1989 AIME Problems/Problem 10

Problem

Let $a_{}^{}$ , $b_{}^{}$ , $c_{}^{}$ be the three sides of a triangle, and let $\alpha_{}^{}$ , $\beta_{}^{}$ , $\gamma_{}^{}$ , be the angles opposite them. If $a^2+b^2=1989^{}_{}c^2$ , find

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

Solution

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 therefore $\cot{\alpha}+\cot{\beta}=\frac{c^2}{2A}$ .

Now we evaluate the numerator:

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

From the Law of Cosines ( $R$ is the circumradius),

$\begin{eqnarray*}\cos{\gamma}&=&\frac{1988c^2}{2ab}\\ \sin{\gamma}&=&\frac{c}{2R}\\ \cot{\gamma}&=&\frac{1988cR}{ab}\end{eqnarray*}$

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

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
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1989 AIME Problems/Problem 10

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