2010 AIME II Problems/Problem 15

Problem 15.

In triangle $ABC$ , $AC = 13$ , $BC = 14$ , and $AB=15$ . Points $M$ and $D$ lie on $AC$ with $AM=MC$ and $\angle ABD = \angle DBC$ . Points $N$ and $E$ lie on $A$ B with $AN=NB$ and $\angle ACE = \angle ECB$ . Let $P$ be the point, other than $A$ , of intersection of the circumcircles of $\triangle AMN$ and $\triangle ADE$ . Ray $AP$ meets $BC$ at $Q$ . The ratio $\frac{BQ}{CQ}$ can be written in the form $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m-n$ .

Solution.

Let $Y = MN \cap AQ$ . $\frac {BQ}{QC} = \frac {NY}{MY$ (Error compiling LaTeX. Unknown error_msg)} since $\triangle AMN \sim \triangle ACB$ . Since quadrilateral $AMYN$ is cyclic, $\triangle MYA \sim \triangle PYN$ and $\triangle MYP \sim \triangle AYN$ , yielding $\frac {YM}{YA} = \frac {MP}{AN}$ and $\frac {YA}{YN} = \frac {AM}{PN}$ . Multiplying these together yields $\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)$ .

$\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}$ . Also, $P$ is the center of spiral similarity of segments $MD$ and $FN$ , so $\triangle PMD \sim \triangle PNF$ . Therefore, $\frac {PN}{PM} = \frac {NE}{MD}$ , which can easily be computed by the angle bisector theorem to be $\frac {145}{117}$ . It follows that $\frac {BQ}{CQ} = \frac {13}{15} \cdot \frac {145}{117} = \frac {725}{507}$ , giving us an answer of $725 - 507 = \boxed{218}$ .

Note: Spiral similarities may sound complex, but they're really not. The fact that $\triangle PMD \sim \triangle PNF$ is really just a result of simple angle chasing.

Source: [1] by Zhero

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2010 AIME II Problems/Problem 15

Problem 15.

Solution.