Difference between revisions of "1989 AIME Problems/Problem 10"
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(solution (2) by k18o7) |
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<center><math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}</math></center> | <center><math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}</math></center> | ||
+ | __TOC__ | ||
== Solution == | == Solution == | ||
− | We can draw the [[altitude]] <math>h</math> to <math>c</math>, to get two [[right triangle]]s. <math>\cot{\alpha}+\cot{\beta}=\frac{c}{h}</math>, from the definition of the [[cotangent]]. From the definition of area, <math>h=\frac{2A}{c}</math>, so | + | === Solution 1 === |
+ | We can draw the [[altitude]] <math>h</math> to <math>c</math>, to get two [[right triangle]]s. <math>\cot{\alpha}+\cot{\beta}=\frac{c}{h}</math>, from the definition of the [[cotangent]]. From the definition of area, <math>h=\frac{2A}{c}</math>, so <math>\cot{\alpha}+\cot{\beta}=\frac{c^2}{2A}</math>. | ||
Now we evaluate the numerator: | Now we evaluate the numerator: | ||
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<cmath>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</cmath> | <cmath>\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}</cmath> | ||
− | From the [[Law of Cosines]] | + | From the [[Law of Cosines]] and the sine area formula, |
− | <cmath>\begin{ | + | <cmath>\begin{align*}\cos{\gamma}&=\frac{1988c^2}{2ab}\ |
− | \sin{\gamma}&= | + | \sin{\gamma}&= \frac{2A}{ab}\ |
− | \cot{\gamma}&= | + | \cot{\gamma}&= \frac{\cos \gamma}{\sin \gamma} = \frac{1988c^2}{4A} \end{align*}</cmath> |
− | + | Then <math>\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\boxed{994}</math>. | |
+ | |||
+ | === Solution 2 === | ||
+ | <cmath>\begin{align*} | ||
+ | \cot{\alpha} + \cot{\beta} &= \frac {\cos{\alpha}}{\sin{\alpha}} + \frac {\cos{\beta}}{\sin{\beta}} = \frac {\sin{\alpha}\cos{\beta} + \cos{\alpha}\sin{\beta}}{\sin{\alpha}\sin{\beta}}\ &= \frac {\sin{(\alpha + \beta)}}{\sin{\alpha}\sin{\beta}} = \frac {\sin{\gamma}}{\sin{\alpha}\sin{\beta}} | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | By the Law of Cosines, | ||
+ | |||
+ | <cmath>a^2 + b^2 - 2ab\cos{\gamma} = c^2 = 1989c^2 - 2ab\cos{\gamma} \implies ab\cos{\gamma} = 994c^2</cmath> | ||
+ | |||
+ | Now | ||
+ | |||
+ | <cmath>\begin{align*}\frac {\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac {\cot{\gamma}\sin{\alpha}\sin{\beta}}{\sin{\gamma}} = \frac {\cos{\gamma}\sin{\alpha}\sin{\beta}}{\sin^2{\gamma}} = \frac {ab}{c^2}\cos{\gamma} = \frac {ab}{c^2} \cdot \frac {994c^2}{ab}\ &= \boxed{994}\end{align*}</cmath> | ||
== See also == | == See also == |
Revision as of 15:31, 3 August 2008
Problem
Let , , be the three sides of a triangle, and let , , , be the angles opposite them. If , find
Contents
[hide]Solution
Solution 1
We can draw the altitude to , to get two right triangles. , from the definition of the cotangent. From the definition of area, , so .
Now we evaluate the numerator:
From the Law of Cosines and the sine area formula,
Then .
Solution 2
By the Law of Cosines,
Now
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 9 |
Followed by Problem 11 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |