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> | ||
| − | == Solution | + | == 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> | ||
| − | Let <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}} | ||
Latest revision as of 14:13, 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:
Solution
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.
![\[(10\sqrt{3})^2+r^2=PR^2\]](http://latex.artofproblemsolving.com/4/2/5/425eb6be1f5c44418f0947eea5e343bef27181ce.png)
![\[(\sqrt{r^2-10^2})^2+PE^2=r^2-10^2+PE^2=PR^2\]](http://latex.artofproblemsolving.com/6/1/7/617c9d83ded67ea4a34c02357f203c0c59a060bd.png)
Set them equal to each other and solve.
![\[(10\sqrt{3})^2+r^2=r^2-10^2+PE^2\]](http://latex.artofproblemsolving.com/d/d/9/dd9477954a9bfbcf1e4a5b973c117f1650cc3408.png)
![\[(10\sqrt{3})^2=-10^2+PE^2\]](http://latex.artofproblemsolving.com/5/a/4/5a4551b657da9370dd56c2ddd74394d51b31f041.png)
![\[400=PE^2\]](http://latex.artofproblemsolving.com/0/1/4/01459f317a9b452f31fd2040b3a26523cc99b00c.png)
![\[20=PE\]](http://latex.artofproblemsolving.com/1/c/a/1caef441b7ba01c1c8106ccb51f1be085649f51c.png)
We know 
, so 
, which is 
.
~Revised by MC413551
See Also
| 1956 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
