2007 AIME I Problems/Problem 9

Problem

In right triangle $ABC$ with right angle $C$ , $CA = 30$ and $CB = 16$ . Its legs $CA$ and $CB$ are extended beyond $A$ and $B$ . Points $O_1$ and $O_2$ lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center $O_1$ is tangent to the hypotenuse and to the extension of leg $CA$ , the circle with center $O_2$ is tangent to the hypotenuse and to the extension of leg $CB$ , and the circles are externally tangent to each other. The length of the radius either circle can be expressed as $p/q$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$ .

Solution

Let the point where CB's extension hits the circle be D, and the point where the hypotenuse hits that circle be E. Clearly EB=DB. Let $EB=x$ . Draw the two perpendicular radii to D and E. Now we have a cyclic quadrilateral. Let the radius be length $r$ . We see that since the cosine of angle ABC is $\frac{15}{17}$ the cosine of angle EBD is $-\frac{15}{17}$ . Since the measure of the angle opposite to EBD is the complement of this one, its cosine is $\frac{15}{17}$ . Using the law of cosines, we see that $x^{2}+x^{2}+\frac{30x^{2}}{17}=r^{2}+r^{2}-\frac{30r^{2}}{17}$ This tells us that $r=4x$ .

Now look at the other end of the hypotenuse. Call the point where CA hits the circle F and the point where the hypotenuse hits the circle G. Draw the radii to F and G and we have cyclic quadrilaterals once more. Using the law of cosines again, we find that the length of our tangents is $2.4x$ . Note that if we connect the centers of the circles we have a rectangle with sidelengths 8x and 4x. So, 8x+2.4x+x=34. Solving we find that $4x=\frac{680}{57}$ so our answer is 737.

2007 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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2007 AIME I Problems/Problem 9

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