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 1

Let $A$ , $B$ , and $C$ be the vertices of the right triangle, where $C$ is the right angle. Let $a$ , $b$ , and $c$ be the lengths of the sides opposite $A$ , $B$ , and $C$ , respectively. Place $\triangle ABC$ in the coordinate plane such that $C$ is at the origin, $B$ lies on the positive $x$ -axis, and $A$ lies on the positive $y$ -axis. Then $A$ has coordinates $(0, b)$ and $B$ has coordinates $(a, 0)$ .

Let $M$ and $N$ be the trisection points of $\overline{AB}$ , with $M$ closer to $A$ and $N$ closer to $B$ . Then $M$ has coordinates $\left(\frac{1}{3}a, \frac{2}{3}b\right)$ and $N$ has coordinates $\left(\frac{2}{3}a, \frac{1}{3}b\right)$ .

Let $m = CM$ and $n = CN$ . Then $m^2 = \frac{1}{9}a^2 + \frac{4}{9}b^2$ and $n^2 = \frac{4}{9}a^2 + \frac{1}{9}b^2$ . Adding these two equations yields $m^2 + n^2 = \frac{5}{9}(a^2+b^2) = \frac{5}{9}c^2$ .

Either $m = \sin x$ and $n = \cos x$ , or $m = \cos x$ and $n = \sin x$ . In both cases, $m^2 + n^2 = \sin^2 x + \cos^2 x = 1$ . Therefore, $\frac{5}{9}c^2 = 1$ and $c = \sqrt{\frac{9}{5}} = \boxed{(\textbf{C})\ \frac{3\sqrt{5}}{5}}$ .

-j314andrews

Solution 2 (Stewart's Theorem)

Let $A$ , $B$ , and $C$ be the vertices of the right triangle, where $C$ is the right angle. Let $a$ , $b$ , and $c$ be the lengths of the sides opposite $A$ , $B$ , and $C$ , respectively. Let $M$ and $N$ be the trisection points of $\overline{AB}$ . Let $m = CM$ and $n = CN$ .

Using Stewart's Theorem on $\triangle ABC$ with segment $CM$ yields:

$m^2 \cdot c + \frac{2}{3}c \cdot \frac{1}{3}c \cdot c = a^2 \cdot \frac{1}{3}c + b^2 \cdot \frac{2}{3}c$

Dividing both sides by $c$ and isolating $m^2$ yields:

$m^2 = \frac{1}{3}a^2 + \frac{2}{3}b^2 - \frac{2}{9}c^2\ (\textrm{i})$

Using Stewart's Theorem on $\triangle ABC$ with segment $CN$ yields:

$n^2 \cdot c + \frac{2}{3}c \cdot \frac{1}{3}c \cdot c =a^2 \cdot \frac{2}{3}c + b^2 \cdot \frac{1}{3}c$

Dividing both sides by $c$ and isolating $n^2$ yields:

$n^2 = \frac{2}{3}a^2 + \frac{1}{3}b^2 - \frac{2}{9}c^2\ (\textrm{ii})$

Adding equations $(\textrm{i})$ and $(\textrm{ii})$ yields:

$m^2 + n^2 = a^2 + b^2 - \frac{4}{9}c^2$ .

Either $m = \sin x$ and $n = \cos x$ , or $m = \cos x$ and $n = \sin x$ . In both cases, $m^2 + n^2 = \sin^2 x + \cos^2 x = 1$ . Also, by the Pythagorean theorem, $a^2 + b^2 = c^2$ .

Therefore, $1=c^2 - \frac{4}{9}c^2 = \frac{5}{9}c^2$ , so $c^2 = \frac{9}{5}$ and $c = \sqrt{\frac{9}{5}} = \boxed{(\textbf{C})\ \frac{3\sqrt{5}}{5}}$ .

1980 AHSME (Problems • Answer Key • Resources)
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1980 AHSME Problems/Problem 23

Contents

Problem

Solution 1

Solution 2 (Stewart's Theorem)

See also