Difference between revisions of "2008 AIME II Problems/Problem 11"
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pathpen=black;pointpen=black;pen f=fontsize(9); | pathpen=black;pointpen=black;pen f=fontsize(9); | ||
real r=44-6*35^.5; | real r=44-6*35^.5; | ||
− | pair A=(0,96),B=(- | + | pair A=(0,96),B=(-208,0),C=(28,0),X=C-(64/3,0),Y=B+(4*r/3,0),P=X+(0,16),Q=Y+(0,r),M=foot(Q,X,P); |
path PC=CR(P,16),QC=CR(Q,r); | path PC=CR(P,16),QC=CR(Q,r); | ||
D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed); | D(A--B--C--cycle); D(Y--Q--P--X); D(Q--M); D(P--C,dashed); |
Revision as of 11:42, 11 February 2018
Problem
In triangle , , and . Circle has radius and is tangent to and . Circle is externally tangent to and is tangent to and . No point of circle lies outside of . The radius of circle can be expressed in the form , where , , and are positive integers and is the product of distinct primes. Find .
Solution
Let and be the feet of the perpendiculars from and to , respectively. Let the radius of be . We know that . From draw segment such that is on . Clearly, and . Also, we know is a right triangle.
To find , consider the right triangle . Since is tangent to , then bisects . Let ; then . Dropping the altitude from to , we recognize the right triangle, except scaled by .
So we get that . From the half-angle identity, we find that . Therefore, . By similar reasoning in triangle , we see that .
We conclude that .
So our right triangle has sides , , and .
By the Pythagorean Theorem, simplification, and the quadratic formula, we can get , for a final answer of .
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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.