1959 AHSME Problems/Problem 40

Problem

On the same side of a straight line three circles are drawn as follows: a circle with a radius of $4$ inches is tangent to the line, the other two circles are equal, and each is tangent to the line and to the other two circles. The radius of the equal circles is: $\textbf{(A)}\ 24 \qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 12$

Solution

Make a line DG which is parallel to FC. We know that GF = BF = 5, since BFE is similar to BGD.

Since ADG is similar to ACB, we know that AG is 10.

Thus, AF = 10 + 5 = 15, which is answer $\fbox{\textbf{(C)}}$ .

1959 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 39	Followed by Problem 41
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1959 AHSME Problems/Problem 40

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