2015 AMC 12B Problem 24

Since every center is equidistant from $P$ and $Q,$ all four centers and $R$ lie on the perpendicular bisector of $PQ,$ with $PR=24.$ First consider two centers whose radii are in the ratio $5{:}8$ and whose distance apart is $39.$ If $R$ lies between them, let $y = AR$ and $x = \tfrac15$ of circle $A$’s radius. Then $y^2 + 24^2 = 25x^2$ and $(39 - y)^2 + 24^2 = 64x^2.$ Subtracting gives $x^2 = 39 - 2y,$ so $y^2 + 50y - 399 = 0$ and $y=7.$ Here $x=5,$ so the two center distances from $R$ are $7$ and $32,$ and the radii are $25$ and $40.$ If instead the two centers lie on the same side of $R,$ their distances are $w$ and $w+39.$ The analogous equations give $w^2-50w-399=0,$ hence $w=57,$ and the distances are $57$ and $96.$ In this case the radii are $5\sqrt{153}$ and $8\sqrt{153}.$ Using the same placement for both pairs $(A,B)$ and $(C,D)$ would give two congruent circles of each radius, contrary to the hypothesis. Thus one pair uses each placement. Their distance sums are $7+32=39$ and $57+96=153,$ so the requested total is $39+153=192.$ Thus, the correct answer is D.