# Serendipitous Proofs Of The Pythagorean Theorem

It often happens, in mathematics in particular, that in pursuit of one goal a person unexpectedly attains another. Serendipity is an occurrence of such a fortunate discovery, usually by accident. In the case at hand, I think that the chance played a minor role, if any at all. To discover a novel proof of the Pythagorean theorem, John Molokach needed and demonstrated uncommon perseverance and the power of observation.

Theorem

In a right triangle with legs a and b and hypotenuse c

a^{2}+b^{2}=c^{2}.

Proof

John was solving the following problem:

Consider a right triangle with legs a and b so that a \lt b. The smaller acute angle is doubled as in the image below to form another right triangle with legs b and m and hypotenuse n. Find m and n in terms of a, b and c. There were two solution to the problem. One was based on the construction depicted below: It is not hard to see (the main tools are the similarity of triangles and the formula for the sum of a geometric series), this construction leads to

m = a + {\frac{c^2}{a}}\sum_{n=1}^{\infty}{(\frac{a}{b})}^{2n} = \frac{a(b^{2}-a^{2}+c^{2})}{b^{2}-a^{2}},
n = {\frac{c^2}{b}}\sum_{n=1}^{\infty}{(\frac{a}{b})}^{2n} = \frac{bc^{2}}{b^{2}-a^{2}}.

The other - a more straightforward - solution was based on a simpler diagram: and directly led to a pair of equations:

\frac{m-a}{n}=\frac{n-b}{m}=\frac{a}{b},

from which one obtains

m=\frac{2ab^{2}}{b^{2}-a^{2}},
n=\frac{b(a^{2}+b^{2})}{b^{2}-a^{2}}.

Now, the "lucky part" was in John's insight to compare the expressions for m and n in the two solutions. Do that and observe that both comparisons lead to the Pythagorean formula.

At this point one may stop and congratulate John for a nice proof, but John himself did not stop to pat himself on a shoulder. He went on digging.

If \alpha is the angle opposite leg a, then from the first set of formulas,

\sin 2\alpha = \frac{a(b^{2}-a^{2}+c^{2})}{bc^{2}}.

By adding a second copy of the original triangle and computing the area of the shaded portion in two ways one obtains

2ab = c^{2}\sin 2\alpha

from which

\sin 2\alpha = \frac{2ab}{c^{2}}.

The two expressions for \sin 2\alpha can, too, be equated, and - as before - the result is the Pythagorean identity.