuredangle$ . We directly give a super awesome claim which is just enough to annihilate the problem.**<span style=\"color:red\">Claim:**</span> $DF \\parallel D_1F_1$ and $D_1E_1 \\parallel DE$ and $I_1$ is the intersection of $CF_1$ and $BE_1$ .\n\n*Proof:* This will be angle chase\n\\[\\measuredangle F_1D_1A = \\measuredangle F_1CA = \\measuredangle FCA = \\measuredangle FDA \\implies DF \\parallel D_1F_1\\]\nOne can analogously chase to prove that other pair of lines are parallel as well. Now, we show that $F_1C$ bisects angle $\\angle D_1F_1E_1$ . Just see that \n\\[\\measuredangle CF_1E_1 = \\measuredangle CBE_1 = \\measuredangle DBE =\\measuredangle DAE = \\measuredangle D_1AC = \\measuredangle D_1F_1C\\]\nand the other one follows symmetrically. $\\square$ ------------------------------\nWith the above claim, we can deduce that $\\angle F_1D_1E = \\angle FDE$ . By a well-known lemma, we know that\n\\[\\angle F_1I_1E_1 = \\angle FI_1E = \\angle 90^\\circ + \\frac{\\angle F_1D_1E_1}{2} = \\angle FIE \\implies \\angle EI_1F = \\angle EIF\\]\nand thus we are done. $\\blacksquare$ ",
"We first prove that $AEIF$ is concyclic. Notice that $\\angle FIE =90°+\\dfrac{1}{2}\\angle FDE=90°+\\dfrac{1}{2}(\\angle BDE -\\angle BDF)= 90°+\\dfrac{1}{2}(180° - \\angle BAC -\\angle BAC)=180° -\\angle BAC$ Now $\\angle AF_1C =\\angle AD_1C$ and $\\angle F_1FA =180°-\\angle CFA =180°-\\angle CDA=\\angle D_1DA$ implies that $\\triangle AF_1F$ and $\\triangle CD_1D$ are similar. So $$ \\angle F_1 AB= \\angle BCD_1=\\angle BAD_1 $$ So B is the midpoint of arc $F_1D_1$ that doesn't contain $E_1$ . So $BE_1$ is the angle bisector of $\\angle F_1E_1D_1$ . Similarly $CF_1$ is the angle bisector of $\\angle E_1F_1D_1$ .So, $B,I_1,E$ and $E_1$ lie in a line. It's easy to see that $\\angle F_1AE_1=2\\angle BAC$ due to the bisectors. So, $\\angle FI_1E= 90°+\\dfrac{1}{2}\\angle F_1D_1E_1= 90°+\\dfrac{1}{2}( 180°-\\angle F_1AE_1)=90°+\\dfrac{1}{2}(180°-2\\angle BAC)=180°-\\angle BAC$ Therfore $AFI_1E$ are concyclic and thus I lies in the circle as proved before and the claim is proved. $\\blacksquare$ ",
"Check the video solution at \nhttps://www.youtube.com/watch?v=Bi7BVaIRlVQ",
"Well for an hour I just thought $D$ is inside of $ABC$ :D $\\angle D'F'C = \\angle D'AC = \\angle DAE= \\angle DBE = \\angle CDE' = \\angle E'F'C \\implies CF'$ is angle bisector of $\\angle E'F'D'$ . $\\angle D'E'B = \\angle D'AB = \\angle DAF = \\angle DCF = \\angle BCF' = \\angle F'E'B \\implies BE'$ is angle bisector of $\\angle F'E'D'$ . $\\angle FI'E = \\angle F'I'E' = \\angle 180 - \\angle I'E'F' - \\angle I'F'E' = \\angle 180 - \\angle A$ . $\\angle FIE = \\angle 90 + \\frac{\\angle FDE}{2} = \\angle 90 + \\angle FCA + \\angle EBA = \\angle 90 + \\frac{\\angle 180 - \\angle 2A}{2} = \\angle 180 - \\angle A$ .\nNow we have $\\angle FI'E = \\angle FIE$ which implies $II'E'F'$ is cyclic.",
"anyone with inversion at A? :maybe: ",
"@above, could you share",
"Quite a nice problem.. It is very easy to get intimidated by the sheer amount of lines and absurd sounding points defined in the diagram..\n",
