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=== | ||
+ | |||
+ | |||
+ | Let <math>\gamma</math> be <math>(180-\alpha-\beta)</math> | ||
+ | |||
+ | <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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | We see that our derived equation equals to <math>\tan^2 \alpha</math> as <math>\tan \beta</math> approaches infinity. | ||
+ | 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 , , be the three sides of a triangle, and let , , , be the angles opposite them. If , find
Contents
[hide]Solution
Solution 1
We can draw the altitude to , to get two right triangles. , from the definition of the cotangent. From the definition of area, , so .
Now we evaluate the numerator:
From the Law of Cosines and the sine area formula,
Then .
Solution 2
By the Law of Cosines,
Now
Solution 3
Use Law of cosines to give us or therefore . Next, we are going to put all the sin's in term of . We get . Therefore, we get .
Next, use Law of Cosines to give us . Therefore, . Also, . Hence, .
Lastly, . Therefore, we get .
Now, . After using , we get .
Solution 4
Let be
WLOG, assume that and are legs of right triangle with and
By Pythagorean theorem, we have , and the given . Solving the equations gives us and . We see that , and .
We see that our derived equation equals to as approaches infinity. Evaluating , we get .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.