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 $AB$ 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 $Y = MN \cap AQ$ . $\frac {BQ}{QC} = \frac {NY}{MY}$ since $\triangle AMN \sim \triangle ACB$ . Since quadrilateral $AMPN$ 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 * $\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)$ .

$\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}$ .

Now we claim that $\triangle PMD \sim \triangle PNE$ . To prove this, we can use cyclic quadrilaterals.

From $AMPN$ , $\angle PNE \cong \angle PAM$ and $\angle ANM \cong \angle APM$ . So, $m\angle PNA = m\angle PNE + m\angle ANM = m\angle PAM + m\angle APM = 180-m\angle PMA$ and $\angle PNA \cong \angle PMD$ .

From $ADPE$ , $\angle PDE \cong \angle PAE$ and $\angle EDA \cong \angle EPA$ . Thus, $m\angle MDP = m\angle PDE + m\angle EDA = m\angle PAE + m\angle EPA = 180-m\angle PEA$ and $\angle PDM \cong \angle PEN$ .

Thus, from AA similarity, $\triangle PMD \sim \triangle PNE$ .

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 * $\frac {BQ}{CQ} = \frac {15}{13} \cdot \frac {145}{117} = \frac {725}{507}$ , giving us an answer of $725 - 507 = \boxed{218}$ .

These two ratios are the same thing and can also be derived from the Ratio Lemma.

Ratio Lemma : $\frac{BD}{DC} = \frac{AB}{AC} \cdot \frac{\sin \angle BAD}{\sin \angle CAD}$ , for any cevian AD of a triangle ABC. For the sine ratios use Law of Sines on triangles APM and APN, $\frac{PM}{\sin \angle PAM}=\frac{AP}{\sin \angle AMP}=\frac{AP}{\sin \angle ANP}=\frac{PN}{\sin \angle PAN}$ . The information needed to use the Ratio Lemma can be found from the similar triangle section above.

Source: [1] by Zhero

Extension

The work done in this problem leads to a nice extension of this problem:

Given a $\triangle ABC$ and points $A_1$ , $A_2$ , $B_1$ , $B_2$ , $C_1$ , $C_2$ , such that $A_1$ , $A_2$ $\in BC$ , $B_1$ , $B_2$ $\in AC$ , and $C_1$ , $C_2$ $\in AB$ , then let $\omega_1$ be the circumcircle of $\triangle AB_1C_1$ and $\omega_2$ be the circumcircle of $\triangle AB_2C_2$ . Let $A'$ be the intersection point of $\omega_1$ and $\omega_2$ distinct from $A$ . Define $B'$ and $C'$ similarly. Then $AA'$ , $BB'$ , and $CC'$ concur.

This can be proven using Ceva's theorem and the work done in this problem, which effectively allows us to compute the ratio that line $AA'$ divides the opposite side $BC$ into and similarly for the other two sides.

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2010 AIME II Problems/Problem 15

Contents

Problem 15

Solution

Extension

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