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, 2023

Geometry Difficulty 5.2 AIME, harder Prove it United States

Problem:

Let $A$ and $B$ be points in space for which $AB = 1$ . Let $\mathcal{R}$ be the region of points $P$ for which $AP \leq 1$ and $BP \leq 1$ . Compute the largest possible side length of a cube contained within $\mathcal{R}$ .

Solution

Solution:

Let $h$ be the distance between the center of one sphere and the center of the opposite face of the cube. Let $x$ be the side length of the cube. Then we can draw a right triangle by connecting the center of the sphere, the center of the opposite face of the cube, and one of the vertices that make up that face. This gives us $h^{2} + \left(\frac{\sqrt{2} x}{2}\right)^{2} = 1$ .

Because the centers of the spheres are $1$ unit apart, $h = \frac{1}{2} x + \frac{1}{2}$ , giving us the quadratic
$\left(\frac{1}{2} x + \frac{1}{2}\right)^{2} + \left(\frac{\sqrt{2} x}{2}\right)^{2} = 1.$
Solving yields $x = \frac{\sqrt{10} - 1}{3}$ .

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