Difference between revisions of "2005 AIME II Problems/Problem 15"

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Let <math> w_1 </math> and <math> w_2 </math> denote the [[circle]]s <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest positive value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally [[tangent (geometry)|tangent]] to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math>
 
Let <math> w_1 </math> and <math> w_2 </math> denote the [[circle]]s <math> x^2+y^2+10x-24y-87=0 </math> and <math> x^2 +y^2-10x-24y+153=0, </math> respectively. Let <math> m </math> be the smallest positive value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally [[tangent (geometry)|tangent]] to <math> w_2 </math> and internally tangent to <math> w_1. </math> Given that <math> m^2=\frac pq, </math> where <math> p </math> and <math> q </math> are relatively prime integers, find <math> p+q. </math>
  
== Solution ==
+
== Solution 1 ==
 
Rewrite the given equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>.
 
Rewrite the given equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>.
  
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Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is <math>\boxed{169}</math>.
 
Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is <math>\boxed{169}</math>.
  
 +
== Solution 2 ==
 +
As above, we rewrite the equations as <math>(x+5)^2 + (y-12)^2 = 256</math> and <math>(x-5)^2 + (y-12)^2 = 16</math>. Let <math>F_1=(-5,12)</math> and <math>F_2=(5,12)</math>. If a circle with center <math>C=(a,b)</math> and radius <math>r</math> is externally tangent to <math>w_2</math> and internally tangent to <math>w_1</math>, then <math>CF_1=16-r</math> and <math>CF_2=4+r</math>. Therefore, <math>CF_1+CF_2=20</math>. In particular, the locus of points <math>C</math> that can be centers of circles must be an ellipse with foci <math>F_1</math> and <math>F_2</math> and major axis <math>20</math>.
 +
 +
Clearly, the minimum value of the slope <math>a</math> will occur when the line <math>y=ax</math> is tangent to this ellipse. Suppose that this point of tangency is denoted by <math>T</math>, and the line <math>y=ax</math> is denoted by <math>\ell</math>. Then we reflect the ellipse over <math>\ell</math> to a new ellipse with foci <math>F_1'</math> and <math>F_2'</math> as shown below.
 +
<center><asy>
 +
size(220);
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pair F1 = (-5, 12), F2 = (5, 12),C=(0,12);
 +
draw(circle(F1,16));
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draw(circle(F2,4));
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draw(ellipse(C,10,5*sqrt(3)));
 +
xaxis("$x$",Arrows);
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yaxis("$y$",Arrows);
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dot(F1^^F2^^C);
 +
 +
real l(real x) {return sqrt(69)*x/10;}
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path g=graph(l,-7,14);
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draw(g);
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draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3)));
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pair T=intersectionpoint(ellipse(C,10,5*sqrt(3)),(0,0)--(10,l(10)));
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dot(T);
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pair F1P=reflect((0,0),(10,l(10)))*F1;
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pair F2P=reflect((0,0),(10,l(10)))*F2;
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dot(F1P^^F2P);
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dot((0,0));
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label("$F_1$",F1,N,fontsize(9));
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label("$F_2$",F2,N,fontsize(9));
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label("$F_1'$",F1P,SE,fontsize(9));
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label("$F_2'$",F2P,SE,fontsize(9));
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label("$O$",(0,0),NW,fontsize(9));
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label("$\ell$",(13,l(13)),SE,fontsize(9));
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label("$T$",T,NW,fontsize(9));
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draw((0,0)--F1--F2--F2P--F1P--cycle);
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draw(F1--F2P^^F2--F1P);
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</asy></center>
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By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that <math>F_1</math>, <math>T</math>, and <math>F_2'</math> are collinear, and similarly, <math>F_2</math>, <math>T</math> and <math>F_1'</math> are collinear. Therefore, <math>OF_1F_2F_2'F_1'</math> is a pentagon with <math>OF_1=OF_2=OF_1'=OF_2'=13</math>, <math>F_1F_2=F_1'F_2'=10</math>, and <math>F_1F_2'=F_1'F_2=20</math>. Note that <math>\ell</math> bisects <math>\angle F_1'OF_1</math>. We can bisect this angle by bisecting <math>\angle F_1'OF_2</math> and <math>F_2OF_1</math> separately.
 +
 +
We proceed using complex numbers. Triangle <math>F_2OF_1'</math> is isosceles with side lengths <math>13,13,20</math>. The height of this from the base of <math>20</math> is <math>\sqrt{69}</math>. Therefore, the complex number <math>\sqrt{69}+10i</math> represents the bisection of \angle <math>F_1'OF_2</math>.
 +
 +
Similarly, using the 5-12-13 triangles, we easily see that <math>12+5i</math> represents the bisection of the angle <math>F_2OF_1</math>. Therefore, we can add these two angles together by multiplying the complex numbers, finding
 +
<cmath>\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.</cmath>
 +
Now the point <math>F_1</math> is given by the complex number <math>-5+12i</math>. Therefore, to find a point on line <math>\ell</math>, we simply subtract <math>\frac{1}{2}\angle F_1'OF_1</math>, which is the same as multiplying <math>-5+12i</math> by the conjugate of <math>(\sqrt{69}+10i)(12+5i)</math>. We find
 +
<cmath>(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).</cmath>
 +
In particular, note that the tangent of the argument of this complex number is <math>\sqrt{69}/10</math>, which must be the slope of the tangent line. Hence <math>a^2=69/100</math>, and the answer is <math>\boxed{169}</math>.
 
