1958 AHSME Problems/Problem 42

Problem

In a circle with center $O$ , chord $\overline{AB}$ equals chord $\overline{AC}$ . Chord $\overline{AD}$ cuts $\overline{BC}$ in $E$ . If $AC = 12$ and $AE = 8$ , then $AD$ equals:

$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 20\qquad \textbf{(E)}\ 18$

Solution

Let $X$ be a point on $BC$ so $AX \perp BC$ . Let $AX = h$ , $EX = \sqrt{64 - h^2}$ and $BX = \sqrt{144 - h^2}$ . $CE = CX - EX = \sqrt{144 - h^2} - \sqrt{64 - h^2}$ . Using Power of a Point on $E$ , $(BE)(EC) = (AE)(ED)$ (there isn't much information about the circle so I wanted to use PoP).

$(\sqrt{144 - h^2} - \sqrt{64 - h^2})(\sqrt{144 - h^2} + \sqrt{64 - h^2}) = 8(ED)$

$(144 - h^2) - (64 - h^2) = 8(ED)$

$80 = 8(ED)$

$ED = 10$

Adding up $AD$ and $ED$ we get $\fbox{E}$ .

1958 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 41	Followed by Problem 43
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1958 AHSME Problems/Problem 42

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