Problem:
Four circles Ο,ΟAβ,ΟBβ, and ΟCβ with the same radius are drawn in the interior of triangle ABC such that ΟAβ is tangent to sides AB and AC,ΟBβ to BC and BA,ΟCβ to CA and CB, and Ο is externally tangent to ΟAβ,ΟBβ, and ΟCβ. If the sides of triangle ABC are 13,14, and 15, the radius of Ο can be represented in the form nmβ, where m and n are relatively prime positive integers. Find m+n.
Solution:
Let OAβ,OBβ, and OCβ be the centers of ΟAβ,ΟBβ, and ΟCβ, respectively. Then OAβOBβββ₯β₯β₯βAB,OBβOCβββ₯β₯β₯βBC, and OCβOAβββ₯CA. Also, the lines AOAβ,BOBβ, and COCβ are concurrent at I, the incenter of triangle ABC, and therefore there is a dilation D centered at I that sends triangle OAβOBβOCβ to triangle ABC. Let R and r be the circumradius and inradius of triangle ABC, respectively, and let R1β and r1β be the circumradius and inradius of triangle OAβOBβOCβ, respectively. Then rRβ=r1βR1ββ. By Heron's formula,
implying that r=4 and R=865β. Let x be the radius of Ο. Because I is the center of D,r1β=rβx. Let S be the center of Ο. Then S is equidistant from OAβ,OBβ, and OCβ, that is, S is the circumcenter of triangle OAβOBβOCβ. Thus R1β=SOAβ=2x. Therefore
4βx2xβ=rβx2xβ=r1βR1ββ=rRβ=3265β
Solving the last equation gives x=129260β, and m+n=389β.