== See also ==
 
== See also ==
 
{{AIME box|year=2005|n=II|num-b=14|after=Last Question}}
 
{{AIME box|year=2005|n=II|num-b=14|after=Last Question}}

Revision as of 15:55, 1 August 2015

Problem

Let $w_1$ and $w_2$ denote the circles $x^2+y^2+10x-24y-87=0$ and $x^2 +y^2-10x-24y+153=0,$ respectively. Let $m$ be the smallest positive value of $a$ for which the line $y=ax$ contains the center of a circle that is externally tangent to $w_2$ and internally tangent to $w_1.$ Given that $m^2=\frac pq,$ where $p$ and $q$ are relatively prime integers, find $p+q.$

Solution 1

Rewrite the given equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$.

Let $w_3$ have center $(x,y)$ and radius $r$. Now, if two circles with radii $r_1$ and $r_2$ are externally tangent, then the distance between their centers is $r_1 + r_2$, and if they are internally tangent, it is $|r_1 - r_2|$. So we have

\begin{align*} r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\ 16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{align*}

Solving for $r$ in both equations and setting them equal, then simplifying, yields

\begin{align*} 20 - \sqrt{(x+5)^2 + (y-12)^2} &= \sqrt{(x-5)^2 + (y-12)^2} \\ 20+x &= 2\sqrt{(x+5)^2 + (y-12)^2} \end{align*}

Squaring again and canceling yields $1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.$

So the locus of points that can be the center of the circle with the desired properties is an ellipse.

[asy] size(220); pointpen = black; pen d = linewidth(0.7); pathpen = d;  pair A = (-5, 12), B = (5, 12), C = (0, 0); D(CR(A,16));D(CR(B,4));D(shift((0,12)) * yscale(3^.5 / 2) * CR(C,10), linetype("2 2") + d + red); D((0,30)--(0,-10),Arrows(4));D((15,0)--(-25,0),Arrows(4));D((0,0)--MP("y=ax",(14,14 * (69/100)^.5),E),EndArrow(4));  void bluecirc (real x) {  pair P = (x, (3 * (25 - x^2 / 4))^.5 + 12); dot(P, blue);  D(CR(P, ((P.x - 5)^2 + (P.y - 12)^2)^.5 - 4) , blue + d + linetype("4 4")); }  bluecirc(-9.2); bluecirc(-4); bluecirc(3); [/asy]

Since the center lies on the line $y = ax$, we substitute for $y$ and expand: \[1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.\]

We want the value of $a$ that makes the line $y=ax$ tangent to the ellipse, which will mean that for that choice of $a$ there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is $0$, so $(-96a)^2 - 4(3+4a^2)(276) = 0$.

Solving yields $a^2 = \frac{69}{100}$, so the answer is $\boxed{169}$.

