2012 AIME I Problems/Problem 12
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
Solution 1
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 .
Solution 2
(This solution does not use the Angle Bisector Theorem but uses more trig.)
Find values for all angles in terms of . , , , , and .
Use the law of sines on and :
In , . This simplifies to .
In , . This simplifies to .
Solve for and equate them so that you get .
From this, .
Use a trig identity on the denominator on the right to obtain:
This simplifies to $\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cosB + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}$ (Error compiling LaTeX. ! Undefined control sequence.)
This gives $\frac{15}{8} = \frac{\sqrt{3}\cos B + \sin B}{2\sin B} = \frac{\sqrt{3}\cos B}{2\sinB} + \frac{1}{2} = \frac{\sqrt{3}\cot B}{2} + \frac{1}{2}.$ (Error compiling LaTeX. ! Undefined control sequence.)
Thus: and .
Since , . Our final answer is .
See also
2012 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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