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Difference between revisions of "2010 AIME II Problems/Problem 15"

(Created page with '== '''Problem 15.''' == In triangle <math>ABC</math>, <math>AC = 13</math>, <math>BC = 14</math>, and <math>AB=15</math>. Points <math>M</math> and <math>D</math> lie on <math…')
 
m (→‎Solution.)
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== Solution. ==
 
== Solution. ==
  
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Let <math>Y = MN \cap AQ</math>. <math>\frac {BQ}{QC} = \frac {NY}{MY</math>} since <math>\triangle AMN \sim \triangle ACB</math>. Since quadrilateral <math>AMYN</math> is cyclic, <math>\triangle MYA \sim \triangle PYN</math> and <math>\triangle MYP \sim \triangle AYN</math>, yielding <math>\frac {YM}{YA} = \frac {MP}{AN}</math> and <math>\frac {YA}{YN} = \frac {AM}{PN}</math>. Multiplying these together yields <math>\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)</math>.
+
Let <math>Y = MN \cap AQ</math>. <math>\frac {BQ}{QC} = \frac {NY}{MY}</math> since <math>\triangle AMN \sim \triangle ACB</math>. Since quadrilateral <math>AMYN</math> is cyclic, <math>\triangle MYA \sim \triangle PYN</math> and <math>\triangle MYP \sim \triangle AYN</math>, yielding <math>\frac {YM}{YA} = \frac {MP}{AN}</math> and <math>\frac {YA}{YN} = \frac {AM}{PN}</math>. Multiplying these together yields <math>\frac {YN}{YM} = \left(\frac {AN}{AM}\right) \left(\frac {PN}{PM}\right)</math>.
  
 
<math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>. Also, <math>P</math> is the center of spiral similarity of segments <math>MD</math> and <math>FN</math>, so <math>\triangle PMD \sim \triangle PNF</math>. Therefore, <math>\frac {PN}{PM} = \frac {NE}{MD}</math>, which can easily be computed by the angle bisector theorem to be <math>\frac {145}{117}</math>. It follows that <math>\frac {BQ}{CQ} = \frac {13}{15} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>.
 
<math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>. Also, <math>P</math> is the center of spiral similarity of segments <math>MD</math> and <math>FN</math>, so <math>\triangle PMD \sim \triangle PNF</math>. Therefore, <math>\frac {PN}{PM} = \frac {NE}{MD}</math>, which can easily be computed by the angle bisector theorem to be <math>\frac {145}{117}</math>. It follows that <math>\frac {BQ}{CQ} = \frac {13}{15} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>.

Revision as of 21:06, 21 May 2010

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 $Y = MN \cap AQ$. $\frac {BQ}{QC} = \frac {NY}{MY}$ 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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