"<blockquote>Well for an hour I just thought $D$ is inside of $ABC$ </blockquote>\n\nwhoa same\n\nObserve that $\\angle EDA=\\angle EBA=\\angle E_1BA=\\angle E_1D_1A$ , and likewise $\\angle FDA=\\angle F_1D_1A$ , so $\\angle EDF=\\angle E_1D_1F_1 \\implies \\angle EIF=\\angle E_1I_1F_1$ . Thus it suffices to show that $E,E_1,I_1$ and $F,F_1,I_1$ are collinear. By symnmetry, it suffices to show that $\\overline{E_1C}$ bisects $\\angle F_1E_1D_1$ . But this follows since $\\angle F_1E_1C=\\angle EBD=\\angle EAD=\\angle CE_1D_1$ . $\\blacksquare$ no i dont care about config issues",
"Just notice that $\\angle F_1E_1B = \\angle F_1CB = \\angle BAD_1 = \\angle BE_1D_1$ , hence $I_1 = \\overline{BE_1} \\cap \\overline{CF_1}$ . Then $\\angle EI_1F = \\angle EIF$ as $\\overline{FD} \\parallel \\overline{F_1D_1}$ by Reim's theorem and similar.",
"The key claim is that $AEFII_1$ is cyclic.\n\n\n- To show $I \\in (AEF)$ , notice $\\angle FDB = \\angle EDC = \\angle A$ , so\n\\[\\angle EIF = 90 + \\frac{\\angle EDF}{2} = 90 + \\frac{180-2\\angle A}{2} = 180-\\angle A.\\]\n- The angle bisector of $\\angle D_1E_1F_1$ is $E_1B$ , as\n\\[\\angle D_1E_1B = \\angle DAF = \\angle DCF = \\angle BE_1F_1.\\]\nSimilarily, the angle bisector $\\angle D_1F_1E_1$ is $F_1C$ , making $I_1 = BE \\cap CF$ . From here, it's easy to find $AEI_1F$ cyclic, as desired. $\\blacksquare$\n",
"<details><summary>storage</summary>By angle chase, we can show that $BE_1$ and $CF_1$ are angle bisectors in $\\Delta D_1E_1F_1$ and that $\\angle E_1D_1F_1= \\angle EDF$ and now it is trivial to see the required result.</details>\n\n<details><summary>Remark</summary>The only remarkable thing about this problem, for me, is that it was the first subjective geo i tried in a contest and that it is my 69th post lel. Otherwise solved in a passport sized's photo's worth of page.</details>",
"Let $I_{1}^{'} $ be the intersection of $BE_1$ and $CF_1$ By quick angle chasing we get that triangles $\\triangle DBF, \\triangle DEC$ are simmilar $\\iff \\triangle DBF \\sim \\triangle DEC \\implies \\frac{DB}{DE}=\\frac{DF}{DC}$ combining with $\\angle BDE=\\angle CDE$ we get that triangles $\\triangle DBE , \\triangle DFC$ are also simmilar $\\iff \\triangle DBE \\sim \\triangle DFC$ So now we angle chase and we get that $E_1I_{1}^{'}$ is the angle bisector of $\\angle F_1E_1D_1$ we also get that $F_1I_{1}^{'} $ is the angle bisector of $\\angle F_1E_1D_1$ so $I_{1}^{'}=I_1$ hence $BE_1 \\cap CF_1 \\cap AD_1 = \\{I_1\\}$ Also know by $\\triangle DBE \\sim \\triangle DFC$ we get that quadrilaterals $BDF_1I_1$ and $CDEI_1$ are cyclic $\\implies$ By Miquel Theorem that the quadrilateral $AFEI_1$ is also cyclic\n\nWe also have that $\\angle FIE=90 + \\frac{\\angle FDE}{2}=90+\\frac{180-\\angle FDB-\\angle EDC}{2}=90+\\frac{180-2 \\cdot \\angle A}{2}=180-\\angle A \\implies \\angle FIE=180- \\angle A$ which means that $AFIE$ is cyclic hence $I \\in \\odot (AFEI_1) \\implies EFII_1$ is cyclic\n\n*My 70th post :bye:*",
"*Can't believe in $2022$ you solve this problem and you are $INMO$ $AWARDEE$* $\\angle FCA = \\angle FDA$ but $\\angle FCA = \\angle F_1CA = \\angle F_1DA$ implies $\\angle FDA = \\angle F_1D_1A$ \nSimilarly $\\angle EDA = \\angle E_1D_1A$ \nCombining both, we get $\\angle FDE = \\angle F_1D_1E_1$ \nNow, $$ \\angle EI_1F = \\angle E_1I_1F_1 = 90^{\\circ} + \\frac{\\angle E_1D_1F_1}{2} = 90^{\\circ} + \\frac{\\angle EDF}{2} = \\angle EIF $$ Hence proved!",
