# 1980 AHSME Problems/Problem 23

## Problem

Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths $\sin x$ and $\cos x$, where $x$ is a real number such that $0. The length of the hypotenuse is $\text{(A)} \ \frac{4}{3} \qquad \text{(B)} \ \frac{3}{2} \qquad \text{(C)} \ \frac{3\sqrt{5}}{5} \qquad \text{(D)}\ \frac{2\sqrt{5}}{3}\qquad \text{(E)}\ \text{not uniquely determined}$

## Solution

Consider right triangle $ABC$ with hypotenuse $BC$. Let points $D$ and $E$ trisect $BC$. WLOG, let $AD=cos(x)$ and $AE=sin(x)$ (the proof works the other way around as well).

Applying Stewart's theorem on $\bigtriangleup ABC$ with point $D$, we obtain the equation $$cos^{2}(x)\cdot c=a^2 \cdot \frac{c}{3} + b^2 \cdot \frac{2c}{3} - \frac{2c}{3}\cdot \frac{c}{3} \cdot c$$

Similarly using point $E$, we obtain $$sin^{2}(x)\cdot c=a^2 \cdot \frac{2c}{3} + b^2 \cdot \frac{c}{3} - \frac{2c}{3}\cdot \frac{c}{3} \cdot c$$ $$(sin^{2}(x) + cos^{2}(x)) \cdot c=a^{2} c + b^{2} c - \frac{4c^3}{9}$$ $$c=(a^2 + b^2) c - \frac{4c^3}{9}$$ $$c=c^2 \cdot c -\frac{4c^3}{9}$$ $$c=\pm \frac{3\sqrt{5}}{5}, 0$$

As the other 2 answers yield degenerate triangles, we see that the answer is $$\boxed{(C) \frac{3\sqrt{5}}{5}}$$ $\fbox{}$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 