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>
  
== Solution Sketch==
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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>
Draw the diagram. (someone else link it please)
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Let <math>AR = r</math> and <math>PE = a</math>. We now have a system of equations.
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<cmath>(10\sqrt{3})^2+r^2=PR^2</cmath>
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<cmath>(\sqrt{r^2-10^2})^2+PE^2=r^2-10^2+PE^2=PR^2</cmath>
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Set them equal to each other and solve.
  
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>  
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<cmath>(10\sqrt{3})^2+r^2=r^2-10^2+PE^2</cmath>
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<cmath>(10\sqrt{3})^2=-10^2+PE^2</cmath>
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<cmath>400=PE^2</cmath>
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<cmath>20=PE</cmath>
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We know <math>BE = 10</math>, so <math>PE = 20 - 10 = 10</math>, which is <math>\boxed{B}</math>.
  
Let <math>PB = x</math> and the radius of the circle be <math>r.</math> We have a system of equations, and we can solve for <math>x</math> and <math>r.</math>
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~Revised by MC413551
  
 
==See Also==
 
==See Also==
 
{{AHSME 50p box|year=1956|num-b=13|num-a=15}}
 
{{AHSME 50p box|year=1956|num-b=13|num-a=15}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:12, 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:

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

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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