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Problem 1033

AMC 12 late, AIME early

Geometry Difficulty 4.9 Prove it Harvard-MIT Mathematics Tournament · United States

A cube with side length $2$ is inscribed in a sphere. A second cube, with faces parallel to the first, is inscribed between the sphere and one face of the first cube. What is the length of a side of the smaller cube?

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:
First note that the long diagonal of the cube has length $2\sqrt{3}$ , so the radius of the sphere is $\sqrt{3}$ . Let $x$ be the side length of the smaller cube. Then the distance from the center of the sphere to the far face of the smaller cube is $1 + x$ , while the distance from the center of the far face to a vertex lying on the sphere is $\frac{x\sqrt{2}}{2}$ . Therefore, the square of the radius is
$3 = (1 + x)^2 + \frac{x^2}{2},$
or
$3x^2 + 4x - 4 = (3x - 2)(x + 2) = 0,$
so $x = \frac{2}{3}$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.