2002 AIME I Problem 2

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Concepts:tangent circlesequilateral trianglerationalizing denominator

Difficulty rating: 2020

2.

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 12(pq),\frac{1}{2}\left(\sqrt{p} - q\right), where pp and qq are positive integers. Find p+q.p + q.

Solution:

Let rr be the common radius. The longer side holds a row of seven circles, so it equals 14r.14r. The centers of three mutually tangent circles in adjacent rows form an equilateral triangle with side 2r,2r, whose height is r3,r\sqrt{3}, so the two gaps between rows of centers contribute 2r3,2r\sqrt{3}, and the shorter side is r+2r3+r=2r+2r3.r + 2r\sqrt{3} + r = 2r + 2r\sqrt{3}.

The ratio is 14r2r(1+3)=71+3=7(31)2=12(1477),\frac{14r}{2r\left(1 + \sqrt{3}\right)} = \frac{7}{1 + \sqrt{3}} = \frac{7\left(\sqrt{3} - 1\right)}{2} = \frac{1}{2}\left(\sqrt{147} - 7\right), so p=147,p = 147, q=7,q = 7, and p+q=154.p + q = 154.

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