AMC10 2013 10A, Problem 22

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?

Solution: Set up an isosceles triangle between the center of the $8$th sphere and two opposite ends of the hexagon. Then set up another triangle between the point of tangency of the $7$th and $8$th spheres, and the points of tangency between the $7$th sphere and $2$ of the original spheres on opposite sides of the hexagon. Express each side length of the triangles in terms of $r$ (the radius of sphere $8$) and $h$ (the height of the first triangle). You can then use Pythagorean Theorem to set up two equations for the two triangles, and find the values of $h$ and $r$. $(1+r)^2=2^2+h^2$ $(3\sqrt{2})^2=3^2+(h+r)^2$ $r = \fbox{\frac{3}{2}}$

Answer: \frac{3}{2}