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

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\text{(E)}\ \text{not uniquely determined}</math>     
 
\text{(E)}\ \text{not uniquely determined}</math>     
 
    
 
    
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== Solution ==
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== Solution 1 ==
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Let <math>A</math>, <math>B</math>, and <math>C</math> be the vertices of the right triangle, where <math>C</math> is the right angle.  Let <math>a</math>, <math>b</math>, and <math>c</math> be the lengths of the sides opposite <math>A</math>, <math>B</math>, and <math>C</math>, respectively.  Place <math>\triangle ABC</math> in the coordinate plane such that <math>C</math> is at the origin, <math>B</math> lies on the positive <math>x</math>-axis, and <math>A</math> lies on the positive <math>y</math>-axis.  Then <math>A</math> has coordinates <math>(0, b)</math> and <math>B</math> has coordinates <math>(a, 0)</math>.
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Let <math>M</math> and <math>N</math> be the trisection points of <math>\overline{AB}</math>, with <math>M</math> closer to <math>A</math> and <math>N</math> closer to <math>B</math>.  Then <math>M</math> has coordinates <math>\left(\frac{1}{3}a, \frac{2}{3}b\right)</math> and <math>N</math> has coordinates <math>\left(\frac{2}{3}a, \frac{1}{3}b\right)</math>.
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Let <math>m = CM</math> and <math>n = CN</math>.  Then <math>m^2 = \frac{1}{9}a^2 + \frac{4}{9}b^2</math> and <math>n^2 = \frac{4}{9}a^2 + \frac{1}{9}b^2</math>.  Adding these two equations yields <math>m^2 + n^2 = \frac{5}{9}(a^2+b^2) = \frac{5}{9}c^2</math>.
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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>.  Therefore, <math>\frac{5}{9}c^2 = 1</math> and <math>c = \sqrt{\frac{9}{5}} = \boxed{(\textbf{C})\ \frac{3\sqrt{5}}{5}}</math>.
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-j314andrews
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== Solution 2 ==
 
Consider right triangle <math>ABC</math> with hypotenuse <math>BC</math>. Let points <math>D</math> and <math>E</math> trisect <math>BC</math>. WLOG, let <math>AD=cos(x)</math> and <math>AE=sin(x)</math> (the proof works the other way around as well).  
 
Consider right triangle <math>ABC</math> with hypotenuse <math>BC</math>. Let points <math>D</math> and <math>E</math> trisect <math>BC</math>. WLOG, let <math>AD=cos(x)</math> and <math>AE=sin(x)</math> (the proof works the other way around as well).  
  

Revision as of 14:51, 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

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 \[\boxed{(C) \frac{3\sqrt{5}}{5}}\]

$\fbox{}$

See also

1980 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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