"First, observe that $\\angle BE_1D_1 = \\angle BAD = \\angle F_1CB = \\angle F_1E_1B,$ so $BE_1$ bisects $\\angle F_1E_1D_1.$ \nThus, $I_1=BE_1\\cap CF_1.$ We get $\\angle EIF = 90 - \\frac{\\angle EDF}2 = 90 + \\frac{180 - 2\\angle A}2 = 180-\\angle A = 180 - \\angle BAD -\\angle DAC = 180 - \\angle I_1BC-\\angle I_1CB = \\angle EI_1B,$ which finishes.",
"Main claims: \n1) AEFI cyclic\n2) I_1 = BE cap CF\n3) AEI_1F cyclic\nand we're done!!\n[asy]\nimport graph; size(28.259587460059326cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.647195849593462,xmax=24.612391610465863,ymin=-7.509267023422269,ymax=5.573532154808459; \npen qqwuqq=rgb(0.,0.39215686274509803,0.); \npair A=(7.04,2.68), B=(5.56,-2.08), C=(11.84,-1.9), D=(8.406260425428417,-1.9984192871692494), F=(6.625518995492593,1.3469394719896899), E_1=(11.51418510673573,1.598202569865462), F_1=(5.933180208263576,1.7780431876423513), D_1=(8.968848272394865,-3.9248615561455518), I=(8.352856644003863,-0.13522069080146587), I_1=(8.857582314416177,-0.04291591111145887); \ndraw(A--B--C--cycle,linewidth(2.)+red); draw(D_1--E_1--F_1--cycle,linewidth(2.)+qqwuqq); draw(D--(9.486520842231533,0.34561136303741335)--F--cycle,linewidth(2.)+blue); \ndraw(A--B,linewidth(2.)+red); draw(B--C,linewidth(2.)+red); draw(C--A,linewidth(2.)+red); draw(circle((6.9216226068641475,0.10672238273971847),2.575999010616734),linewidth(2.)); draw(circle((8.655353603542784,-0.4323368347149838),3.5065663885348504),linewidth(2.)); draw(circle((10.03811586962979,1.0168463262495604),3.4285243050378944),linewidth(2.)); draw(D_1--E_1,linewidth(2.)+qqwuqq); draw(E_1--F_1,linewidth(2.)+qqwuqq); draw(F_1--D_1,linewidth(2.)+qqwuqq); draw(circle((8.30438487295115,1.5559055437042604),1.6918207510320091),linewidth(2.)+linetype(\"4 4\")); draw(D--(9.486520842231533,0.34561136303741335),linewidth(2.)+blue); draw((9.486520842231533,0.34561136303741335)--F,linewidth(2.)+blue); draw(F--D,linewidth(2.)+blue); draw(B--E_1,linewidth(2.)); draw(C--F_1,linewidth(2.)); draw(A--D_1,linewidth(2.)); \ndot(A,ds); label(\" $A$ \",(7.119243700763566,2.8680241094519623),NE*lsf); dot(B,ds); label(\" $B$ \",(5.451464768694492,-2.5429919812610295),NE*lsf); dot(C,ds); label(\" $C$ \",(11.714901202465018,-1.709102515226493),NE*lsf); dot(D,ds); label(\" $D$ \",(8.24962719916594,-2.48739935019206),NE*lsf); dot((9.486520842231533,0.34561136303741335),linewidth(4.pt)+ds); label(\" $E$ \",(9.63944297589017,0.19957781814144593),NE*lsf); dot(F,linewidth(4.pt)+ds); label(\" $F$ \",(6.693033529234802,1.4967392097507248),NE*lsf); dot(E_1,linewidth(4.pt)+ds); label(\" $E_1$ \",(11.51106155521213,1.9600111353254672),NE*lsf); dot(F_1,linewidth(4.pt)+ds); label(\" $F_1$ \",(5.710897047016347,2.0156037663944364),NE*lsf); dot(D_1,linewidth(4.pt)+ds); label(\" $D_1$ \",(9.102047542223467,-4.247832667376081),NE*lsf); dot(I,linewidth(4.pt)+ds); label(\" $I$ \",(8.286688953211918,-0.02279270613443046),NE*lsf); dot(I_1,linewidth(4.pt)+ds); label(\" $I_1$ \",(8.8611461409246,0.1810469411184562),NE*lsf); \nclip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); \n[/asy]",
