Problem:
Four circles ω,ωA​,ωB​, and ωC​ with the same radius are drawn in the interior of triangle ABC such that ωA​ is tangent to sides AB and AC,ωB​ to BC and BA,ωC​ to CA and CB, and ω is externally tangent to ωA​,ωB​, and ωC​. If the sides of triangle ABC are 13,14, and 15, the radius of ω can be represented in the form nm​, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let OA​,OB​, and OC​ be the centers of ωA​,ωB​, and ωC​, respectively. Then OA​OB​​∥∥∥​AB,OB​OC​​∥∥∥​BC, and OC​OA​​∥CA. Also, the lines AOA​,BOB​, and COC​ are concurrent at I, the incenter of triangle ABC, and therefore there is a dilation D centered at I that sends triangle OA​OB​OC​ to triangle ABC. Let R and r be the circumradius and inradius of triangle ABC, respectively, and let R1​ and r1​ be the circumradius and inradius of triangle OA​OB​OC​, respectively. Then rR​=r1​R1​​. By Heron's formula,
implying that r=4 and R=865​. Let x be the radius of ω. Because I is the center of D,r1​=r−x. Let S be the center of ω. Then S is equidistant from OA​,OB​, and OC​, that is, S is the circumcenter of triangle OA​OB​OC​. Thus R1​=SOA​=2x. Therefore
4−x2x​=r−x2x​=r1​R1​​=rR​=3265​
Solving the last equation gives x=129260​, and m+n=389​.