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 draw the altitude to , to get two right triangles.
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
Solution 2
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.
Solution 3
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,
Then .
Solution 4
By the Law of Cosines,
Now
Solution 5
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 6
Let be
WLOG, assume that and are legs of right triangle with and
By the Pythagorean theorem, we have , and the given . Solving the equations gives us and . We see that , and .
Our derived equation equals as approaches infinity. Evaluating , we get .
Solution 7
As in Solution 1, drop an altitude to . Let meet at , and let .
Then, , . We can calculate using the tangent addition formula, after noticing that . So, we find that \begin{align*} \cot{\gamma} &= \frac{1}{\tan{\gamma}} \\ &= \frac{1}{\frac{\frac{x}{h} + \frac{y}{h}}{1 - \frac{xy}{h^2}}} \\ &= \frac{1}{\frac{(x+y)h}{h^2 - xy}} \\ &= \frac{h^2 - xy}{(x+y)h}. \end{align*}
So now we can simplify our original expression: \begin{align*} \frac{\cot{\gamma}}{\cot{\alpha} + \cot{\beta}} &= \frac{\frac{h^2 - xy}{(x+y)h}}{\frac{x + y}{h}} \\ &= \frac{h^2 - xy}{(x+y)^2}. \end{align*}
But notice that , so this becomes Now note that we can use the Pythagorean theorem to calculate , we get that So our expression simplifies to since from the problem and that there is another after the in our expression. Again note that , so it again simplifies to , or .
Solution 8 (Quick and Easy)
Since no additional information is given, we can assume that triangle ABC is right with the right angle at B. We can use the Pythagorean theorem to say We can now solve for in terms of
Using the definition of cotangent
Plugging into our desired expression, 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 |
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