1989 AIME Problems/Problem 10
Problem
Let , , be the three sides of a triangle, and let , , , be the angles opposite them. If , find
Contents
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.