Congruent AND Concurrent

Let $D$ , $E$ , and $F$ be the centers of the three congruent circles with radius $r$ , and draw in and label the diagram as follows:

By Heron's formula, the area of $\triangle ABC$ is $A_{\triangle ABC} = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = 84$ .

Since each circle is tangent to the sides of $\triangle ABC$ , the centers $D$ , $E$ , and $F$ are on the angle bisectors of $A$ , $B$ , and $C$ .

A $13$ - $14$ - $15$ triangle is a $5$ - $12$ - $13$ triangle combined with a $9$ - $12$ - $15$ , so $AG = DG \cot \angle DAG = r \cot (\frac{1}{2} \angle CAB) = r \cot (\frac{1}{2} \cos^{-1} (\frac{5}{13})) = \frac{3}{2}r$ and $HB = EH \cot \angle EBH = r \cot (\frac{1}{2} \angle ABC) = r \cot (\frac{1}{2} \cos^{-1} (\frac{9}{15})) = 2r$ , so that $DE = GH = AB - AG - HB = 14 - \frac{3}{2}r - 2r = 14 - \frac{7}{2}r$ .

Since the sides of $\triangle DEF$ are parallel to the sides of $\triangle ABC$ , the two triangles are similar, and have a side ratio of $\cfrac{DE}{AB} = \cfrac{14 - \frac{7}{2}r}{14} = 1 - \frac{1}{4}r$ .

Since $ZD = ZE = ZF = r$ , $Z$ is the circumcenter of $\triangle DEF$ , so $2r = \cfrac{DE \cdot EF \cdot DF}{2 A_{\triangle DEF}} = \cfrac{(1 - \frac{1}{4}r)AB \cdot (1 - \frac{1}{4}r)BC \cdot (1 - \frac{1}{4}r)AC}{(1 - \frac{1}{4}r)^2 A_{\triangle ABC}} = \cfrac{14 \cdot 15 \cdot 13 \cdot (1 - \frac{1}{4}r)}{84}$ , and $2r = \cfrac{14 \cdot 15 \cdot 13 \cdot (1 - \frac{1}{4}r)}{84}$ solves to $r = \cfrac{260}{97}$ .

Therefore, $p = 260$ , $q = 97$ , and $p + q = \boxed{357}$ .