# 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 $a^2+b^2=1989c^2$, find $\frac{\cot \gamma}{\cot \alpha+\cot \beta}$

## 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}{\sin \gamma} = \frac{1988c^2}{4A} \end{align*}

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 {\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*}

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}} = \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*}

### Solution 3

Use Law of cosines to give us $c^2=a^2+b^2-2ab\cos(\gamma)$ or therefore $\cos(\gamma)=\frac{994c^2}{ab}$. Next, we are going to put all the sin's in term of $\sin(a)$. We get $\sin(\gamma)=\frac{c\sin(a)}{a}$. Therefore, we get $\cot(\gamma)=\frac{994c}{b\sin a}$.

Next, use Law of Cosines to give us $b^2=a^2+c^2-2ac\cos(\beta)$. Therefore, $\cos(\beta)=\frac{a^2-994c^2}{ac}$. Also, $\sin(\beta)=\frac{b\sin(a)}{a}$. Hence, $\cot(\beta)=\frac{a^2-994c^2}{bc\sin(a)}$.

Lastly, $\cos(\alpha)=\frac{b^2-994c^2}{bc}$. Therefore, we get $\cot(\alpha)=\frac{b^2-994c^2}{bc\sin(a)}$.

Now, $\frac{\cot(\gamma)}{\cot(\beta)+\cot(\alpha)}=\frac{\frac{994c}{b\sin a}}{\frac{a^2-994c^2+b^2-994c^2}{bc\sin(a)}}$. After using $a^2+b^2=1989c^2$, we get $\frac{994c*bc\sin a}{c^2b\sin a}=\boxed{994}$.

### Solution 4

Let $\gamma$ be $(180-\alpha-\beta)$ $\frac{\cot \gamma}{\cot \alpha+\cot \beta} = \frac{\frac{-\tan \alpha \tan \beta}{\tan(\alpha+\beta)}}{\tan \alpha + \tan \beta} = \frac{(\tan \alpha \tan \beta)^2-\tan \alpha \tan \beta}{\tan^2 \alpha + 2\tan \alpha \tan \beta +\tan^2 \beta}$

WLOG, assume that $a$ and $c$ are legs of right triangle $abc$ with $\beta = 90^o$ and $c=1$

By Pythagorean theorem, we have $b^2=a^2+1$, and the given $a^2+b^2=1989$. Solving the equations gives us $a=\sqrt{994}$ and $b=\sqrt{995}$. We see that $\tan \beta = \infty$, and $\tan \alpha = \sqrt{994}$.

We see that our derived equation equals to $\tan^2 \alpha$ as $\tan \beta$ approaches infinity. Evaluating $\tan^2 \alpha$, we get $\boxed{994}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 