"Noice! Grinding the art school unit in OTIS right now! :)\n\n<span style=\"color:#f00\"> Claim - </span> $A F E I_1$ are concyclic $\\angle F D B = a = \\angle E D C$ and $\\angle E D F = 180 - 2a $ and since $\\angle F I_1 E = 90 + \\angle F D E/2$ we have $\\angle F I_1 E = 180 - a$ as required.\n\n<span style=\"color:#f00\"> Claim - </span> $F C$ and $B E$ intersect at $I_2$ Just note that $\\angle C F_1 E_1 = \\angle C B E_1 = \\angle D_1 A C = \\angle D_1 F_1 C$ so, $F_1 C$ is the angle bisector, similarly, $B E_1$ is the other angle bisector and they intersect at $I_2$ as required.\n\n<span style=\"color:#f00\"> Claim - </span> $I_2$ lies on the circumcircle of $A F E$ just note that $\\angle F I_2 E = 90 + \\angle F_1 D_1 E_1/2 = 180 - a$ $\\square$ :starwars:",
" By Reim's Theorem, $DF\\parallel D_1F_1$ and $DE\\parallel D_1E_1$ . This means $\\angle EDF=\\angle E_1D_1F_1$ and so $\\angle EIF=\\angle EI_1F$ as desired. ",
"Notice that $\\angle F_1E_1B=\\angle F_1CB=\\angle FCD=\\angle FAD=\\angle BAD_1=\\angle BE_1D_1$ By symmetry, $BE_1\\cap CF_1=EE_1\\cap FF_1=I_1$ Finally, \\[\\angle EIF=90+\\frac{\\angle EDF}{2}=90+\\frac{\\angle EDA+\\angle ADF}{2}=90+\\frac{\\angle EBA+\\angle ACF}{2}=90+\\frac{\\angle E_1BA+\\angle ACF_1}{2}=90+\\frac{\\angle E_1D_1A+\\angle AD_1F_1}{2}=90+\\frac{\\angle E_1D_1F_1}{2}=\\angle EI_1F\\]\nSo $EII_1F$ are concyclic as desired. $\\square$ ",
"We begin with a claim:\n\n-----**<span style=\"color:#f00\">Claim:</span>** $I_1$ lies on $\\overline{BE}$ and $\\overline{CF}$ *Proof:* We show that $\\overline{BE}$ bisects $\\angle D_1E_1F_1$ , and the other part will follow analogously. Simple angle chasing yields\n\n\\begin{align*}\n\\angle BE_1D_1 = \\angle BAD_1 &= \\angle FAD \n&= \\angle FCD = \\angle F_1CB = \\angle F_1E_1B. \\ \\square\n\\end{align*}\n\n-----\n\nAlso, we have\n\n\\begin{align*}\n\\angle EDF &= \\angle ADE + \\angle ADF \n&= \\angle ABE + \\angle ACF \n&= \\angle ABE_1 + \\angle ACF_1 \n&= \\angle AD_1E_1 + \\angle AD_1F_1 = \\angle E_1D_1F_1.\n\\end{align*}\n\nFinally, we can put everything together to get\n\n\\[\\angle EIF = 90^\\circ + \\frac{1}{2} \\angle EDF = 90^\\circ + \\frac{1}{2}\\angle E_1D_1F_1 = \\angle E_1I_1F_1 = \\angle EI_1F,\\]\n\nimplying that $E,F,I,I_1$ are concyclic, as desired. $\\blacksquare$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 India National Olympiad"
] |
{
"answer_score": 170,
"boxed": false,
"end_of_proof": true,
"n_reply": 40,
"path": "Contest Collections/2022 Contests/2022 India National Olympiad/2795412.json"
}
|
For a positive integer $N$ , let $T(N)$ denote the number of arrangements of the integers $1, 2, \cdots N$ into a sequence $a_1, a_2, \cdots a_N$ such that $a_i > a_{2i}$ for all $i$ , $1 \le i < 2i \le N$ and $a_i > a_{2i+1}$ for all $i$ , $1 \le i < 2i+1 \le N$ . For example, $T(3)$ is $2$ , since the possible arrangements are $321$ and $312$ (a) Find $T(7)$ (b) If $K$ is the largest non-negative integer so that $2^K$ divides $T(2^n - 1)$ , show that $K = 2^n - n - 1$ .
(c) Find the largest non-negative integer $K$ so that $2^K$ divides $T(2^n + 1)$
|
For the sake of completeness, here's a full solution.
Consider a graph with vertices as $1,2,\cdots,N$ and draw an arrow $i \rightarrow j$ if $j = 2i$ or $j = 2i+1$ . For $N = 2^n - 1$ , this actually is a binary tree with $n-1$ levels with all numbers in the range $[2^k, 2^{k+1})$ on the $k$ th level.
