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

m (Solution)
(fmt + asy)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
Let <math> w_1 </math> and <math> w_2 </math> denote the circles <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 possible value of <math> a </math> for which the line <math> y=ax </math> contains the center of a circle that is externally 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 possible 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 ==
Line 7: Line 7:
 
Let <math>w_3</math> have center <math>(x,y)</math> and radius <math>r</math>.  Now, if two circles with radii <math>r_1</math> and <math>r_2</math> are externally tangent, then the distance between their centers is <math>r_1 + r_2</math>, and if they are internally tangent, it is <math>|r_1 - r_2|</math>.  So we have
 
Let <math>w_3</math> have center <math>(x,y)</math> and radius <math>r</math>.  Now, if two circles with radii <math>r_1</math> and <math>r_2</math> are externally tangent, then the distance between their centers is <math>r_1 + r_2</math>, and if they are internally tangent, it is <math>|r_1 - r_2|</math>.  So we have
  
<math>r + 4 = \sqrt{(x-5)^2 + (y-12)^2}</math>
+
<cmath>\begin{align*}
 +
r + 4 &= \sqrt{(x-5)^2 + (y-12)^2} \\
 +
16 - r &= \sqrt{(x+5)^2 + (y-12)^2} \end{align*} </cmath>
  
<math>16 - r = \sqrt{(x+5)^2 + (y-12)^2}</math>
+
Solving for <math>r</math> in both equations and setting them equal, then simplifying, yields
  
Solving for <math>r</math> in both equations and setting them equal yields
+
<cmath>\begin{align*}
<math>20 - \sqrt{(x+5)^2 + (y-12)^2} = \sqrt{(x-5)^2 + (y-12)^2}</math>
+
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*} </cmath>
  
Squaring both sides, canceling common terms, and rearranging yields
+
Squaring again and canceling yields <math>1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.</math>
  
<math>20+x = 2\sqrt{(x+5)^2 + (y-12)^2}</math>
+
So the locus of points that can be the center of the circle with the desired properties is an [[ellipse]].
 
 
Squaring again and canceling yields <math>1 = \frac{x^2}{100} + \frac{(y-12)^2}{75}.</math>
 
  
So the locus of points that can be the center of the circle with the desired properties is an ellipse.
+
<center><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));
  
Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math>:
+
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"));
 +
}
  
<math>1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75}</math>
+
bluecirc(-9.2); bluecirc(-4); bluecirc(3);
 +
</asy></center>
  
Expanding yields <math>(3+4a^2)x^2 - 96ax + 276 = 0</math>.
+
Since the center lies on the line <math>y = ax</math>, we substitute for <math>y</math> and expand:
 +
<cmath>1 = \frac{x^2}{100} + \frac{(ax-12)^2}{75} \Longrightarrow (3+4a^2)x^2 - 96ax + 276 = 0.</cmath>
  
We want the value of <math>a</math> that makes the line <math>y=ax</math> tangent to the ellipse, which will mean that for that choice of <math>a</math> there is only one solution to the most recent equation. But a quadratic has one solution iff its discriminant is 0, so <math>(-96a)^2 - 4(3+4a^2)(276) = 0</math>.
+
We want the value of <math>a</math> that makes the line <math>y=ax</math> tangent to the ellipse, which will mean that for that choice of <math>a</math> there is only one solution to the most recent equation. But a quadratic has one solution [[iff]] its discriminant is <math>0</math>, so <math>(-96a)^2 - 4(3+4a^2)(276) = 0</math>.
  
Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is <math>169</math>.
+
Solving yields <math>a^2 = \frac{69}{100}</math>, so the answer is <math>\boxed{169}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:10, 25 July 2008

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 possible 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

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 $\boxed{169}$.

See also

2005 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Last Question
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
Invalid username
Login to AoPS