Difference between revisions of "2004 AIME II Problems/Problem 12"
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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>. | 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>. | ||
− | 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 | + | 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 Pythagorean Theorem, |
<math>3^2 + y^2 = (r + 3)^2 \rightarrow y = \sqrt {r^2 + 6r}</math>. | <math>3^2 + y^2 = (r + 3)^2 \rightarrow y = \sqrt {r^2 + 6r}</math>. | ||
Revision as of 21:07, 8 December 2007
Problem
Let be an isosceles trapezoid, whose dimensions are
and
Draw circles of radius 3 centered at
and
and circles of radius 2 centered at
and
A circle contained within the trapezoid is tangent to all four of these circles. Its radius is
where
and
are positive integers,
is not divisible by the square of any prime, and
and
are relatively prime. Find
Solution
Let the radius of the center circle be and its center be denoted as
. The center obviously has an x-coordinate of
.
So line passes through the point of tangency of circle
and circle
. Now let
be height from the base of trapezoid to O. Now, from Pythagorean Theorem,
.
We use a similar argument with the line , and find the height from the top of the trapezoid to
,
, to be
.
Now , so we solve the equation
Solving this, we get
So .