Clearly, we must have $a_1$ to be greater than any of the others and so $a_1 = 2^n - 1$ . Note that to arrange the $a_i$ , only the relative order matters, and not the fact that the numbers are $1,2,\cdots, N$ . So we choose $2^{n-1} - 1$ numbers for each subtree, which can be done in $\binom{2^n - 2}{2^{n-1} - 1}$ ways. Now, each the numbers of each subtree can be arranged in $T(2^{n-1} - 1)$ ways, so we have the recurrence $$ \boxed{T(2^n - 1) = \binom{2^n - 2}{2^{n-1} - 1}T(2^{n-1} - 1)^2} $$ Let $v(N) = v_2(T(N))$ , so we have $v(2^n - 1) = 2v(2^{n-1} -1) + v_2 \left(\binom{2^n - 2}{2^{n-1} - 1} \right)$ But we have $v_2(N!) = N - s_2(N)$ so the second term is just $2s_2(2^{n-1} - 1) - s_2(2^n - 2) = 2(n-1) - (n-1) = n-1$ , so we have $v(2^n - 1) = 2v(2^{n-1} - 1) + n-1 = 2(2^{n-1} - (n-1) - 1) + n-2 = 2^n - n - 1$ , by induction (base case being $n = 2$ ), so part (b) done.
Now consider $N = 2^n +1$ , this is the same as the previous case except we have two edges extra on one subtree, but the subtrees have sizes $2^{n-1} - 1$ and $2^{n-1} + 1$ so recursing again seems like it would work. In this case, once again, $a_1 = N$ and we choose numbers in $\binom{2^n}{2^{n-1} - 1}$ ways and arrange them in $T(2^{n-1}-1)$ and $T(2^{n-1}+1)$ ways respectively, so we have $$ \boxed{T(2^n + 1) = \binom{2^n}{2^{n-1}-1} T(2^{n-1} -1)T(2^{n-1}+1)} $$ We have $v_2$ of the binomial in this case is $s_2(2^{n-1}+1) + s(2^{n-1}-1) - s(2^n) = 2 + n-1 - 1 = n$ and using the previous result we had, we have $v(2^n + 1) = n + (2^{n-1} - (n-1) - 1) + T(2^{n-1} + 1) = 2^{n-1} + T(2^{n-1} + 1)$ , I claim $T(2^n +1) = 2^n - 1$ and this just follows by induction with base case $n = 1$ and using the previous line., that finishes part (c).
Finally for part (a), we have $T(7) = \binom{6}{3} T(3)^2 = 20(2)^2 = 80$ , done. $\blacksquare$
|
[
"Legendre Formula , and Counting works... I am not sure that my solution works , but I got a general formula for $\\nu _{2} ( T(2k+1))$ .. ( and this was the only problem I could solve in Test :stink:",
" $T(7)=80$ . Could not formally write the complete proof of second part but it required to be deal with powers. \n\n@below I did the same thing for the part 2 but handling the combinatorial calculations was kinda the tricky part. How did you proceed after it like \n\n\\[\\frac{2^n(2^n-3)(2^n-5)\\cdots}{(2^{n-1}-1)!}\\]",
"I got $T(2^n-1)=\\binom{2^n-1}{2^{n-1}-1}(T(2^{n-1}-1})^2$ iirc",
"Even I had the same subtree logic, got the same recurrence , unfortunately was unable to write it in a detailed manner because I got the idea in the last 5 minutes :wallbash: ",
"T(7) is 80 and thats all i know ;-;\n\nand i fakesolved (b) i think smh",
"I used induction to prove (b).\n\nBasically, the same as above. I utilized the perfect binary tree structure for (b) to prove the recurrence.\n\nThe largest value had to be at the root of the tree and the two subtrees are independent of each other.\n\nProblem (c) involved using (b)\n\nHere is my recurrence formula for (c): $$ g(n) = \\binom{2 ^ n}{2 ^ {n - 1} + 1} \\cdot g(n - 1) \\cdot f(n - 1) $$ where $f(n) = T(2 ^ n - 1)$ and $g(n) = T(2 ^ n + 1)$ . Now use legendre's (just like (b)) and you get the difference of two adjacent elements.\n\nLet $h(x)$ denote the maximal power of $2$ dividing $x$ $$ h(g(n)) = h(\\binom{2 ^ n}{2 ^ {n - 1} + 1}) + h(g(n - 1)) + h(f(n - 1)) $$ We now get $$ h(g(n)) - h(g(n - 1)) = h((2 ^ n)!) - h((2 ^ {n - 1} - 1)!) - h((2 ^ {n - 1} + 1)!) + h(f(n - 1)) $$ $$ h(g(n)) - h(g(n - 1)) = 2 ^ {n - 1} $$ $$ h(g(n)) - (h(g(n - 2)) + 2 ^ {n - 2} - 1) = 2 ^ {n - 1} $$ $$ h(g(n)) - h(g(n - 2)) = 2 ^ {n - 2} + 2 ^ {n - 1} $$ $$ h(g(n)) - h(g(1)) = 2 ^ {1} + \\cdots + 2 ^ {n - 1} $$ $$ h(g(n)) - h(g(1)) = 2 ^ n - 2 $$ $$ h(g(n)) = 2 ^ n - 1 $$ ",
