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2010 AIME II Problems/Problem 15

Revision as of 17:20, 5 January 2020 by Vvluo (talk | contribs) (→‎Solution)

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.

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.

See Also

2010 AIME II (Problems • Answer Key • Resources)
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