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Difference between revisions of "1980 AHSME Problems/Problem 23"

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Either <math>m = \sin x</math> and <math>n = \cos x</math>, or <math>m = \cos x</math> and <math>n = \sin x</math>.  In both cases, <math>m^2 + n^2 = \sin^2 x + \cos^2 x = 1</math>.  Also, by the Pythagorean theorem, <math>a^2 + b^2 = c^2</math>.
 
Either <math>m = \sin x</math> and <math>n = \cos x</math>, or <math>m = \cos x</math> and <math>n = \sin x</math>.  In both cases, <math>m^2 + n^2 = \sin^2 x + \cos^2 x = 1</math>.  Also, by the Pythagorean theorem, <math>a^2 + b^2 = c^2</math>.
  
−
Therefore, <math>1=c^2 - \frac{4}{9}c^2 = \frac{5}{9}c^2</math>, so <math>c^2 = \frac{9}{5}</math> and <math>c = \sqrt{\frac{9}{5}} = \boxed{(\textbf{C}) \frac{3\sqrt{5}}{5}}</math>.
+
Therefore, <math>1=c^2 - \frac{4}{9}c^2 = \frac{5}{9}c^2</math>, so <math>c^2 = \frac{9}{5}</math> and <math>c = \sqrt{\frac{9}{5}} = \boxed{(\textbf{C})\ \frac{3\sqrt{5}}{5}}</math>.
  
 
== See also ==
 
== See also ==

Latest revision as of 16:04, 16 August 2025

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 $c = \sqrt{\frac{9}{5}} = \boxed{(\textbf{C})\ \frac{3\sqrt{5}}{5}}$.

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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All AHSME Problems and Solutions

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