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← Problem 224: (2015 AMC 12b, 24) →
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Problem 224: (2015 AMC 12b, 24)
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 $\tfrac{5}{8}$ times the radius of circle $B$, and the radius of circle $C$ is $\tfrac{5}{8}$ times the radius of circle $D$. Furthermore, $AB = CD = 39$ and $PQ = 48$. Let $R$ be the midpoint of $\overline{PQ}$. What is $AR+BR+CR+DR$ ? $\textbf{(A)}\; 180 \qquad$ $\textbf{(B)}\; 184 \qquad$ $\textbf{(C)}\; 188 \qquad$ $\textbf{(D)}\; 192\qquad$ $\textbf{(E)}\; 196$
Solution
Exercise 224: Solution
Four circles (no two congruent) have centers $A,B,C,D$, with $P,Q$ on all four. Circle $A$'s radius is $\frac59$ circle $B$'s, and circle $C$'s is $\frac59$ circle $D$'s. Also $AB=CD=56$ and $PQ=54$. Let $R$ be the midpoint of $\overline{PQ}$. What is $AR+BR+CR+DR$?
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