M25-08

Official Solution:

A sphere is inscribed in a cube with an edge of \(x\) centimeters. In terms of \(x\) what is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?

A. \(x(\sqrt{3}-1)\)


B. \(\frac{x}{2}\)


C. \(x(\sqrt{2}-1)\)


D. \(\frac{x}{2}(\sqrt{3}-1)\)


E. \(\frac{x}{2}(\sqrt{2}-1)\)


Consider an image of a sphere inscribed in a cube:



Let's assume that \(x=10\) centimeters.

Given that a sphere is inscribed in the cube, the edge length of the cube is equal to the sphere's diameter. Hence, \(Diameter=10\).

The cube's diagonal is then calculated as \(Diagonal=\sqrt{10^2+10^2+10^2}=10\sqrt{3}\).

The distance between the vertex of the cube and the surface of the sphere is equal to half of the difference between the diagonal and the diameter. Therefore, \(gap=\frac{Diagonal-Diameter}{2}=\frac{10*\sqrt{3}-10}{2}=5(\sqrt{3}-1)\).

Given that \(x=10\), we can express the gap as \(5(\sqrt{3}-1)=\frac{x}{2}(\sqrt{3} - 1)\).


Answer: D