Difference between revisions of "1989 AIME Problems/Problem 10"
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=== Solution 3=== | === Solution 3=== | ||
− | Use Law of cosines to give us <math>c^2=a^2+b^2-2ab\cos(\gamma)</math> or therefore <math>\cos(\gamma)=\frac{994c^2}{ab}</math>. Next, we are going to put all the sin's in term of <math>\sin(a)</math>. We get <math>\sin(\gamma)=\frac{c\sin(a)}{a}</math>. Therefore, we get <math>\cot(\gamma)=\frac{994c}{b\ | + | Use Law of cosines to give us <math>c^2=a^2+b^2-2ab\cos(\gamma)</math> or therefore <math>\cos(\gamma)=\frac{994c^2}{ab}</math>. Next, we are going to put all the sin's in term of <math>\sin(a)</math>. We get <math>\sin(\gamma)=\frac{c\sin(a)}{a}</math>. Therefore, we get <math>\cot(\gamma)=\frac{994c}{b\sin a}</math>. |
Next, use Law of Cosines to give us <math>b^2=a^2+c^2-2ac\cos(\beta)</math>. Therefore, <math>\cos(\beta)=\frac{a^2-994c^2}{ac}</math>. Also, <math>\sin(\beta)=\frac{b\sin(a)}{a}</math>. Hence, <math>\cot(\beta)=\frac{a^2-994c^2}{bc\sin(a)}</math>. | Next, use Law of Cosines to give us <math>b^2=a^2+c^2-2ac\cos(\beta)</math>. Therefore, <math>\cos(\beta)=\frac{a^2-994c^2}{ac}</math>. Also, <math>\sin(\beta)=\frac{b\sin(a)}{a}</math>. Hence, <math>\cot(\beta)=\frac{a^2-994c^2}{bc\sin(a)}</math>. | ||
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Lastly, <math>\cos(\alpha)=\frac{b^2-994c^2}{bc}</math>. Therefore, we get <math>\cot(\alpha)=\frac{b^2-994c^2}{bc\sin(a)}</math>. | Lastly, <math>\cos(\alpha)=\frac{b^2-994c^2}{bc}</math>. Therefore, we get <math>\cot(\alpha)=\frac{b^2-994c^2}{bc\sin(a)}</math>. | ||
− | Now, <math>\frac{\cot(\gamma)}{\cot(\beta)+\cot(\alpha)}=\frac{\frac{994c}{b\ | + | 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>. |
== See also == | == See also == |
Revision as of 15:51, 13 March 2015
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
.
Next, use Law of Cosines to give us . Therefore,
. Also,
. Hence,
.
Lastly, . Therefore, we get
.
Now, . After using
, 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.