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Difference between revisions of "1989 AIME Problems/Problem 10"

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Now, <math>\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)}}</math>.  After using <math>a^2+b^2=1989c^2</math>, we get <math>\frac{994c*bc\sin a}{c^2b\sin a}=\boxed{994}</math>.
 
Now, <math>\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)}}</math>.  After using <math>a^2+b^2=1989c^2</math>, we get <math>\frac{994c*bc\sin a}{c^2b\sin a}=\boxed{994}</math>.
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=== Solution 4===
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Let <math>\gamma</math> be <math>(180-\alpha-\beta)</math>
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<math>\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}</math>
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WLOG, assume that <math>a</math> and <math>c</math> are legs of right triangle <math>abc</math> with <math>\beta = 90^o</math> and <math>c=1</math>
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By Pythagorean theorem, we have <math>b^2=a^2+1</math>, and the given <math>a^2+b^2=1989</math>. Solving the equations gives us <math>a=\sqrt{994}</math> and <math>b=\sqrt{995}</math>. We see that <math>\tan \beta = \infty</math>, and <math>\tan \alpha = \sqrt{994}</math>.
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We see that our derived equation equals to <math>\tan^2 \alpha</math> as <math>\tan \beta</math> approaches infinity.
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Evaluating <math>\tan^2 \alpha</math>, we get <math>\boxed{994}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:55, 11 August 2017

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 $\boxed{994}$.

See also

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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