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1989 AIME Problems/Problem 10

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Problem

Let $a$ , $b$ , $c$ be the three sides of a triangle, and let $\alpha$ , $\beta$ , $\gamma$ , be the angles opposite them. If $a^2+b^2=1989c^2$ , find

$\frac{\cot \gamma}{\cot \alpha+\cot \beta}$

Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See also

Solution

Solution 1

We can draw the altitude $h$ to $c$ , to get two right triangles. $\cot{\alpha}+\cot{\beta}=\frac{c}{h}$ , from the definition of the cotangent. From the definition of area, $h=\frac{2A}{c}$ , so $\cot{\alpha}+\cot{\beta}=\frac{c^2}{2A}$ .

Now we evaluate the numerator:

$\cot{\gamma}=\frac{\cos{\gamma}}{\sin{\gamma}}$

From the Law of Cosines and the sine area formula,

$\begin{align*}\cos{\gamma}&=\frac{1988c^2}{2ab}\\\sin{\gamma}&= \frac{2A}{ab}\\\cot{\gamma}&= \frac{\cos \gamma}{...$

Then $\frac{\cot \gamma}{\cot \alpha+\cot \beta}=\frac{\frac{1988c^2}{4A}}{\frac{c^2}{2A}}=\frac{1988}{2}=\boxed{994}$ .

Solution 2

$\begin{align*}\cot{\alpha} + \cot{\beta} &= \frac {\cos{\alpha}}{\sin{\alpha}} + \frac {\cos{\beta}}{\sin{\beta}} = \frac...$

By the Law of Cosines,

$a^2 + b^2 - 2ab\cos{\gamma} = c^2 = 1989c^2 - 2ab\cos{\gamma} \implies ab\cos{\gamma} = 994c^2$

Now

$\begin{align*}\frac {\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac {\cot{\gamma}\sin{\alpha}\sin{\beta}}{\sin{\gamma...$

See also

1989 AIME (Problems • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1989_AIME_Problems/Problem_10"

Categories: Intermediate Geometry Problems | Intermediate Trigonometry Problems

Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.

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