The situation is shown in the picture below. The radius we seek is $SD = AD + AS$. Clearly $AD=1$. The point $S$ is the center of the equilateral triangle $ABC$, thus $AS$ is $2/3$ of the altitude of this triangle. We get that $AS = \frac23 \cdot \sqrt 3$. Therefore the radius we seek is
$1 + \frac23 \cdot \sqrt 3 = \boxed{\mathrm{(D)\ }\frac{3+2\sqrt{3}}3}$.
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?
$\mathrm{(A) \ } \frac{2 + \sqrt{6}}{3} \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } \frac{2 + 3\sqrt{2}}{2} \qquad \mathrm{(D) \ } \frac{3 + 2\sqrt{3}}{3} \qquad \mathrm{(E) \ } \frac{3 + \sqrt{3}}{2}$