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Difference between revisions of "2015 AIME II Problems/Problem 15"
(→‎Solution)
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==Solution==
 
==Solution==
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<asy>
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unitsize(35);
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draw(Circle((-1,0),1));
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draw(Circle((4,0),4));
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pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y;
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A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A);
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label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4));
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draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed);
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dot(O_1);dot(O_2);
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draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X);
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</asy>
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Call <math>O_1</math> and <math>O_2</math> the centers of circles <math>\mathcal{P}</math> and  <math>\mathcal{Q}</math>, respectively, and call <math>K</math> and <math>L</math> the feet of the altitudes from <math>B</math> to <math>O_1N</math> and <math>C</math> to <math>O_2N</math>, respectively. Extend <math>CB</math> and <math>O_2O_1</math> to meet at point <math>N</math>. Using the fact that <math>\triangle{O_1BN} \sim \triangle{O_2CN}</math> and setting <math>NO_1 = k</math>, we have that <math>\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}</math>. We can do some more length chasing using triangles similar to <math>OBN</math> to get that <math>AK = AL = \frac{24}{15}</math>, <math>BK = \frac{12}{15}</math>, and <math>CL = \frac{48}{15}</math>. Now, consider the circles <math>\mathcal{P}</math> and <math>\mathcal{Q}</math> on the coordinate plane, where <math>A</math> is the origin. If the line <math>\ell</math> through <math>A</math> intersects <math>\mathcal{P}</math> at <math>D</math> and <math>\mathcal{Q}</math> at <math>E</math>, then <math>4 \cdot DA = AE</math>. To verify this, notice that <math>\triangle{AO_1D} \sim \triangle{EO_2A}</math> from the fact that both triangles are isosceles with <math>\angle{O_1AD} \cong \angle{O_2AE}</math>, which are corresponding angles. Since <math>O_2A = 4\cdot O_1A</math>, we can conclude that <math>\triangle{AO_1D} \sim \triangle{EO_2A}</math>.
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 +
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Hence, we need to find the slope <math>m</math> of line <math>\ell</math> such that the perpendicular distance <math>n</math> from <math>B</math> to <math>AD</math> is four times the perpendicular distance <math>p</math> from <math>C</math> to <math>AE</math>. This will mean that the product of the bases and heights of triangles <math>ACE</math> and <math>DBA</math> will be equal, which in turn means that their areas will be equal. Let the line <math>\ell</math> have the equation <math>y = -mx \implies m + yx = 0</math>. Then, the coordinates of <math>B</math> are <math>\left(\frac{-24}{15}, \frac{-12}{15}\right)</math>, and the coordinates of <math>C</math> are <math>\left(\frac{24}{15}, \frac{-48}{15}\right)</math>. Using the point-to-line distance formula and the fact that <math>n = 4p</math>, we have <cmath>\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}</cmath> <cmath>\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.</cmath> Since <math>m</math> takes on a positive value, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have <cmath>\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.</cmath> Thus, the equation of <math>\ell</math> is <math>y = -\frac{3}{2}x</math>.
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 +
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The we can find the coordinates of <math>D</math> by finding the point <math>(x,y)</math> other than <math>A = (0,0)</math> where the circle <math>\mathcal{P}</math> intersects <math>\ell</math>. <math>\mathcal{P}</math> can be represented with the equation <math>(x + 1)^2 + y^2 = 1</math>, and substituting <math>y = -\frac{3}{2}x</math> into this equation yields <cmath>x^2 + 2x + 1 + \frac{9}{4}x^2 = 1 \implies \frac{13}{4}x^2 + 2x = 0 \implies x = 0, -\frac{8}{13}.</cmath> Discarding <math>x = 0</math>, the <math>y</math>-coordinate of <math>D</math> is  <math>-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}</math>. The distance from <math>D</math> to <math>A</math> is then <math>\sqrt{\left(\frac{-8}{13}\right)^2 + \left(\frac{12}{13}\right)^2} = \frac{4}{\sqrt{13}}.</math> The perpendicular distance from <math>B</math> to <math>AD</math> or the height of <math>\triangle{DBA}</math> is <math>\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.</math> Finally, the common area is <math>\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}</math>, and <math>m + n = 64 + 65 = \boxed{129}</math>.
  
 
==See also==
 
==See also==
 
{{AIME box|year=2015|n=II|num-b=14|after=Last Problem}}
 
{{AIME box|year=2015|n=II|num-b=14|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:44, 28 March 2015

Problem

Circles $\mathcal{P}$ and $\mathcal{Q}$ have radii $1$ and $4$, respectively, and are externally tangent at point $A$. Point $B$ is on $\mathcal{P}$ and point $C$ is on $\mathcal{Q}$ so that line $BC$ is a common external tangent of the two circles. A line $\ell$ through $A$ intersects $\mathcal{P}$ again at $D$ and intersects $\mathcal{Q}$ again at $E$. Points $B$ and $C$ lie on the same side of $\ell$, and the areas of $\triangle DBA$ and $\triangle ACE$ are equal. This common area is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution

