1956 AHSME Problems/Problem 14

Problem 14

The points $A,B,C$ are on a circle $O$ . The tangent line at $A$ and the secant $BC$ intersect at $P, B$ lying between $C$ and $P$ . If $\overline{BC} = 20$ and $\overline{PA} = 10\sqrt {3}$ , then $\overline{PB}$ equals:

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 10\sqrt {3} \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 30$

Solution Sketch

Draw the diagram. (someone else link it please)

Because $PA$ is a tangent line, angle $\angle OAP$ is a right angle. Drop a perpendicular from $O$ to $BC$ at $E.$ We find that $BE = EC = 10.$

Let $PB = x$ and the radius of the circle be $r.$ We have a system of equations, and we can solve for $x$ and $r.$

1956 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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1956 AHSME Problems/Problem 14

Problem 14

Solution Sketch

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