Dynamic Geometry: P43

Let the variable triangle be $ABC$ such that $AB = c = 15$ (red) and $BC = a = 8$ (green). Since the circumradius $R$ of $\triangle ABC$ is given by:

$2R = \frac a{\sin A} = \frac b{\sin B} = \frac c{\sin C}$

and that $c = 15$ is fixed, $R$ is minimum when $C = 90^\circ$ . (Note that $a < c$ and $a$ cannot be the hypotenuse and $A < 90^\circ$ .) When $C = 90^\circ$ , $b = \sqrt{c^2 - a^2} = \sqrt{15^2 - 8^2} = \sqrt{161}$ and the inradius $r$ is given by $A_\triangle = sr$ , where $A_\triangle = \dfrac {ab}2$ and $s = \dfrac {a+b+c}2$ are the area and semiperimeter of $\triangle ABC$ . Then we have:

$\begin{aligned} \frac {ab}2 & = \frac {a+b+c}2 \cdot r \\ 8 \sqrt{161} & = (8 + \sqrt{161} + 15) r \\ \implies r & = \frac {8\sqrt{161}}{23 + \sqrt{161}} = \frac {\sqrt{161}-7}2 \end{aligned}$

Therefore the required answer $\sqrt{2+161+7 - 1} = \sqrt{169} = \boxed {13}$ .