Question

Six spheres of radius $1$ are positioned so that their centers are at the vertices of a regular hexagon of side length $2.$ The six spheres are internally tangent to a larger sphere whose center is the center of the hexagon. An eighth sphere is externally tangent to the six smaller spheres and internally tangent to the larger sphere. What is the radius of this eighth sphere?

$\sqrt{2}$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$\sqrt{3}$

$2$

Solution:

Each small center is $2$ from the center $O,$ and the small spheres have radius $1,$ so the large sphere has radius $3.$ Let the eighth sphere have radius $r$ and center $G$ at distance $x$ from $O;$ then $x + r = 3.$

Since $G$ is equidistant from two opposite hexagon vertices, $GO$ is perpendicular to the line to a vertex, and the Pythagorean Theorem gives $(r + 1)^2 = 2^2 + x^2 = 4 + (3 - r)^2.$

This simplifies to $2r + 1 = 13 - 6r,$ so $r = \tfrac32.$

Thus, the correct answer is B.

2013 AMC 12A Problem 18

Solution:

Problem 18 in Other Years