Difference between revisions of "2012 AIME I Problems/Problem 12"
Algebra2000 (talk | contribs) (→Solution) |
Algebra2000 (talk | contribs) (→Solution) |
||
| Line 7: | Line 7: | ||
Now, we have <math>cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}</math> by another application of the Law of Cosines to triangle <math>DCB</math>, so <math>cos \angle B = \frac{11}{13}</math>. In addition, <math>sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}</math>, so <math>tan \angle B = \frac{4\sqrt{3}}{11}</math>. | Now, we have <math>cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}</math> by another application of the Law of Cosines to triangle <math>DCB</math>, so <math>cos \angle B = \frac{11}{13}</math>. In addition, <math>sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}</math>, so <math>tan \angle B = \frac{4\sqrt{3}}{11}</math>. | ||
| − | Our final answer is <math>4+3+11 = | + | Our final answer is <math>4+3+11 = \boxed{018}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2012|n=I|num-b=11|num-a=13}} | {{AIME box|year=2012|n=I|num-b=11|num-a=13}} | ||
Revision as of 10:54, 17 March 2012
Problem 12
Let 
be a right triangle with right angle at 
Let 
and 
be points on 
with between 
and such that 
and 
trisect 
If 
then 
can be written as 
where 
and 
are relatively prime positive integers, and 
is a positive integer not divisible by the square of any prime. Find 
Solution
Without loss of generality, set 
. Then, by the Angle Bisector Theorem on triangle 
, we have 
. We apply the Law of Cosines to triangle to get 
, which we can simplify to get 
.
Now, we have 
by another application of the Law of Cosines to triangle , so 
. In addition, 
, so 
.
Our final answer is 
.
See also
| 2012 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
