Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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>...")
 
(→‎Solution)
 
(One intermediate revision by the same user not shown)
Line 10: Line 10:
 
    
 
    
 
== Solution ==
 
== Solution ==
 +
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).
 +
 +
Applying Stewart's theorem on <math>\bigtriangleup ABC</math> with point <math>D</math>, we obtain the equation
 +
 +
<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>
 +
 +
Similarly using point <math>E</math>, we obtain
 +
 +
<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
 +
 +
<cmath>(sin^{2}(x) + cos^{2}(x)) \cdot c=a^{2} c + b^{2} c - \frac{4c^3}{9}</cmath>
 +
 +
<cmath>c=(a^2 + b^2) c - \frac{4c^3}{9}</cmath>
 +
 +
<cmath>c=c^2 \cdot c -\frac{4c^3}{9}</cmath>
 +
 +
<cmath>c=\pm \frac{3\sqrt{5}}{5}, 0</cmath>
 +
 +
As the other 2 answers yield degenerate triangles, we see that the answer is <cmath>\boxed{(C) \frac{3\sqrt{5}}{5}}</cmath>
 +
 
<math>\fbox{}</math>
 
<math>\fbox{}</math>
  

Latest revision as of 21:06, 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 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1980_AHSME_Problems/Problem_23&oldid=72896"