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

(Created page with "== Problem == Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths <math>\sin x</math> and <math>...")
 
(Added Solution)
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== Solution ==
 
== Solution ==
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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).
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 +
Applying Stewart's theorem on <math>\bigtriangleup ABC</math> with point <math>D</math>, we obtain the equation
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<cmath>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</cmath>
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 +
Similarly using point <math>E</math>, we obtain
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<cmath>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</cmath>
 +
 +
Adding these equations, we get
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 +
<cmath>(sin^{2}(x) + cos^{2}(x)) \cdot c=a^{2} c + b^{2} c - \frac{4c^3}{9}</cmath>
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<cmath>c=(a^2 + b^2) c - \frac{4c^3}{9}</cmath>
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<cmath>c=c^2 \cdot c -\frac{4c^3}{9}</cmath>
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<cmath>c=\pm \frac{3\sqrt{5}}{5}, 0</cmath>
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As the other 2 answers are extraneous, we see that the answer is <cmath>\boxed{(C) \frac{3\sqrt{5}}{5}}</cmath>
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<math>\fbox{}</math>
 
<math>\fbox{}</math>
  

Revision as of 21:05, 11 November 2015

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 are extraneous, 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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