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 $\fbox{E}$.

See Also

1958 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 41
Followed by
Problem 43
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