Triangle $A_1B_1C_1$ is an equilateral triangle with sidelength $1$.  For each $n>1$, we construct triangle $A_nB_nC_n$ from $A_{n-1}B_{n-1}C_{n-1}$ according to the following rule: $A_n,B_n,C_n$ are points on segments $A_{n-1}B_{n-1},B_{n-1}C_{n-1},C_{n-1}A_{n-1}$ respectively, and satisfy the following: \[\dfrac{A_{n-1}A_n}{A_nB_{n-1}}=\dfrac{B_{n-1}B_n}{B_nC_{n-1}}=\dfrac{C_{n-1}C_n}{C_nA_{n-1}}=\dfrac1{n-1}\]  So for example, $A_2B_2C_2$ is formed by taking the midpoints of the sides of $A_1B_1C_1$.  Now, we can write $\tfrac{|A_5B_5C_5|}{|A_1B_1C_1|}=\tfrac mn$ where $m$ and $n$ are relatively prime integers.  Find $m+n$.  (For a triangle $\triangle ABC$, $|ABC|$ denotes its area.)

Let's think step by step and output the final answer within \boxed{}.
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