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. ==
+
==Diagram==
  
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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<asy>
 +
size(250);
 +
defaultpen(fontsize(9pt));
 +
picture pic;
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pair A,B,C,D,E,M,N,P,Q;
 +
B=MP("B",origin, SW);
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C=MP("C", (12.5,0), SE);
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A=MP("A", IP(CR(C,10),CR(B,15)), dir(90));
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N=MP("N", (A+B)/2, dir(180));
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M=MP("M", midpoint(C--A), dir(70));
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D=MP("D", extension(B,incenter(A,B,C),A,C), dir(C-B));
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E=MP("E", extension(C,incenter(A,B,C),A,B), dir(90));
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P=MP("P", OP(circumcircle(A,M,N),circumcircle(A,D,E)), dir(-70));
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Q = MP("Q", extension(A,P,B,C),dir(-90));
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draw(B--C--A--B^^M--P--N^^D--P--E^^A--Q);
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draw(circumcircle(A,M,N), gray);
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draw(circumcircle(A,D,E), heavygreen);
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dot(A);dot(B);dot(C);dot(D);dot(E);dot(P);dot(Q);dot(M);dot(N);
 +
</asy>
  
<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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== Official Solution (MAA) ==
  
'''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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The Angle Bisector Theorem implies that <math>E</math> lies on <math>\overline{AN}</math> and <math>D</math> lies on <math>\overline{MC}</math> because <math>AE/EB = AC/BC < 1</math> and <math>AD/DC = AB/CB > 1</math>. The Angle Bisector Theorem furthermore implies
 +
<cmath>NE=AN-AE=\frac 12\cdot AB - \frac{AC}{AC+BC}\cdot AB=\frac 5{18}</cmath>
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and
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<cmath>MD = CM - CD =  \frac 12\cdot AC - \frac{BC}{BC+BA}\cdot AC = \frac{13}{58}.</cmath>
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Because <math>ANPM</math> is cyclic, <math>\angle ENP = \angle ANP = \angle PMD</math>. Because <math>AEPD</math> is cyclic, <math>\angle NEP = 180^\circ-\angle AEP = \angle MDP</math>. Because <math>\angle ENP =\angle PMD</math> and <math>\angle NEP = \angle MDP</math>, triangles <math>NEP</math> and <math>MDP</math> are similar. Hence
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<cmath>\frac{NE}{MD}=\frac{NP}{MP}.</cmath>
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Applying the Law of Sines to <math>\triangle ANP</math> and <math>\triangle AMP</math> gives
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<cmath>\frac{NE}{MD} = \frac{NP}{MP} = \frac{\sin  \angle NAP}{\sin  \angle PAM} = \frac{\sin  \angle BAQ}{\sin  \angle QAC}</cmath>
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and thus
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<cmath>\frac{\sin  \angle BAQ}{\sin  \angle QAC} = \frac{(\frac{5}{18})}{(\frac{13}{58})} = \frac{145}{117}.</cmath>
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Thus
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<cmath>\frac{BQ}{QC} = \frac{\textrm{Area}(ABQ)}{\textrm{Area}(ACQ)}  = \frac{\frac{1}{2}\cdot AB \cdot AQ \cdot \sin \angle BAQ}{\frac{1}{2}\cdot AC \cdot AQ \cdot \sin \angle QAC}= \frac{15}{13} \cdot \frac{145}{117} = \frac{725}{507},</cmath>
 +
and <math>m - n = 218</math>.
 +
 
 +
 
 +
 
 +
== 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>.
 +
 
 +
<math>\frac {AN}{AM} = \frac {\frac {AB}{2}}{\frac {AC}{2}} = \frac {15}{13}</math>.
 +
 
 +
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>.
 +
 
 +
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
 +
 +
== 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
 +
 +
==See Also==
 +
{{AIME box|year=2010|n=II|num-b=14|after=Last Problem}}
 +
{{MAA Notice}}

Revision as of 17:08, 16 January 2022

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 $m-n$.

Diagram

[asy] size(250); defaultpen(fontsize(9pt)); picture pic; pair A,B,C,D,E,M,N,P,Q; B=MP("B",origin, SW); C=MP("C", (12.5,0), SE); A=MP("A", IP(CR(C,10),CR(B,15)), dir(90)); N=MP("N", (A+B)/2, dir(180)); M=MP("M", midpoint(C--A), dir(70)); D=MP("D", extension(B,incenter(A,B,C),A,C), dir(C-B)); E=MP("E", extension(C,incenter(A,B,C),A,B), dir(90)); P=MP("P", OP(circumcircle(A,M,N),circumcircle(A,D,E)), dir(-70)); Q = MP("Q", extension(A,P,B,C),dir(-90)); draw(B--C--A--B^^M--P--N^^D--P--E^^A--Q); draw(circumcircle(A,M,N), gray); draw(circumcircle(A,D,E), heavygreen); dot(A);dot(B);dot(C);dot(D);dot(E);dot(P);dot(Q);dot(M);dot(N); [/asy]

Official Solution (MAA)

The Angle Bisector Theorem implies that $E$ lies on $\overline{AN}$ and $D$ lies on $\overline{MC}$ because $AE/EB = AC/BC < 1$ and $AD/DC = AB/CB > 1$. The Angle Bisector Theorem furthermore implies \[NE=AN-AE=\frac 12\cdot AB - \frac{AC}{AC+BC}\cdot AB=\frac 5{18}\] and \[MD = CM - CD =  \frac 12\cdot AC - \frac{BC}{BC+BA}\cdot AC = \frac{13}{58}.\] Because $ANPM$ is cyclic, $\angle ENP = \angle ANP = \angle PMD$. Because $AEPD$ is cyclic, $\angle NEP = 180^\circ-\angle AEP = \angle MDP$. Because $\angle ENP =\angle PMD$ and $\angle NEP = \angle MDP$, triangles $NEP$ and $MDP$ are similar. Hence \[\frac{NE}{MD}=\frac{NP}{MP}.\] Applying the Law of Sines to $\triangle ANP$ and $\triangle AMP$ gives \[\frac{NE}{MD} = \frac{NP}{MP} = \frac{\sin  \angle NAP}{\sin  \angle PAM} = \frac{\sin  \angle BAQ}{\sin  \angle QAC}\] and thus \[\frac{\sin  \angle BAQ}{\sin  \angle QAC} = \frac{(\frac{5}{18})}{(\frac{13}{58})} = \frac{145}{117}.\] Thus \[\frac{BQ}{QC} = \frac{\textrm{Area}(ABQ)}{\textrm{Area}(ACQ)}  = \frac{\frac{1}{2}\cdot AB \cdot AQ \cdot \sin \angle BAQ}{\frac{1}{2}\cdot AC \cdot AQ \cdot \sin \angle QAC}= \frac{15}{13} \cdot \frac{145}{117} = \frac{725}{507},\] and $m - n = 218$.


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
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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