1989 AIME Problems/Problem 10
Let , , be the three sides of a triangle, and let , , , be the angles opposite them. If , find
Then , from the definition of the cotangent.
Let be the area of Then , so .
By identical logic, we can find similar expressions for the sums of the other two cotangents: Adding the last two equations, subtracting the first, and dividing by 2, we get Therefore
By the law of cosines, So, by the extended law of sines, Identical logic works for the other two angles in the triangle. So, the cotangent of any angle in the triangle is directly proportional to the sum of the squares of the two adjacent sides, minus the square of the opposite side. Therefore We can then finish as in solution 1.
We start as in solution 1, though we'll write instead of for the area. Now we evaluate the numerator:
From the Law of Cosines and the sine area formula,
By the Law of Cosines,
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 .
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 .
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