2004 AMC 10B Problem 16

Question

Three circles of radius

1

are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?

\dfrac{2 + \sqrt{6}}{3}

2

\dfrac{2 + 3\sqrt{2}}{3}

\dfrac{3 + 2\sqrt{3}}{3}

\dfrac{3 + \sqrt{3}}{2}

The three small centers form an equilateral triangle of side

2

The large radius is

1

plus the distance from that triangle’s center to a vertex

The centers of the three unit circles form an equilateral triangle with side

2.

Its center is the center of the large circle.

For an equilateral triangle of side length

2,

the distance from its center to a vertex is

\dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}.

Adding the unit radius, the large radius is

1 + \dfrac{2\sqrt3}{3} = \dfrac{3 + 2\sqrt3}{3}.

Thus, the correct answer is D.

Problem 16 in other years