Problem: Four circles $\omega,$ $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ with the same radius are drawn in the interior of triangle $ABC$ such that $\omega_{A}$ is tangent to sides $AB$ and $AC$ $\omega_{B}$ to $BC$ and $BA$ $\omega_{C}$ to $CA$ and $CB$ , and $\omega$ is externally tangent to $\omega_{A},$ $\omega_{B},$ and $\omega_{C}$ . If the sides of triangle $ABC$ are $13,$ $14,$ and $15,$ the radius of $\omega$ can be represented in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: [asy] defaultpen(fontsize(12)+0.8); size(300);  pair A,B,C,X,Y,Z,P,Q,R; B=origin; C=15*right; A=IP(CR(B,13),CR(C,14)); P=incenter(A,B,C); real r=260/129; Q=r*(rotate(-90)*A/length(A)); X=extension(A,P,Q,Q+A); Y=extension(B,P,Q,Q+A); R=rotate(90,C)*(C+r*(A-C)/length(A-C)); Z=extension(C,P,R,R+A-C); draw(A--B--C--A); draw(CR(X,r)^^CR(Y,r)^^CR(Z,r)^^A--P--B^^P--C, gray); draw(CR(circumcenter(X,Y,Z),r), gray); label("$A$",A,N); label("$B$",B,0.15*(B-P)); label("$C$",C,0.1*(C-P)); draw(X--Y--Z--cycle); draw(Y--(3.5,0),blue); draw(P--(7,0), blue); pen p=fontsize(10)+linewidth(3); dot("$O_A$",X,right,p); dot("$O_B$",Y,left+up,p); dot("$O_C$",Z,right+up,p); dot("$O$",circumcenter(X,Y,Z),right+down,p); dot("$I$",P,left+up,p);  dot("$H$",(7,0),down,p); [/asy]
Let $O_A, O_B, O_C, O$ be the centers of $w_A, w_B, w_C, w$ , respectively. Also, let $I$ be the incenter of $ABC$ and $r$ be the radius of circle $w$ . Since $AB||O_AO_B$ $BC||O_BO_C$ , and $CA||O_CO_A$ , we know that
\[\angle BAI = \angle O_BO_AI, \angle CBI = \angle O_CO_BI, \angle ACI = \angle O_AO_CI \text{ and }\angle CAI = \angle O_CO_AI, \angle ABI = \angle O_AO_BI, \angle BCI = \angle O_BO_CI.\]
That means $\angle ABC = \angle O_AO_BO_C$ $\angle BAC = \angle O_BO_AO_C$ , and $\angle ACB = \angle O_AO_CO_B$ . Thus, $\triangle ABC \sim \triangle O_AO_BO_C$ . We also know that we are scaling each side of $\triangle ABC$ (from $AB$ to $O_AO_B$ for instance), about $I$ (since A,O_A,I are collinear ; same apply with $B$ and $C$ ).
Now, let the homothety $\mathcal{H} (I, x)$ map $\triangle ABC$ to $\triangle O_AO_BO_C$ . To start off, we know the circumradius of $O_AO_BO_C$ is $O$ , since $OO_A = OO_B = OO_C = 2r$ . Since $O_AO_B = 13x$ $O_BO_C = 14x$ $O_CO_A = 15x$ , we can get an relationship involving $x$ and $r$ via another way to find the circumradius:
\[[\triangle O_AO_BO_C ] =\frac{abc}{4R} \Longrightarrow 84x^2 =\frac{13x\cdot 14x\cdot 15x}{4\cdot 2r} \Longrightarrow r=\frac{65x}{16}\]
Take notice of the inradius of $ABC$ . We get the inradius to be $[\triangle ABC ] = sr_0 \Longrightarrow r_0=4$ . Let the tangency point of the incircle and side $BC$ be $H$ . We know $IH = 4$ . We also know that we can cut off the part of $IH$ that is outside of $\triangle O_AO_BO_C$ to get the inradius of $\triangle O_AO_BO_C$ . To part that is outside $\triangle O_AO_BO_C$ turns out just to be the radius of circle $w_B$ (as seen in the picture). That means the inradius of $\triangle O_AO_BO_C$ is just $4-r$ . We can calculate that incradius in another way, though. We know that the inradius of $\triangle ABC$ is $4$ , which means the inradius of $\triangle O_AO_BO_C$ is just $4x$ (by our homethety ratio).
Thus, we have $4x = 4-r = 4-\dfrac{65x}{16} \Longrightarrow x = \dfrac{64}{129} \Longrightarrow r = \dfrac{260}{129}$ . That gives $\boxed{389}$ as our final answer.

Answer:  389