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 O1 and O2 lie in the exterior of the triangle and are the centers of two circles with equal radii. The circle with center O1 is tangent to the hypotenuse and to the extension of leg CA, the circle with center O2 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 of either circle can be expressed as p/q, where p and q are relatively prime positive integers. Find p+q.
Solution:
Let T1 and T2 be the points of tangency on AB so that O1T1 and O2T2 are radii of length r. Let circles O1 and O2 be tangent to each other at point T3, let circle O1 be tangent to the extension of AC at E1, and let circle O2 be tangent to the extension of BC at E2. Let D1 be the foot of the perpendicular from O1 to BC, let D2 be the foot of the perpendicular from O2 to AC, and let O1D1 and O2D2 intersect at P. Let r be the radius of O1 and O2. Note that O1T3=O2T3=r,CD1=O1E1=r, and CD2=O2E2=r. Because triangles ABC and O1O2P are similar, ABO1O2=ACO1P=BCO2P. Because O1O2=2r and AB=302+162=34,O1P=1730r and O2P=1716r. By equal tangents, AE1=AT1=x and BE2=BT2=y. Thus
AB=34=AT1+T1T2+BT2=AT1+O1O2+BT2=x+2r+y
Also, O1P=E1D2, so 1730r=AD2+AE1=30−r+x, and O2P=E2D1, so 1716r=BD1+BE2=16−r+y. Adding these two equations produces 1746r=46−2r+x+y. Substituting x+y=34−2r yields 1746r=80−4r. Thus r=57680, and p+q=737.
OR
Draw O1A and O2B and note that these segments bisect the external angles of the triangle at A and B, respectively. Thus ∠T1O1A=21(∠A) and T1A=rtan∠T1O1A=rtan21(∠A). Similarly T2B=rtan21(∠B). By the half-angle identity for tangent,
tan(21∠A)=cos∠A+1sin∠A=(30/34)+116/34=41
and similarly tan(21(∠B))=53. Then 34=AB=AT1+T1T2+T2B=r/4+2r+3r/5=57r/20 and r=680/57. Thus p+q=737.