# 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

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 $AR = r$ and $PE = a$. We now have a system of equations. $$(10\sqrt{3})^2+r^2=PR^2$$ $$(\sqrt{r^2-10^2})^2+PE^2=r^2-10^2+PE^2=PR^2$$

Set them equal to each other and solve. $$(10\sqrt{3})^2+r^2=r^2-10^2+PE^2$$ $$(10\sqrt{3})^2=-10^2+PE^2$$ $$400=PE^2$$ $$20=PE$$

We know $BE = 10$, so $PE = 20 - 10 = 10$, which is $\boxed{B}$.

~Revised by MC413551

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