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Difference between revisions of "2004 AIME II Problems/Problem 12"

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== Solution ==
 
== Solution ==
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Let the radius of the center circle be <math>r</math> and its center be denoted as <math>O</math>. The center obviously has an x-coordinate of <math>3</math>.  
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Let the radius of the center circle be <math>r</math> and its center be denoted as <math>O</math>.
  
−
So line <math>AO</math> passes through the point of tangency of circle <math>A</math> and circle <math>O</math>. Now let <math>y</math> be height from the base of trapezoid to O. Now, from Pythagoreum Theorem,  
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<center><asy>
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<math>3^2 + y^2 = (r + 3)^2 \rightarrow y = \sqrt {r^2 + 6r}</math>.
+
pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pen f = fontsize(8);
 +
 
 +
real r = (-60 + 48 * 3^.5)/23;
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pair A=(0,0), B=(6,0), D=(1, 24^.5), C=(5,D.y), O = (3,(r^2 + 6*r)^.5);
 +
D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(CR(A,3));D(CR(B,3));D(CR(C,2));D(CR(D,2));D(CR(O,r));
 +
D(O); D((3,0)--(3,D.y),d); D(A--O--D,d); MP("3",(3/2,0),S,f);MP("2",(2,D.y),N,f);
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</asy></center>  
  
−
We use a similar argument with the line <math>AD</math>, and find the height from the top of the trapezoid to <math>O</math>, <math>z</math>, to be <math>z = \sqrt {r^2 + 4r}</math>.
+
Clearly line <math>AO</math> passes through the point of tangency of circle <math>A</math> and circle <math>O</math>. Let <math>y</math> be the height from the base of the trapezoid to <math>O</math>. From the [[Pythagorean Theorem]],
 +
<cmath>3^2 + y^2 = (r + 3)^2 \Longrightarrow y = \sqrt {r^2 + 6r}.</cmath>
  
−
Now <math>y + z = 5</math>, so we solve the equation <math>\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = 5</math>
+
We use a similar argument with the line <math>DO</math>, and find the height from the top of the trapezoid to <math>O</math>, <math>z</math>, to be <math>z = \sqrt {r^2 + 4r}</math>.
  
−
Solving this, we get <math>r = \frac { - 60 + 48\sqrt {3}}{23}</math>  
+
Now <math>y + z</math> is simply the height of the trapezoid. Let <math>D'</math> be the foot of the [[perpendicular]] from <math>D</math> to <math>AB</math>; then <math>AD' = 3 - 2 = 1</math>. By the Pythagorean Theorem, <math>(AD')^2 + (DD')^2 = (AD)^2 \Longrightarrow DD' = \sqrt{24}</math> so we need to solve the equation <math>\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = \sqrt {24}</math>. We can solve this by moving one radical to the other side, and squaring the equation twice to end with a [[quadratic equation]].
  
−
So <math>k + m + n + p = 60 + 48 + 3 + 23 = \fbox{134}</math>.
+
Solving this, we get <math>r = \frac { - 60 + 48\sqrt {3}}{23}</math>, and the answer is <math>k + m + n + p = 60 + 48 + 3 + 23 = \boxed{134}</math>.
  
 
== See also ==
 
== See also ==
−
* [[2004 AIME II Problems/Problem 11 | Previous problem]]
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{{AIME box|year=2004|n=II|num-b=11|num-a=13}}
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* [[2004 AIME II Problems/Problem 13 | Next problem]]
 
−
* [[2004 AIME II Problems]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:21, 4 July 2013

Problem

Let $ABCD$ be an isosceles trapezoid, whose dimensions are $AB = 6, BC=5=DA,$and $CD=4.$ Draw circles of radius 3 centered at $A$ and $B,$ and circles of radius 2 centered at $C$ and $D.$ A circle contained within the trapezoid is tangent to all four of these circles. Its radius is $\frac{-k+m\sqrt{n}}p,$ where $k, m, n,$ and $p$ are positive integers, $n$ is not divisible by the square of any prime, and $k$ and $p$ are relatively prime. Find $k+m+n+p.$

Solution

Let the radius of the center circle be $r$ and its center be denoted as $O$.

[asy] pointpen = black; pathpen = black+linewidth(0.7); pen d = linewidth(0.7) + linetype("4 4"); pen f = fontsize(8); real r = (-60 + 48 * 3^.5)/23; pair A=(0,0), B=(6,0), D=(1, 24^.5), C=(5,D.y), O = (3,(r^2 + 6*r)^.5); D(MP("A",A)--MP("B",B)--MP("C",C,N)--MP("D",D,N)--cycle); D(CR(A,3));D(CR(B,3));D(CR(C,2));D(CR(D,2));D(CR(O,r)); D(O); D((3,0)--(3,D.y),d); D(A--O--D,d); MP("3",(3/2,0),S,f);MP("2",(2,D.y),N,f); [/asy]

Clearly line $AO$ passes through the point of tangency of circle $A$ and circle $O$. Let $y$ be the height from the base of the trapezoid to $O$. From the Pythagorean Theorem, \[3^2 + y^2 = (r + 3)^2 \Longrightarrow y = \sqrt {r^2 + 6r}.\]

We use a similar argument with the line $DO$, and find the height from the top of the trapezoid to $O$, $z$, to be $z = \sqrt {r^2 + 4r}$.

Now $y + z$ is simply the height of the trapezoid. Let $D'$ be the foot of the perpendicular from $D$ to $AB$; then $AD' = 3 - 2 = 1$. By the Pythagorean Theorem, $(AD')^2 + (DD')^2 = (AD)^2 \Longrightarrow DD' = \sqrt{24}$ so we need to solve the equation $\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = \sqrt {24}$. We can solve this by moving one radical to the other side, and squaring the equation twice to end with a quadratic equation.

Solving this, we get $r = \frac { - 60 + 48\sqrt {3}}{23}$, and the answer is $k + m + n + p = 60 + 48 + 3 + 23 = \boxed{134}$.

See also

2004 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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