Let S_1, S_2, and S_3 be three circles of uni | vedclass

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(B) Let the centers of the three circles form an equilateral triangle of side length $2$. The distance from the center of a circle to the side of the

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$2(1+\frac{1}{\sqrt{3}})$

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(B) Let the centers of the three circles form an equilateral triangle of side length $2$. The distance from the center of a circle to the side of the large triangle $ABC$ is $1$. The angle of the triangle $ABC$ is $60^{\circ}$ because the circles are arranged symmetrically.From the geometry,the base $BC$ consists of the distance between the centers of the two bottom circles $(2)$ plus the horizontal projections of the distances from the centers to the vertices $B$ and $C$. Since the radius is $1$ and the angle is $60^{\circ}$,the distance from the projection of the center to the vertex is $1 \cdot \cot(30^{\circ}) = \sqrt{3}$.Thus,$BC = \sqrt{3} + 2 + \sqrt{3} = 2(1+\sqrt{3})$.Since $ABC$ is an equilateral triangle with side $a = 2(1+\sqrt{3})$,the circumradius $R$ is given by $R = \frac{a}{2 \sin(60^{\circ})} = \frac{2(1+\sqrt{3})}{2 \cdot \frac{\sqrt{3}}{2}} = \frac{2(1+\sqrt{3})}{\sqrt{3}} = 2(\frac{1}{\sqrt{3}} + 1)$.

Let $S_1, S_2,$ and $S_3$ be three circles of unit radius which touch each other externally. The common tangents to each pair of circles are drawn and extended so that they intersect and form a triangle $ABC$ with circumradius $R$. Then $R$ is equal to

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