1989 AIME Problems/Problem 10
Problem
Let ,
,
be the three sides of a triangle, and let
,
,
, be the angles opposite them. If
, find
![$\frac{\cot \gamma}{\cot \alpha+\cot \beta}$](http://latex.artofproblemsolving.com/8/5/8/858c49b569dc76c503dbdbfa8d0b78fd2c04fe1d.png)
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 $\cot(\gamma)=\frac{994c}{b\sina}$ (Error compiling LaTeX. Unknown error_msg).
Next, use Law of Cosines to give us . Therefore,
. Also,
. Hence,
.
Lastly, . Therefore, we get
.
Now, $\frac{\cot(\gamma)}{\cot(\beta)+\cot(\alpha)}=\frac{\frac{994c}{b\sina}}{\frac{a^2-994c^2+b^2-994c^2}{bc\sin(a)}}$ (Error compiling LaTeX. Unknown error_msg). After using , we get $\frac{994c*bc\sina}{c^2b\sina}=\boxed{994}$ (Error compiling LaTeX. Unknown error_msg).
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 |