Difference between revisions of "1956 AHSME Problems/Problem 14"
Coolmath34 (talk | contribs) (Created page with "== Problem 14== The points <math>A,B,C</math> are on a circle <math>O</math>. The tangent line at <math>A</math> and the secant <math>BC</math> intersect at <math>P, B</math...") |
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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> | ||
− | == | + | 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> |
− | + | ||
+ | Let <math>AR = r</math> and <math>PE = a</math>. We now have a system of equations. | ||
+ | |||
+ | <cmath>(10\sqrt{3})^2+r^2=PR^2</cmath> | ||
+ | <cmath>(\sqrt{r^2-10^2})^2+PE^2=r^2-10^2+PE^2=PR^2</cmath> | ||
+ | |||
+ | Set them equal to each other and solve. | ||
− | + | <cmath>(10\sqrt{3})^2+r^2=r^2-10^2+PE^2</cmath> | |
+ | <cmath>(10\sqrt{3})^2=-10^2+PE^2</cmath> | ||
+ | <cmath>400=PE^2</cmath> | ||
+ | <cmath>20=PE</cmath> | ||
+ | |||
+ | We know <math>BE = 10</math>, so <math>PE = 20 - 10 = 10</math>, which is <math>\boxed{B}</math>. | ||
− | + | ~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 are on a circle . The tangent line at and the secant intersect at lying between and . If and , then equals:
Because is a tangent line, angle is a right angle. Drop a perpendicular from to at We find that
Let and . We now have a system of equations.
Set them equal to each other and solve.
We know , so , which is .
~Revised by MC413551
See Also
1956 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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