Difference between revisions of "2004 AIME II Problems/Problem 12"
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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>. | 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>. | ||
| − | Now <math>y + z = 5</math>, so we solve the equation <math>\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = \sqrt 24</math> | + | Now <math>y + z = 5</math>, so we solve the equation <math>\sqrt {r^2 + 4r} + \sqrt {r^2 + 6r} = \sqrt (24)</math> |
Solving this, we get <math>r = \frac { - 60 + 48\sqrt {3}}{23}</math> | Solving this, we get <math>r = \frac { - 60 + 48\sqrt {3}}{23}</math> | ||
Revision as of 16:42, 15 March 2008
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 
.
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 | ||
