2002 AIME I Problems/Problem 2
Problem
The diagram shows twenty congruent circles arranged in three rows and enclosed in a rectangle. The circles are tangent to one another and to the sides of the rectangle as shown in the diagram. The ratio of the longer dimension of the rectangle to the shorter dimension can be written as 
where 
and 
are positive integers. Find 
.
Solution
Let the radius of the circles be 
. The longer dimension of the rectangle can be written as 
, and by the Pythagorean Theorem, we find that the shorter dimension is 
.
Therefore, ![$\frac{14r}{2r\left(\sqrt{3}+1\right)}= \frac{7}{\sqrt{3} + 1} \cdot \left[\frac{\sqrt{3}-1}{\sqrt{3}-1}\right] = \frac{1}{2}\left(7\sqrt{3} - 7\right) = \frac{1}{2}\left(\sqrt{p}-q\right)$](http://latex.artofproblemsolving.com/b/a/c/bac1c28eb0b7334f0866a1e9a58d97956ee24c78.png)
. Thus we have 
and 
, so 
.
See also
| 2002 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
