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.