Difference between revisions of "1980 AHSME Problems/Problem 23"
(Added Solution) |
(→Solution) |
||
| Line 30: | Line 30: | ||
<cmath>c=\pm \frac{3\sqrt{5}}{5}, 0</cmath> | <cmath>c=\pm \frac{3\sqrt{5}}{5}, 0</cmath> | ||
| − | As the other 2 answers | + | 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 
and 
, where 
is a real number such that 
. The length of the hypotenuse is
Solution
Consider right triangle 
with hypotenuse 
. Let points 
and 
trisect . WLOG, let 
and 
(the proof works the other way around as well).
Applying Stewart's theorem on 
with point , we obtain the equation
![\[cos^{2}(x)\cdot c=a^2 \cdot \frac{c}{3} + b^2 \cdot \frac{2c}{3} - \frac{2c}{3}\cdot \frac{c}{3} \cdot c\]](http://latex.artofproblemsolving.com/b/4/0/b408335fec0ac71cb332d7962409f2dbc9395c65.png)
Similarly using point , we obtain
![\[sin^{2}(x)\cdot c=a^2 \cdot \frac{2c}{3} + b^2 \cdot \frac{c}{3} - \frac{2c}{3}\cdot \frac{c}{3} \cdot c\]](http://latex.artofproblemsolving.com/3/5/8/358d4499d5e5a8bdf95f2b3ab1335bf482c3200c.png)
Adding these equations, we get
![\[(sin^{2}(x) + cos^{2}(x)) \cdot c=a^{2} c + b^{2} c - \frac{4c^3}{9}\]](http://latex.artofproblemsolving.com/8/5/4/85490501141a08e4acaad2ea893ad9bb02cc8f04.png)
![\[c=(a^2 + b^2) c - \frac{4c^3}{9}\]](http://latex.artofproblemsolving.com/c/6/8/c686cf77046f36b53e5f4f30cd907b13acb8710f.png)
![\[c=c^2 \cdot c -\frac{4c^3}{9}\]](http://latex.artofproblemsolving.com/a/6/6/a66c4d589c65a1b64b9ceb91e1ece29a41081dea.png)
![\[c=\pm \frac{3\sqrt{5}}{5}, 0\]](http://latex.artofproblemsolving.com/7/0/1/701b7d93ffbbdca3dc4fe8f7aca09ccf26800375.png)
As the other 2 answers yield degenerate triangles, we see that the answer is ![\[\boxed{(C) \frac{3\sqrt{5}}{5}}\]](http://latex.artofproblemsolving.com/2/c/3/2c318bab23a10220c2036f8a7d0146addd545bf2.png)
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. 
