1997 AIME Problem 4

The two radius-$5$ centers are $10$ apart, and the other two centers both lie on the perpendicular bisector of that segment

Right triangles give midpoint distances $12$ and $\sqrt{r^2 + 10r};$ rule out opposite sides before subtracting them

Let the radius-$5$ circles have centers $P_1$ and $P_2,$ so $P_1P_2 = 5 + 5 = 10,$ and let $M$ be the midpoint. The radius-$8$ circle’s center $Q$ satisfies $QP_1 = QP_2 = 13,$ so $Q$ lies on the perpendicular bisector of $\overline{P_1P_2}$ at distance $\sqrt{13^2 - 5^2} = 12$ from $M.$ Likewise the fourth circle, of radius $r,$ has its center $R$ on the same perpendicular bisector with $RP_1 = 5 + r,$ so $RM = \sqrt{(5+r)^2 - 25}$ $= \sqrt{r^2 + 10r}.$

The centers $Q$ and $R$ cannot lie on opposite sides of $M:$ that would give $12 + RM = 8 + r,$ hence $RM = r - 4,$ but squaring would yield $18r = 16,$ contrary to $r \ge 4.$ Nor can $R$ lie beyond $Q,$ since then $RM - 12 = 8 + r$ would force $RM = r + 20,$ which is incompatible with $RM^2 = r^2 + 10r.$ Thus $R$ lies between $M$ and $Q,$ and external tangency to the radius-$8$ circle gives $$12 - \sqrt{r^2 + 10r} = 8 + r.$$ Then $\sqrt{r^2 + 10r} = 4 - r,$ and squaring yields $r^2 + 10r = 16 - 8r + r^2,$ so $18r = 16$ and $r = \frac{8}{9}.$

Thus $m + n = 8 + 9 = 17.$