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?

$\sqrt2$

$\dfrac{3}{2}$

$\dfrac{5}{3}$

$\sqrt3$

$2$

Solution:

The centers of the six radius- $1$ spheres form a regular hexagon of side length $2$ , so each is $2$ units from the large sphere's center. Hence the large sphere has radius $3$ .

Let the eighth sphere have radius $r$ , and let its center be distance $x$ from the large sphere's center. Internal tangency gives $x+r=3$ , so $x=3-r$ .

Using the right triangle between the large center, a small-sphere center, and the eighth-sphere center, $(r+1)^2=2^2+(3-r)^2$ . Thus $2r+1=13-6r$ , so $r=\frac32$ .

Thus, B is the correct answer.

2013 AMC 10A Problem 22

Solution:

Problem 22 in Other Years