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Difference between revisions of "2010 AIME II Problems/Problem 14"

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== Problem 14 ==
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Triangle <math>ABC</math> with right angle at <math>C</math>, <math>\angle BAC < 45^\circ</math> and <math>AB = 4</math>. Point <math>P</math> on <math>\overbar{AB}</math> is chosen such that <math>\angle APC = 2\angle ACP</math> and <math>CP = 1</math>. The ratio <math>\frac{AP}{BP}</math> can be represented in the form <math>p + q\sqrt{r}</math>, where <math>p</math>, <math>q</math>, <math>r</math> are positive integers and <math>r</math> is not divisible by the square of any prime. Find <math>p+q+r</math>.
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== Solution ==
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Label the center of the circumcircle of <math>ABC</math> as <math>O</math> and the intersection of <math>CP</math> with the circumcircle as <math>D</math>. It now follows that <math>\angle{DOA} = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</math>.  
 
Label the center of the circumcircle of <math>ABC</math> as <math>O</math> and the intersection of <math>CP</math> with the circumcircle as <math>D</math>. It now follows that <math>\angle{DOA} = \angle{APC} = \angle{DPB}</math>. Hence <math>ODP</math> is isosceles and <math>OD = DP = 2</math>.  
  
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<math>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}</math>
 
<math>\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}</math>
  
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This gives that the answer is <math>007</math>.
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This gives that the answer is <math>\boxed{007}</math>.
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== See also ==
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{{AIME box|year=2010|num-b=14|num-a=15|n=I}}

Revision as of 17:13, 3 April 2010

Problem 14

Triangle $ABC$ with right angle at $C$, $\angle BAC < 45^\circ$ and $AB = 4$. Point $P$ on $\overbar{AB}$ (Error compiling LaTeX. Unknown error_msg) is chosen such that $\angle APC = 2\angle ACP$ and $CP = 1$. The ratio $\frac{AP}{BP}$ can be represented in the form $p + q\sqrt{r}$, where $p$, $q$, $r$ are positive integers and $r$ is not divisible by the square of any prime. Find $p+q+r$.

Solution

Label the center of the circumcircle of $ABC$ as $O$ and the intersection of $CP$ with the circumcircle as $D$. It now follows that $\angle{DOA} = \angle{APC} = \angle{DPB}$. Hence $ODP$ is isosceles and $OD = DP = 2$.

Denote $E$ the projection of $O$ onto $CD$. Now $CD = CP + DP = 3$. By the pythagorean theorem, $OE = \sqrt {2^2 - \frac {3^2}{2^2}} = \sqrt {\frac {7}{4}}$. Now note that $EP = \frac {1}{2}$. By the pythagorean theorem, $OP = \sqrt {\frac {7}{4} + \frac {1^2}{2^2}} = \sqrt {2}$. Hence it now follows that,

$\frac {AP}{BP} = \frac {AO + OP}{BO - OP} = \frac {2 + \sqrt {2}}{2 - \sqrt {2}} = 3 + 2\sqrt {2}$

This gives that the answer is $\boxed{007}$.

See also

2010 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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