2007 AIME I Problems/Problem 9

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 the cosine of angle EBD is . Since the measure of the angle opposite to EBD is the complement of this one, its cosine is . Using the law of cosines, we see that 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 so our answer is 737.