jmc

The two circles described in the problem are shown in the diagram. The circle located inside $△ABC$ is called the incircle; following convention we will label its center $I$. The other circle is known as an excircle, and we label its center $E$. To begin, we may compute the area of triangle $ABC$ using Heron's formula. The side lengths of triangle $△ABC$ are $a=15$, $b=14$, and $c=13$, while the semiperimeter is $s=\frac{1}{2}(a+b+c)=21$, so its area is $$K=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{21\cdot 6\cdot 7\cdot 8}=84.$$We find the inradius $r$ of $△ABC$ by using the fact that $K=rs$, so $84=21r$, giving $r=4$. Next label the points of tangency of the incircle and excircle with ray $\overline{AC}$ as $S$ and $T$, as shown at right. It is a standard fact that $AS=s-a=6$ and $AT=s=21$. (The reader should confirm this. Repeatedly use the fact that tangents from a point to a circle have the same length.) Furthermore, the angle bisector of $\angle A$ passes through $I$ and $E$, and the radii $\overline{SI}$ and $\overline{TE}$ are perpendicular to $\overline{AC}$, so triangles $△ASI$ and $△ATE$ are similar right triangles. By the Pythagorean Theorem we compute $$AI=\sqrt{(AS{)}^2+(SI{)}^2}=\sqrt{36+16}=2\sqrt{13}.$$Using the similar triangles we find that $AI/AE=AS/AT=6/21=2/7$. Therefore $AE=7\sqrt{13}$ and we conclude that $IE=AE-AI=\boxed{5\sqrt{13}}$.