"<blockquote> $T(7)=80$ . Could not formally write the complete proof of second part but it required to be deal with powers. \n\n@below I did the same thing for the part 2 but handling the combinatorial calculations was kinda the tricky part. How did you proceed after it like \n\n\\[\\frac{2^n(2^n-3)(2^n-5)\\cdots}{(2^{n-1}-1)!}\\]</blockquote>\n\nUse Legendre's formula for maximal power of 2 which divides the numerator and denominator, no need to expand the factorials. ~~Expanding the factorials is soo JEE-like~~",
"<blockquote>~~Expanding the factorials is soo JEE-like~~</blockquote>\nIs this a personal attack?\n\n",
"I am getting the same recursion but $2^n-1$ instead of $2^n-n$ for c)\n\nIf $f(n)=\\nu_2(T(2^n+1))$ then from $T(2^n+1)=\\binom{2^n}{2^{n-1}-1} T(2^{n-1}-1)T(2^{n-1}+1)$ we get $f(n)=n+f(n-1)+(2^{n-1}-n)$ , so $f(n)-f(n-1)=2^{n-1}$ I manually checked $n=2$ and think I am right.",
"<blockquote>I am getting the same recursion but $2^n-1$ instead of $2^n-n$ for c)\n\nIf $f(n)=\\nu_2(T(2^n+1))$ then from $T(2^n+1)=\\binom{2^n}{2^{n-1}-1} T(2^{n-1}-1)T(2^{n-1}+1)$ we get $f(n)=n+f(n-1)+(2^{n-1}-n)$ , so $f(n)-f(n-1)=2^{n-1}$ I manually checked $n=2$ and think I am right.</blockquote>\n\nI also got 2^n - 1 for part c",
"<blockquote>I am getting the same recursion but $2^n-1$ instead of $2^n-n$ for c)\n\nIf $f(n)=\\nu_2(T(2^n+1))$ then from $T(2^n+1)=\\binom{2^n}{2^{n-1}-1} T(2^{n-1}-1)T(2^{n-1}+1)$ we get $f(n)=n+f(n-1)+(2^{n-1}-n)$ , so $f(n)-f(n-1)=2^{n-1}$ I manually checked $n=2$ and think I am right.</blockquote>\n\nSorry if i made a calculation mistake. ~~Let me just recheck and edit.~~ Sorry it was a calculation mistake :blush:",
"<blockquote>I lose.\nI could only get parts (a) and (b). What I did was this: draw a digraph with vertices $a_1,\\ldots, a_N$ and draw an edge $a_i\\to a_j$ if $j=2i$ or $j=2i+1$ . Then note that all the vertices in the new layer are different. Using the self-repeating structure of the diagraph. I argued \\[T(2^{n+1}-1)=\\binom{2^{n+1}-2}{2^n-1}T(2^n-1)^2\\]</blockquote>\n\ni did the same\n\n",
"<blockquote><blockquote>~~Expanding the factorials is soo JEE-like~~</blockquote>\nIs this a personal attack?</blockquote>\n\nYes mental and emotional attack :(((",
"I did this by Binary representation, recurrence and Legendre formula $T(7)$ is easy enough to find. \nNow let us find $T(2^n-1)$ $a_1$ should be $2^n-1$ ${a_2,a_3,.... a_{2^n-1}}$ can be divided into sets $A_{10}$ and $A_{11}$ where $A_{10}$ contains numbers whose binary begins with a 10. Now there are $T(2^{n-1}-1)$ ways to arrange $A_{10}$ and similarly $A_{11}$ so we get $T(2^n-1) = \\binom{2^n-2}{2^{n-1}-1}\\cdot {T(2^{n-1}-1)}^2$ and induct. Even I couldnt do part 3, but I did understand the tree solution.",
"Btw, if my INMO paper is checked, will I be exempted from IOQM 2023 or PRMO 2023 if they decide to follow that pattern?",
"[Interesting](https://artofproblemsolving.com/community/c6h287200p1552270).",
"its pretty easy as compared to other two question\n",
"Anybody has any idea about the marks distribution for part (a),(b) and (c)?",
"My solution is same as (19)\na)80\nb)It is trivial by Induction.\nc)Here you get this recursion $$ \\boxed{T(2^n + 1) = \\binom{2^n}{2^{n-1}-1} T(2^{n-1} -1)T(2^{n-1}+1)} $$ Therefore $T(2^n + 1)$ = $2^n -1 $ ",
"Main idea: Obtain the recurrences\n\\[T(2^n-1) = \\binom{2^n-2}{2^{n-1}-1} T(2^{n-1}-1)^2,\\]\n\\[T(2^n+1) = \\binom{2^n}{2^{n-1}-1}T(2^{n-1}-1)T(2^{n-1}+1).\\]\nFrom here use Legendre's formula a bunch to compute the $\\nu_2$ of the binomial coefficients and induct."
