1989 AIME Problem 10

Question

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}.

Answer

Simplify

\cot\alpha+\cot\beta

using

\alpha+\beta=\pi-\gamma

Use the Law of Sines for the resulting sine factors and the Law of Cosines for

\cos\gamma

First,

\begin{aligned}\cot\alpha+\cot\beta&=\frac{\sin(\alpha+\beta)}{\sin\alpha\sin\beta}\\&=\frac{\sin\gamma}{\sin\alpha\sin\beta}.\end{aligned}

Hence the desired ratio is

\frac{\cos\gamma\sin\alpha\sin\beta}{\sin^2\gamma}=\frac{ab\cos\gamma}{c^2},

where the Law of Sines was used in the final equality. By the Law of Cosines,

\begin{aligned}2ab\cos\gamma&=a^2+b^2-c^2\\&=1988c^2.\end{aligned}

Therefore the ratio is

\frac{1988}{2}=994.

Problem 10 in other years