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.
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,
$$\dfrac{c}{q} = \dfrac{\dfrac{2a^2}{c}}{q-c}$$
$$\dfrac{c^2}{q} = \dfrac{2a^2}{q-c}$$
$$c^2q - c^3 = 2a^2q$$
$$c^2q - 2a^2q = c^3$$
$$q = \dfrac{c^3}{c^2-2a^2}$$
Using Law of Sines,
$$\dfrac{\sin 90}{q} = \dfrac{\sin{(90 - 2α)}}{c}$$
$$\dfrac{\sin 90}{q} = \dfrac{\cos{2α}}{c}$$
$$\dfrac{\sin 90}{q} = \dfrac{\cos^2{α} - \sin^2{α}}{c}$$
$$\dfrac{1}{\dfrac{c^3}{c^2-2a^2}} = \dfrac{\dfrac{b^2}{c^2} - \dfrac{a^2}{c^2}}{c}$$
$$\dfrac{c^2-2a^2}{c^3} = \dfrac{b^2 - a^2}{c^3}$$
$$c^2 = a^2 + b^2$$