2010 AIME II Problems/Problem 14

Revision as of 16:13, 2 April 2010 by SnowEverywhere (talk | contribs) (Created page with '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 <m…')
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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 $007$.

Invalid username
Login to AoPS