Question

Four circles, no two of which are congruent, have centers at $A,$ $B,$ $C,$ and $D,$ and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\dfrac58$ times the radius of circle $B,$ and the radius of circle $C$ is $\dfrac58$ times the radius of circle $D.$ Furthermore, $AB = CD = 39$ and $PQ = 48.$ Let $R$ be the midpoint of $\overline{PQ}.$ What is $AR + BR + CR + DR?$

$180$

$184$

$188$

$192$

$196$

Solution:

Since every center is equidistant from $P$ and $Q,$ all four centers and $R$ lie on the perpendicular bisector of $PQ,$ with $PR = 24.$ Suppose $R$ lies between $A$ and $B.$ 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 $y = 7.$ Thus $AR = 7$ and $BR = 32.$

Because the circles are noncongruent, $R$ does not lie between $C$ and $D.$ The analogous equations give $w^2 - 50w - 399 = 0$ with $w = CR = 57,$ so $DR = 96.$ The sum is $7 + 32 + 57 + 96 = 192.$

Thus, the correct answer is D.

2015 AMC 12B Problem 24

Solution:

Problem 24 in Other Years