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

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m (→‎Solution 1)
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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>.
 
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>.
 +
 +
==Diagram==
 +
 +
<asy>
 +
size(300);
 +
defaultpen(fontsize(9pt));
 +
picture pic;
 +
pair A,B,C,D,E,M,N,P,Q;
 +
A=MP("A",origin,SW);
 +
B=MP("B", (15,0), SE);
 +
C=MP("C", IP(CR(A,13),CR(B,14)), dir(90));
 +
M=MP("N", (A+B)/2, SW*1.61);
 +
N=MP("M", (A+C)/2, dir(90)*2);
 +
D=MP("D", extension(B,incenter(A,B,C),A,C), 2*dir(220));
 +
E=MP("E", extension(C,incenter(A,B,C),A,B), SE*1.61);
 +
P=MP("P", IP(circumcircle(A,M,N),circumcircle(A,D,E)), dir(90));
 +
Q = MP("Q", extension(A,P,B,C),dir(45));
 +
draw(B--C--A--B--N--M--C--E--D--B^^A--Q);
 +
draw(circumcircle(A,M,N), purple);
 +
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>
  
 
== Solution 1 ==
 
== Solution 1 ==

Revision as of 13:42, 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(300); defaultpen(fontsize(9pt)); picture pic; pair A,B,C,D,E,M,N,P,Q; A=MP("A",origin,SW); B=MP("B", (15,0), SE); C=MP("C", IP(CR(A,13),CR(B,14)), dir(90)); M=MP("N", (A+B)/2, SW*1.61); N=MP("M", (A+C)/2, dir(90)*2); D=MP("D", extension(B,incenter(A,B,C),A,C), 2*dir(220)); E=MP("E", extension(C,incenter(A,B,C),A,B), SE*1.61); P=MP("P", IP(circumcircle(A,M,N),circumcircle(A,D,E)), dir(90)); Q = MP("Q", extension(A,P,B,C),dir(45)); draw(B--C--A--B--N--M--C--E--D--B^^A--Q); draw(circumcircle(A,M,N), purple); 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]

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 (Problems • Answer Key • Resources)
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