[asy] unitsize(35); draw(Circle((-1,0),1)); draw(Circle((4,0),4)); pair A,O_1, O_2, B,C,D,E,N,K,L,X,Y; A=(0,0);O_1=(-1,0);O_2=(4,0);B=(-24/15,-12/15);D=(-8/13,12/13);E=(32/13,-48/13);C=(24/15,-48/15);N=extension(E,B,O_2,O_1);K=foot(B,O_1,N);L=foot(C,O_2,N);X=foot(B,A,D);Y=foot(C,E,A); label("$A$",A,NE);label("$O_1$",O_1,NE);label("$O_2$",O_2,NE);label("$B$",B,SW);label("$C$",C,SW);label("$D$",D,NE);label("$E$",E,NE);label("$N$",N,W);label("$K$",(-24/15,0.2));label("$L$",(24/15,0.2));label("$n$",(-0.8,-0.12));label("$p$",((29/15,-48/15)));label("$\mathcal{P}$",(-1.6,1.1));label("$\mathcal{Q}$",(6,4)); draw(A--B--D--cycle);draw(A--E--C--cycle);draw(C--N);draw(O_2--N);draw(O_1--B,dashed);draw(O_2--C,dashed); dot(O_1);dot(O_2); draw(rightanglemark(O_1,B,N,5));draw(rightanglemark(O_2,C,N,5));draw(C--L,dashed);draw(B--K,dashed);draw(C--Y);draw(B--X); [/asy]

Call $O_1$ and $O_2$ the centers of circles $\mathcal{P}$ and $\mathcal{Q}$, respectively, and call $K$ and $L$ the feet of the altitudes from $B$ to $O_1N$ and $C$ to $O_2N$, respectively. Extend $CB$ and $O_2O_1$ to meet at point $N$. Using the fact that $\triangle{O_1BN} \sim \triangle{O_2CN}$ and setting $NO_1 = k$, we have that $\frac{k+5}{k} = \frac{4}{1} \implies k=\frac{5}{3}$. We can do some more length chasing using triangles similar to $OBN$ to get that $AK = AL = \frac{24}{15}$, $BK = \frac{12}{15}$, and $CL = \frac{48}{15}$. Now, consider the circles $\mathcal{P}$ and $\mathcal{Q}$ on the coordinate plane, where $A$ is the origin. If the line $\ell$ through $A$ intersects $\mathcal{P}$ at $D$ and $\mathcal{Q}$ at $E$, then $4 \cdot DA = AE$. To verify this, notice that $\triangle{AO_1D} \sim \triangle{EO_2A}$ from the fact that both triangles are isosceles with $\angle{O_1AD} \cong \angle{O_2AE}$, which are corresponding angles. Since $O_2A = 4\cdot O_1A$, we can conclude that $\triangle{AO_1D} \sim \triangle{EO_2A}$.


Hence, we need to find the slope $m$ of line $\ell$ such that the perpendicular distance $n$ from $B$ to $AD$ is four times the perpendicular distance $p$ from $C$ to $AE$. This will mean that the product of the bases and heights of triangles $ACE$ and $DBA$ will be equal, which in turn means that their areas will be equal. Let the line $\ell$ have the equation $y = -mx \implies m + yx = 0$. Then, the coordinates of $B$ are $\left(\frac{-24}{15}, \frac{-12}{15}\right)$, and the coordinates of $C$ are $\left(\frac{24}{15}, \frac{-48}{15}\right)$. Using the point-to-line distance formula and the fact that $n = 4p$, we have \[\frac{|mx_B + 1(y_B) + 0|}{\sqrt{m^2 + 1}} = \frac{4|mx_C + 1(y_C) + 0|}{\sqrt{m^2 + 1}}\] \[\implies |mx_B + y_B| = 4|mx_C + y_C| \implies \left|\frac{-24m}{15} + \frac{-12}{15}\right| = 4\left|\frac{24m}{15} + \frac{-48}{15}\right|.\] Since $m$ takes on a positive value, we must switch the signs of all terms in this equation when we get rid of the absolute value signs. We then have \[\frac{6m}{15} + \frac{3}{15} = \frac{48}{15} - \frac{24m}{15} \implies 2m = 3 \implies m = \frac{3}{2}.\] Thus, the equation of $\ell$ is $y = -\frac{3}{2}x$.


The we can find the coordinates of $D$ by finding the point $(x,y)$ other than $A = (0,0)$ where the circle $\mathcal{P}$ intersects $\ell$. $\mathcal{P}$ can be represented with the equation $(x + 1)^2 + y^2 = 1$, and substituting $y = -\frac{3}{2}x$ into this equation yields \[x^2 + 2x + 1 + \frac{9}{4}x^2 = 1 \implies \frac{13}{4}x^2 + 2x = 0 \implies x = 0, -\frac{8}{13}.\] Discarding $x = 0$, the $y$-coordinate of $D$ is $-\frac{3}{2} \cdot -\frac{8}{13} = \frac{12}{13}$. The distance from $D$ to $A$ is then $\sqrt{\left(\frac{-8}{13}\right)^2 + \left(\frac{12}{13}\right)^2} = \frac{4}{\sqrt{13}}.$ The perpendicular distance from $B$ to $AD$ or the height of $\triangle{DBA}$ is $\frac{|\frac{3}{2}\cdot\frac{-24}{15} + \frac{-12}{15} + 0|}{\sqrt{\frac{3}{2}^2 + 1}} = \frac{\frac{48}{15}}{\frac{\sqrt{13}}{2}} = \frac{32}{5\sqrt{13}}.$ Finally, the common area is $\frac{1}{2}\left(\frac{32}{5\sqrt{13}} \cdot \frac{4}{\sqrt{13}}\right) = \frac{64}{65}$, and $m + n = 64 + 65 = \boxed{129}$.

See also

2015 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Problem
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