Can You 'Waffle' Your Way To A Proof?

Riddler Classic

As you may have heard, two high school students from St. Mary’s Academy in New Orleans, Louisiana — Ne’Kiya Jackson and Calcea Johnson — recently discovered a novel proof of the Pythagorean Theorem.

Their proof applied the law of sines (which itself can be derived from equivalent expressions for a triangle’s area and has no dependency on the Pythagorean theorem, thereby avoiding any circular logic) to the construction below:

Atop the figure are two reflected right triangles with legs a and b (with a < b) and hypotenuse c. Below these triangles are what the students called a “waffle cone” shape, formed between the extensions of one of the top triangle’s hypotenuse and a line that’s perpendicular to the other hypotenuse.

In their proof, they compute distances p and q, where p extends from the leftmost vertex of the two triangles to the intersection of the lines, and q extends from the topmost vertex of the two triangles to the same intersection.

Your challenge is to determine expressions for p and q in terms of a, b and c. However, in doing so, you absolutely cannot use the Pythagorean theorem in any of its forms (e.g., the so-called “distance formula,” etc.). After all, solving for p and q is a key step toward proving the Pythagorean theorem.

Extra credit: Once you’ve determined p and q, try completing a proof of the Pythagorean theorem that makes use of them. Remember, the students used the law of sines at one point.

Solution

Construction

Append an additional triangle similar to the original triangle with sides $a, b, \text{and } c$.

Since the hypotenuse of the new triangle is $2a$, the legs are $\dfrac{2ab}{c}$ and $\dfrac{2a^2}{c}$.

Solving for $p$ and $q$

Looking at the two similar triangles,
$$\dfrac{c}{p} = \dfrac{\dfrac{2a^2}{c}}{p - \dfrac{2ab}{c}}$$
$$\dfrac{c}{p} = \dfrac{2a^2}{cp - 2ab}$$
$$c^2p - 2abc = 2a^2p$$
$$c^2p - 2a^2p = 2abc$$
$$p = \dfrac{2abc}{c^2-2a^2}$$

Again,
$$\dfrac{c}{q} = \dfrac{\dfrac{2a^2}{c}}{q-c}$$
$$\dfrac{c^2}{q} = \dfrac{2a^2}{q-c}$$
$$c^2q - c^3 = 2a^2q$$
$$c^2q - 2a^2q = c^3$$
$$q = \dfrac{c^3}{c^2-2a^2}$$

Extra Credit

Using Law of Sines,

$$\dfrac{\sin 90}{q} = \dfrac{\sin{(90 - 2α)}}{c}$$


$$\dfrac{\sin 90}{q} = \dfrac{\cos{2α}}{c}$$
$$\dfrac{\sin 90}{q} = \dfrac{\cos^2{α} - \sin^2{α}}{c}$$
$$\dfrac{1}{\dfrac{c^3}{c^2-2a^2}} = \dfrac{\dfrac{b^2}{c^2} - \dfrac{a^2}{c^2}}{c}$$
$$\dfrac{c^2-2a^2}{c^3} = \dfrac{b^2 - a^2}{c^3}$$
$$c^2 = a^2 + b^2$$

Rohan Lewis

2023.04.16

Code can be found here.