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 $\boxed{994}$ .

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

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