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 $\dfrac{1}{2}(\sqrt{p}-q)$ where $p$ and $q$ are positive integers. Find $p+q$ .

Solution

Let the radius of the circles is $r$ . The longer dimension can be written as $14r$ , and by Pythagorean Theorem, we can find that the shorter dimension: $2r\sqrt{3}+2r$ .

Therefore, $\frac{14r}{2r\sqrt{3}+2r}=\frac{1}{2}(\sqrt{p}-q)$

After simplifying, we get $7\sqrt{3}-7=\sqrt{p}-q$ , which gives $p=147$ and $q=7$ . So, $p+q=154$

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
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2002 AIME I Problems/Problem 2

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