On the theorem of Pythagoras

For the theorem of Pythagoras, I start from Coxeter's formulation ("Introduction to Geometry", p.8)

"In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the two catheti."

Let us play a little bit with that formulation. In a triangle with sides a , b , and c —different from 0 so as to make its angles well-defined— we introdue the usual nomenclature α , β , and γ for their respective opposite angles. (We introduced one angle name so as to be able to express right-angledness, and the other two for reasons of symmetry.)

A formal expression of Coxeter's formulation is



γ = π/2 ⇒ a 2 + b 2 = c 2 .

Besides the nomenclature we introduced, this formulation contains the (transcendantal!) constant π. Fortunately we can eliminate it thanks to

π = α + β + γ .

α + β = γ ⇒ a 2 + b 2 = c 2 .

(0) α + β = γ ≡ a 2 + b 2 = c 2 .

α + β ≠ γ ≡ a 2 + b 2 ≠ c 2 .

Elementary arithmetic yields the equivalent formulationIsn't that nicely symmetric? It immediately suggests —at least to me— the strengthening(This will turn out to be a theorem.) We get an equivalent formulation by negating both sides:

But x ≠ y ≡ x < y ∨ x > y , and the latter disjuncts are mutually exclusive. Remembering that the larger angle is opposite to the larger side, is it bold to guess

α

β

γ

a

b

c

α

β

γ

a

b

c



Bold perhaps, but not unreasonable.

(1)and(2)

Note that (0), (1) and (2) are not independent: from any two of them, the third can be derived. They can be jointly formulated in terms of the function sgn —read "signum"— given by

x

x

x

x

α

β

γ

a

b

c

sgn.0 = 0 ∧ (sgn.=1 ≡> 0) ∧ (sgn.= -1 &equiv< 0) ,viz. sgn.() = sgn.() .

Consider now the following figure. We have drawn





the case α + β < γ , in which the triangles Δ CKB and Δ AHC , of disjoint areas, don't cover the whole of Δ ACB ; denoting the area of Δ XYZ by " XYZ " we have in this case

CKB + AHC < ACB .

In the case α + β = γ , H and K coincide and we have

CKB + AHC = ACB ,

α

β

γ

CKB + AHC > ACB .

sgn.( α + β - γ ) = sgn.( CKB + AHC - ACB ) .

and in the case, the two triangles overlap and we haveIn summary

The three areas of the right-hand side are those of similar triangles and hence have the same ratios as the squares of corresponding lines, in particular

CKB / a 2 = AHC / b 2 = ACB / c 2 > 0 ;

sgn.( CKB + AHC - ACB ) = sgn.( a 2 + b 2 - c 2) ,

henceHence we have proved

sgn.( α + β - γ ) = sgn.( a 2 + b 2 - c 2) ,





* *

*

a theorem, say, 4 times as rich as the one we quoted from Coxeter.

The title of this note could make one wonder why I would waste my time flogging a horse as dead as Pythagoras's Theorem. So let us try to summarize what we could learn from this exercise.

a

b

c

( α + β ) R γ ≡ ( a 2 + b 2) R c 2

R

α

β

γ

K

A

A

A

H

B

CKB + AHC < ACB ⇒ α + β < γ and

CKB + AHC = ACB ⇒ &alpha + β = γ ,

α

β

γ

α

β

γ

α

β

Epilogue.

Austin, 7 September 1986

Transcription by Joel Hockey

Last revised on Thu, 27 May 2010 .

 Three cheers for formalization! Instead of setting out to provefor a right-angled triangle, we included the antecedent γ = π/2 in the formal statement of what was to be proved. It was only after the introduction of π that we could eliminate it and met the "nicely symmetric" formulation. Three cheers for the equivalence! It makes quite clear that the theorem is not about right-angled triangles, but about triangles in general. Three cheers for the notational device captured in sgn. If we had not been careful, we would have ended up provingforany of the six relations =, ≠, , and ≥ No cheers at all for that stage of the argument in which lack of axiomatization forced us to resort to a picture. Pictures are almost unavoidably overspecific and therby often force a case analysis upon you. Note that I carefully avoided the pictures for; there are 9 of them:to the right of, coincident with, and to the left of, and similarly for the pairand. For the argument these distinctions are irrelevant but, when drawing a picture, you can hardly avoid making them. One of these days I would like to find a convincing explanation of the circumstance that youngsters continue to be educated with the theorem of Pythagoras in its diluted form as quoted from Coxeter. Notice that the 9 figures could have been avoided by also provingi.e. proving (0) and (1) in full. Notice that our figure was not pulled out of a magicians hat! As soon as sgn.() occurs in the demonstrandum, it is sweetly reasonable to construct that difference. In order not to destroy the symmetry betweenand, one starts withand subtractsat the one side andat the other:and this is the germ of the figure we drew.I am in a paradoxical situation. I am convinced that of the people knowing the theorem of Pythagoras, almost no one can read the above without being surprised at least once. Furthermore, I think that all those surprises relevant (because telling about their education in reasoning). Yet I don't know of a single respectable journal in which I could flog this dead horse.prof.dr.Edsger W.DijkstraDepartment of Computer SciencesThe University of Texas at AustinAustin, TX 78712-1188, USA