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 $0<x<\frac{\pi}{2}$ . 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

$sin^{2}(x)\cdot c=a^2 \cdot \frac{2c}{3} + b^2 \cdot \frac{c}{3} - \frac{2c}{3}\cdot \frac{c}{3} \cdot c$

Adding these equations, we get

$(sin^{2}(x) + cos^{2}(x)) \cdot c=a^{2} c + b^{2} c - \frac{4c^3}{9}$

$c=(a^2 + b^2) c - \frac{4c^3}{9}$

$c=c^2 \cdot c -\frac{4c^3}{9}$

$c=\pm \frac{3\sqrt{5}}{5}, 0$

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

$\fbox{}$

1980 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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1980 AHSME Problems/Problem 23

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