] |
[
"origin:aops",
"2022 Contests",
"2022 India National Olympiad"
] |
{
"answer_score": 1178,
"boxed": false,
"end_of_proof": false,
"n_reply": 21,
"path": "Contest Collections/2022 Contests/2022 India National Olympiad/2795422.json"
}
|
Given a monic quadratic polynomial $Q(x)$ , define \[ Q_n (x) = \underbrace{Q(Q(\cdots(Q(x))\cdots))}_{\text{compose $n$ times}} \]
for every natural number $n$ . Let $a_n$ be the minimum value of the polynomial $Q_n(x)$ for every natural number $n$ . It is known that $a_n > 0$ for every natural number $n$ and there exists some natural number $k$ such that $a_k \neq a_{k+1}$ .
(a) Prove that $a_n < a_{n+1}$ for every natural number $n$ .
(b) Is it possible to satisfy $a_n < 2021$ for every natural number $n$ ?
*Proposed by Fajar Yuliawan*
|
<details><summary>Official Solution (by Fajar Yuliawan)</summary><details><summary>(a)</summary>Let $Q(a) = a_1$ , then $Q(x) = (x - a)^2 + a_1$ .
Claim 1: $a_1 > a$ Proof. Suppose (FTSOC) $a \geq a_1$ . Notice that $Q_1(a) = a_1$ . Assume that $Q_n(b_n) = a_1$ where $b_n \geq a$ for some $n \geq 1$ . Let $b_{n+1} = a + \sqrt{b_n - a_1} \geq a$ , hence $Q(b_{n+1}) = b_n$ and
\[ Q_{n+1} (b_{n+1}) = Q_n (Q(b_{n+1})) = Q_n(b_n) = a_1. \]
Since $Q_n(x) \geq a_1$ for every natural number $n$ , we obtain that $a_n = a_1$ for all $n$ , a contradiction!
Claim 2: $a_n = Q_n(a)$ Proof. For $n=1$ it's true since $a_1 = Q_1(a)$ . Assume that $a_n = Q_n(a)$ for some $n \geq 1$ . Thus, $Q_n(x) \geq Q_n(a) \geq a_1 > a$ , so
\[ Q_{n+1} (x) = Q(Q_n(x)) = (Q_n(x) - a)^2 + a_1 \geq (Q_n(a) - a)^2 + a_1 = Q_{n+1} (a). \] So $a_{n+1} = Q_{n+1} (a)$ . Claim 2 is proven by induction.
Claim 3: $a_{n+1} > a_n$ for every natural number $n$ .
Proof. Since $a_1 > a$ , then
\[ a_2 = Q_2(a) = Q(a_1) = (a_1 - a)^2 + a_1 > a_1 > a. \]
Suppose it holds that $a_{n+1} > a_n > a$ for some natural number $n$ , then
\[ a_{n+2} = Q(a_{n+1}) = (a_{n+1} - a)^2 + a_1 > (a_n - a)^2 + a_1 = a_{n+1} > a. \]
So claim 3 is proven by induction. $\blacksquare$</details>
<details><summary>(b)</summary>It is possible. Take $Q_1(x) = x^2 + \frac{1}{4}$ . Notice that $a_1 = \frac{1}{4}$ and
\[ a_{n+1} = Q_{n+1} (0) = Q(Q_n(0)) = Q(a_n) = a_n^2 + \frac{1}{4}. \]
Claim. $a_n < \frac{1}{2}$ for every $n \geq 1$ .
