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Difference between revisions of "1956 AHSME Problems/Problem 14"

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<math>\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 10\sqrt {3} \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 30 </math>
 
<math>\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 10 \qquad\textbf{(C)}\ 10\sqrt {3} \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 30 </math>
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== Solution==
  
 
Because <math>PA</math> is a tangent line, angle <math>\angle OAP</math> is a right angle. Drop a perpendicular from <math>O</math> to <math>BC</math> at <math>E.</math> We find that <math>BE = EC = 10.</math>  
 
Because <math>PA</math> is a tangent line, angle <math>\angle OAP</math> is a right angle. Drop a perpendicular from <math>O</math> to <math>BC</math> at <math>E.</math> We find that <math>BE = EC = 10.</math>  

Latest revision as of 15:13, 8 April 2023

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

See Also

1956 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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All AHSME Problems and Solutions

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