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

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== Problem 15 ==
 
== 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>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>A</math>B with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m-n</math>.
+
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>AC</math> with <math>AM=MC</math> and <math>\angle ABD = \angle DBC</math>. Points <math>N</math> and <math>E</math> lie on <math>AB</math> with <math>AN=NB</math> and <math>\angle ACE = \angle ECB</math>. Let <math>P</math> be the point, other than <math>A</math>, of intersection of the circumcircles of <math>\triangle AMN</math> and <math>\triangle ADE</math>. Ray <math>AP</math> meets <math>BC</math> at <math>Q</math>. The ratio <math>\frac{BQ}{CQ}</math> can be written in the form <math>\frac{m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m-n</math>.
  
== Solution ==
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== Solution 1 ==
  
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>AMPN</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>.
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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>AMPN</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>NE</math>, so <math>\triangle PMD \sim \triangle PNE</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 {15}{13} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>.
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<math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>.  
  
'''Note:''' Spiral similarities may sound complex, but they're really not. The fact that <math>\triangle PMD \sim \triangle PNE</math> is really just a result of simple angle chasing.
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Now we claim that <math>\triangle PMD \sim \triangle PNE</math>. To prove this, we can use cyclic quadrilaterals.
 +
 
 +
From <math>AMPN</math>, <math>\angle PNY \cong \angle PAM</math> and <math>\angle ANM \cong \angle APM</math>. So, <math>m\angle PNA = m\angle PNY + m\angle ANM = m\angle PAM + m\angle APM = 180-m\angle PMA</math> and <math>\angle PNA \cong \angle PMD</math>.
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 +
From <math>ADPE</math>, <math>\angle PDE \cong \angle PAE</math> and <math>\angle EDA \cong \angle EPA</math>. Thus, <math>m\angle MDP = m\angle PDE + m\angle EDA =  m\angle PAE + m\angle EPA = 180-m\angle PEA</math> and <math>\angle PDM \cong \angle PEN</math>.
 +
 
 +
Thus, from AA similarity, <math>\triangle PMD \sim \triangle PNE</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 {15}{13} \cdot \frac {145}{117} = \frac {725}{507}</math>, giving us an answer of <math>725 - 507 = \boxed{218}</math>.
 +
 
 +
*These two ratios are the same thing and can also be derived from the Ratio Lemma.
 +
Ratio Lemma :<math>\frac{BD}{DC} = \frac{AB}{AC} \cdot \frac{\sin \angle BAD}{\sin \angle CAD}</math>, for any cevian AD of a triangle ABC.
 +
For the sine ratios use Law of Sines on triangles APM and APN, <cmath>\frac{PM}{\sin \angle PAM}=\frac{AP}{\sin \angle AMP}=\frac{AP}{\sin \angle ANP}=\frac{PN}{\sin \angle PAN}</cmath>. The information needed to use the Ratio Lemma can be found from the similar triangle section above.
  
 
Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero
 
Source: [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1831745#p1831745] by Zhero
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 +
== Extension ==
 +
The work done in this problem leads to a nice extension of this problem:
 +
 +
Given a <math>\triangle ABC</math> and points <math>A_1</math>, <math>A_2</math>, <math>B_1</math>, <math>B_2</math>, <math>C_1</math>, <math>C_2</math>, such that <math>A_1</math>, <math>A_2</math> <math>\in BC</math>, <math>B_1</math>, <math>B_2</math> <math>\in AC</math>, and <math>C_1</math>, <math>C_2</math> <math>\in AB</math>, then let <math>\omega_1</math> be the circumcircle of <math>\triangle AB_1C_1</math> and <math>\omega_2</math> be the circumcircle of <math>\triangle AB_2C_2</math>. Let <math>A'</math> be the intersection point of <math>\omega_1</math> and <math>\omega_2</math> distinct from <math>A</math>. Define <math>B'</math> and <math>C'</math> similarly. Then <math>AA'</math>, <math>BB'</math>, and <math>CC'</math> 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 <math>AA'</math> divides the opposite side <math>BC</math> into and similarly for the other two sides.
 +
 +
== Solution 2 ==
 +
This problem can be solved with barycentric coordinates. Let triangle <math>ABC</math> be the reference triangle with <math>A=(1,0,0)</math>, <math>B=(0,1,0)</math>, and <math>C=(0,0,1)</math>. Thus, <math>N=(1:1:0)</math> and <math>M=(1:0:1)</math>. Using the Angle Bisector Theorem, we can deduce that <math>D=(14:0:15)</math> and <math>(14:13:0)</math>. Plugging the coordinates for triangles <math>ANM</math> and <math>AED</math> into the circle formula, we deduce that the equation for triangle <math>ANM</math> is <math>-a^2yz-b^2zx-c^2xy+(\frac{c^2}{2}y+\frac{b^2}{2}z)(x+y+z)=0</math> and the equation for triangle <math>AED</math> is <math>-a^2yz-b^2zx-c^2xy+(\frac{14c^2}{27}y+\frac{14b^2}{29}z)(x+y+z)=0</math>. Solving the system of equations, we get that <math>\frac{c^2y}{54}=\frac{b^2z}{58}</math>. This equation determines the radical axis of circles <math>ANM</math> and <math>AED</math>, on which points <math>P</math> and <math>Q</math> lie. Thus, solving for <math>\frac{z}{y}</math> gets the desired ratio of lengths, and <math>\frac{z}{y}=\frac{58c^2}{54b^2}</math> and plugging in the lengths <math>b=13</math> and <math>c=15</math> gets <math>\frac{725}{507}</math>. From this we get the desired answer of <math>725-507=\boxed{218}</math>.
 +
-wertguk
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 +
==See Also==
 +
{{AIME box|year=2010|n=II|num-b=14|after=Last Problem}}
 +
{{MAA Notice}}

Revision as of 14:40, 15 February 2021

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 1

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

Solution 2

This problem can be solved with barycentric coordinates. Let triangle $ABC$ be the reference triangle with $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$. Thus, $N=(1:1:0)$ and $M=(1:0:1)$. Using the Angle Bisector Theorem, we can deduce that $D=(14:0:15)$ and $(14:13:0)$. Plugging the coordinates for triangles $ANM$ and $AED$ into the circle formula, we deduce that the equation for triangle $ANM$ is $-a^2yz-b^2zx-c^2xy+(\frac{c^2}{2}y+\frac{b^2}{2}z)(x+y+z)=0$ and the equation for triangle $AED$ is $-a^2yz-b^2zx-c^2xy+(\frac{14c^2}{27}y+\frac{14b^2}{29}z)(x+y+z)=0$. Solving the system of equations, we get that $\frac{c^2y}{54}=\frac{b^2z}{58}$. This equation determines the radical axis of circles $ANM$ and $AED$, on which points $P$ and $Q$ lie. Thus, solving for $\frac{z}{y}$ gets the desired ratio of lengths, and $\frac{z}{y}=\frac{58c^2}{54b^2}$ and plugging in the lengths $b=13$ and $c=15$ gets $\frac{725}{507}$. From this we get the desired answer of $725-507=\boxed{218}$. -wertguk

See Also

2010 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Problem
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