Proof. It obviously holds for $n = 1$ . If $a_n < \frac{1}{2}$ , then
\[ a_{n+1} = a_n^2 + \frac{1}{4} < \frac{1}{4} + \frac{1}{4} = \frac{1}{2}. \]
So the claim is proven by induction. $\blacksquare$</details></details>
Translated by yours truly, from Indonesian. :)
|
[
"Nice one. Who is the author?",
"We do this more generally for a strictly convex twice differentiable function $f$ with a single minimum in $x_0$ .\nWe classify $3$ possible behaviours for $a_n$ . $I$ ) $f(x)>x$ $\\forall x\\in\\mathbb{R}$ . Therefore we have $t=inf(im(f(x)-x))=min(im(f(x)-x))>0$ , or in other words $f(x)\\geq x+t$ $\\forall x\\in \\mathbb{R}$ . In this case $f_{n+1}(x)\\geq f_n(x)+t\\geq a_n+t$ , which implies $a_{n+1}\\geq a_n+t$ , and thus $a_n$ is a strictly increasing unbounded sequence. $II$ ) $f(x)$ has two points of intersection $x_1< x_2$ and $x_1\\leq x_0<x_2$ . Considering the function $f_n$ , we have $f_n(x_2)=2$ and $f_n(x_1)=x_1$ , and so by intermediate value theorem there is a $c$ such that $f_n(c)=x_0$ , and thus $f_{n+1}(x)= f(f_n(x))\\geq f(x_0)=a_1$ , and so $a_n$ is constant. $III$ ) $f(x)$ has two points (they may be coincident, which couldn't happen before) $x_1\\leq x_2$ , and $x_0<x_1$ . For $x<x_1$ we have $f(x)>x$ . Since $f_n'=f'(f_{n-1})\\cdots f'(x)$ , to achieve minimum we must have $f'(f_{k}(x))=0$ for some $k\\geq 0$ which implies $f_k(x)=x_0$ . However if $k>0$ we have $f_k(x)\\geq f(x_0)>x_0$ , and so the only minimum is always in $x_0$ . Now we note by induction that, since $f$ is increasing in $[x_0,x_1]$ and since $f(x)>x$ $\\forall x\\in [x_0,x_1)$ , $x_1=f_{n+1}(x_1)>f_{n+1}(x_0)>f_n(x_0)=a_n>a_{n-1}>\\cdots >a_1>x_0$ , and this $a_n$ is a strictly increasing bounded function, which by monotone convergence theorem must converge, and it must in fact converge to $x_1$ .\n\nNow we finally return to our problem:\nWe couldn't have case $I$ since $a_n$ is constant. In both cases $II$ and $III$ part a) was proved.\nFor part b) it suffices to take a function which satisfies the hypothesis $x_0<x_1\\leq x_2$ . For example, $Q(x)=(x-\\frac{1}{4})^2+0.01$ works since $a_1>0$ and $x_1<2021$ .",
"This example very nice for part (b) $$ x^2+\\frac{1}{4}x+\\frac{1}{8} $$ "
] |
[
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] |
{
"answer_score": 168,
"boxed": false,
"end_of_proof": false,
"n_reply": 4,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735502.json"
}
|
Five numbers are chosen from $\{1, 2, \ldots, n\}$ . Determine the largest $n$ such that we can always pick some of the 5 chosen numbers so that they can be made into two groups whose numbers have the same sum (a group may contain only one number).
|
I think,answer is a $$ n
\leq15 $$ {1,2,4,8,16} for $$ n\geq 16 $$
|
[
"<blockquote>I think,answer is a15 ${1,2,4,8,16}$ for $n>=16$ </blockquote>\n\nWrong:7,8,9,11,14",
"I think answer is 12\n6,9,11,12,13 for n >=13",
"Bump this ",
"Bump bump bump",
"bump bump\n",
"Soooo, this is actually a pretty nice problem. \nFirst we show that n>=13 is impossible. \nWe can easily check that 6,9,11,12,13 is impossible to be partitioned into 2 groups with the same sum. \nNow, if n<=12 \nLet's say our 5 numbers are a>b>c>d>e \nWe can imagine a b c d e and fitting in coefficients of either 1,-1, or 0 to represent partitioning it, if the sum is equal to 0 without all the terms being 0 then we are done. This is the main motivation for the following step. If we want to use PHP, we want to minimize the possibilities of the sum. And so a natural idea is to consider a-b+c-d+e and then putting in coefficients of 0. If any 2 distinct configurations have the same sum we can subtract both to get a configuration that's equal to 0, satisfying the problem requirement. \nNow, consider the 12 numbers\na\na-b\na-d\nc\nc-d\ne\na+c-b\na+c-b-d\na+e-b\na+e-d\nc+e-d\na+c+e-b-d\nAll of these numbers are bigger than 0 but less than 13, since there are 12 numbers with 12 possible values, if no 2 numbers are equal (In which cse subtract them and we are done) then all those numbers are a permutation of 1,2,3...12. Now the maximum of those values is a=12. \nNow there are 2 options, we can either case bash this to oblivion (Which is still somewhat viable, it will probably take some time but definitely doable in <=1,5 hours. But there's a nicer way to do this. If we sum all the numbers we will get \n8a-5b+6c-6d+5e=78\n6(c-d)+5(e-b)=-18 \nThen 6 | e-b which means b-e=6 and c-d=2. \nNow we are pretty much done, doing the rest is just trivial casework which requires probably less than 10 minutes to do. \n"
] |
[
"origin:aops",
"2022 Contests",
"2022 Indonesia TST"
] |
{
"answer_score": 8,
"boxed": false,
"end_of_proof": false,
"n_reply": 7,
"path": "Contest Collections/2022 Contests/2022 Indonesia TST/2735505.json"
}
|
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