2010 AIME II Problems/Problem 14

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 $\boxed{007}$ .

2010 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 15
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2010 AIME II Problems/Problem 14

Problem 14

Solution

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