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

Latest revision as of 20:00, 1 February 2025

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

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

Solution 1 (Linearity)

Define the function $f:\mathbb{R}^{2}\rightarrow\mathbb{R}$ by $f(X)=\text{Pow}_{(AMN)}(X)-\text{Pow}_{(ADE)}(X)$ for points $X$ in the plane. Then $f$ is linear, so $\frac{BQ}{CQ}=\frac{f(B)-f(Q)}{f(Q)-f(C)}$ . But $f(Q)=0$ since $Q$ lies on the radical axis of $(AMN)$ , $(ADE)$ thus $\frac{BQ}{CQ}=-\frac{f(B)}{f(C)}=-\frac{BN\cdot BA-BE\cdot BA}{CM\cdot CA-CD\cdot CA}$ Let $AC=b$ , $BC=a$ and $AB=c$ . Note that $BN=\tfrac{c}{2}$ and $CM=\tfrac{b}{2}$ because they are midpoints, while $BE=\frac{ac}{a+b}$ and $CD=\frac{ab}{a+c}$ by Angle Bisector Theorem. Thus we can rewrite this expression as $\begin{align*}&-\frac{\tfrac{c^{2}}{2}-\tfrac{ac^{2}}{a+b}}{\tfrac{b^{2}}{2}-\tfrac{ab^{2}}{a+c}} \\ =&-\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{\tfrac{1}{2}-\tfrac{a}{a+b}}{\tfrac{1}{2}-\tfrac{a}{a+c}}\right) \\ =&\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{a+c}{a+b}\right) \\ =&\left(\frac{225}{169}\right)\left(\frac{29}{27}\right) \\ =&~\frac{725}{507}\end{align*}$ so $m-n=\boxed{218}$ .

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 2

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 3

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 $E = (14:13:0)$ . 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

Solution 4 (Fast)

Observe that $P$ is the center of spiral symmetry of segments $DM$ and $EN$ .

Using the angle bisector theorem, we compute $DM = \frac{13}{58}$ and $EN = \frac{5}{18}$ . Hence, it follows that the side-length ratio of triangles $DMP$ to $ENP$ is $\frac{117}{145}$ (note that the two triangles are similar by the spiral symmetry). This implies that the ratio of the height from $P$ to $AC$ to the height from $P$ to $AB$ is $\frac{117}{145}$ , so we compute the area ratio $\frac{APC}{APB} = \frac{507}{725}$ .

From the above, we see that the barycentric coordinates of $P$ are of the form $(x : 507 : 725)$ . Hence, it follows that the point $Q$ has the coordinates $(0 : 507 : 725)$ , so $\frac{BQ}{CQ} = \frac{725}{507}$ and our answer is $\boxed{218}$ .

~hgomamogh

Video Solution by the SpreadTheMathLove

https://www.youtube.com/watch?v=gzmY6nbbphw

@@ Line 1: / Line 1: @@
-==Problem==
 == 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>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==
+<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>
+== Solution 1 (Linearity) ==
+Define the function <math>f:\mathbb{R}^{2}\rightarrow\mathbb{R}</math> by <cmath>f(X)=\text{Pow}_{(AMN)}(X)-\text{Pow}_{(ADE)}(X)</cmath> for points <math>X</math> in the plane. Then <math>f</math> is linear, so <math>\frac{BQ}{CQ}=\frac{f(B)-f(Q)}{f(Q)-f(C)}</math>. But <math>f(Q)=0</math> since <math>Q</math> lies on the radical axis of <math>(AMN)</math>, <math>(ADE)</math> thus <cmath>\frac{BQ}{CQ}=-\frac{f(B)}{f(C)}=-\frac{BN\cdot BA-BE\cdot BA}{CM\cdot CA-CD\cdot CA}</cmath> Let <math>AC=b</math>, <math>BC=a</math> and <math>AB=c</math>. Note that <math>BN=\tfrac{c}{2}</math> and <math>CM=\tfrac{b}{2}</math> because they are midpoints, while <math>BE=\frac{ac}{a+b}</math> and <math>CD=\frac{ab}{a+c}</math> by Angle Bisector Theorem. Thus we can rewrite this expression as <cmath>\begin{align*}&-\frac{\tfrac{c^{2}}{2}-\tfrac{ac^{2}}{a+b}}{\tfrac{b^{2}}{2}-\tfrac{ab^{2}}{a+c}} \\ =&-\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{\tfrac{1}{2}-\tfrac{a}{a+b}}{\tfrac{1}{2}-\tfrac{a}{a+c}}\right) \\ =&\left(\frac{c^{2}}{b^{2}}\right)\left(\frac{a+c}{a+b}\right) \\ =&\left(\frac{225}{169}\right)\left(\frac{29}{27}\right) \\ =&~\frac{725}{507}\end{align*}</cmath> so <math>m-n=\boxed{218}</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>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>.
+== Official Solution (MAA) ==
+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>
+and
+<cmath>MD = CM - CD =  \frac 12\cdot AC - \frac{BC}{BC+BA}\cdot AC = \frac{13}{58}.</cmath>
+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
+<cmath>\frac{NE}{MD}=\frac{NP}{MP}.</cmath>
+Applying the Law of Sines to <math>\triangle ANP</math> and <math>\triangle AMP</math> gives
+<cmath>\frac{NE}{MD} = \frac{NP}{MP} = \frac{\sin  \angle NAP}{\sin  \angle PAM} = \frac{\sin  \angle BAQ}{\sin  \angle QAC}</cmath>
+and thus
+<cmath>\frac{\sin  \angle BAQ}{\sin  \angle QAC} = \frac{(\frac{5}{18})}{(\frac{13}{58})} = \frac{145}{117}.</cmath>
+Thus
+<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 2 ==
+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>.
@@ Line 13: / Line 52: @@
 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 PNE \cong \angle PAM</math> and <math>\angle ANM \cong \angle APM</math>. So, <math>m\angle PNA = m\angle PNE + m\angle ANM = m\angle PAM + m\angle APM = 180-m\angle PMA</math> and <math>\angle PNA \cong \angle PMD</math>.
+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>.
@@ Line 19: / Line 58: @@
 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>.
+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
@@ Line 30: / Line 73: @@
 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 3 ==
+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>E = (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
+== Solution 4 (Fast) ==
+Observe that <math>P</math> is the center of spiral symmetry of segments <math>DM</math> and <math>EN</math>.
+Using the angle bisector theorem, we compute <math>DM = \frac{13}{58}</math> and <math>EN = \frac{5}{18}</math>. Hence, it follows that the side-length ratio of triangles <math>DMP</math> to <math>ENP</math> is <math>\frac{117}{145}</math> (note that the two triangles are similar by the spiral symmetry). This implies that the ratio of the height from <math>P</math> to <math>AC</math> to the height from <math>P</math> to <math>AB</math> is <math>\frac{117}{145}</math>, so we compute the area ratio <math>\frac{APC}{APB} = \frac{507}{725}</math>.
+From the above, we see that the barycentric coordinates of <math>P</math> are of the form <math>(x : 507 : 725)</math>. Hence, it follows that the point <math>Q</math> has the coordinates <math>(0 : 507 : 725)</math>, so <math>\frac{BQ}{CQ} = \frac{725}{507}</math> and our answer is <math>\boxed{218}</math>.
+~hgomamogh
+== Video Solution by the SpreadTheMathLove ==
+https://www.youtube.com/watch?v=gzmY6nbbphw
 ==See Also==
 {{AIME box|year=2010|n=II|num-b=14|after=Last Problem}}
+[[Category:Intermediate Geometry Problems]]
 {{MAA Notice}}

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