Solution 2

As above, we rewrite the equations as $(x+5)^2 + (y-12)^2 = 256$ and $(x-5)^2 + (y-12)^2 = 16$. Let $F_1=(-5,12)$ and $F_2=(5,12)$. If a circle with center $C=(a,b)$ and radius $r$ is externally tangent to $w_2$ and internally tangent to $w_1$, then $CF_1=16-r$ and $CF_2=4+r$. Therefore, $CF_1+CF_2=20$. In particular, the locus of points $C$ that can be centers of circles must be an ellipse with foci $F_1$ and $F_2$ and major axis $20$.

Clearly, the minimum value of the slope $a$ will occur when the line $y=ax$ is tangent to this ellipse. Suppose that this point of tangency is denoted by $T$, and the line $y=ax$ is denoted by $\ell$. Then we reflect the ellipse over $\ell$ to a new ellipse with foci $F_1'$ and $F_2'$ as shown below.

[asy] size(220);  pair F1 = (-5, 12), F2 = (5, 12),C=(0,12); draw(circle(F1,16)); draw(circle(F2,4)); draw(ellipse(C,10,5*sqrt(3))); xaxis("$x$",Arrows); yaxis("$y$",Arrows); dot(F1^^F2^^C);  real l(real x) {return sqrt(69)*x/10;} path g=graph(l,-7,14); draw(g); draw(reflect((0,0),(10,l(10)))*ellipse(C,10,5*sqrt(3))); pair T=intersectionpoint(ellipse(C,10,5*sqrt(3)),(0,0)--(10,l(10))); dot(T); pair F1P=reflect((0,0),(10,l(10)))*F1; pair F2P=reflect((0,0),(10,l(10)))*F2; dot(F1P^^F2P); dot((0,0)); label("$F_1$",F1,N,fontsize(9)); label("$F_2$",F2,N,fontsize(9)); label("$F_1'$",F1P,SE,fontsize(9)); label("$F_2'$",F2P,SE,fontsize(9)); label("$O$",(0,0),NW,fontsize(9)); label("$\ell$",(13,l(13)),SE,fontsize(9)); label("$T$",T,NW,fontsize(9)); draw((0,0)--F1--F2--F2P--F1P--cycle); draw(F1--F2P^^F2--F1P); [/asy]

By the reflection property of ellipses (i.e., the angle of incidence to a tangent line is equal to the angle of reflection for any path that travels between the foci), we know that $F_1$, $T$, and $F_2'$ are collinear, and similarly, $F_2$, $T$ and $F_1'$ are collinear. Therefore, $OF_1F_2F_2'F_1'$ is a pentagon with $OF_1=OF_2=OF_1'=OF_2'=13$, $F_1F_2=F_1'F_2'=10$, and $F_1F_2'=F_1'F_2=20$. Note that $\ell$ bisects $\angle F_1'OF_1$. We can bisect this angle by bisecting $\angle F_1'OF_2$ and $F_2OF_1$ separately.

We proceed using complex numbers. Triangle $F_2OF_1'$ is isosceles with side lengths $13,13,20$. The height of this from the base of $20$ is $\sqrt{69}$. Therefore, the complex number $\sqrt{69}+10i$ represents the bisection of \angle $F_1'OF_2$.

Similarly, using the 5-12-13 triangles, we easily see that $12+5i$ represents the bisection of the angle $F_2OF_1$. Therefore, we can add these two angles together by multiplying the complex numbers, finding \[\text{arg}\left((\sqrt{69}+10i)(12+5i)\right)=\frac{1}{2}\angle F_1'OF_1.\] Now the point $F_1$ is given by the complex number $-5+12i$. Therefore, to find a point on line $\ell$, we simply subtract $\frac{1}{2}\angle F_1'OF_1$, which is the same as multiplying $-5+12i$ by the conjugate of $(\sqrt{69}+10i)(12+5i)$. We find \[(-5+12i)(\sqrt{69}-10i)(12-5i)=169(10+i\sqrt{69}).\] In particular, note that the tangent of the argument of this complex number is $\sqrt{69}/10$, which must be the slope of the tangent line. Hence $a^2=69/100$, and the answer is $\boxed{169}$.

See also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Question
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All AIME Problems and Solutions

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