Difference between revisions of "1980 AHSME Problems/Problem 23"
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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). | ||
+ | |||
+ | 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 are extraneous, we see that the answer is <cmath>\boxed{(C) \frac{3\sqrt{5}}{5}}</cmath> | ||
+ | |||
<math>\fbox{}</math> | <math>\fbox{}</math> | ||
Revision as of 22:05, 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 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
Similarly using point , we obtain
Adding these equations, we get
As the other 2 answers are extraneous, we see that the answer is
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.