Maths Olympiad Prep

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Geometry Difficulty 5.2 AIME, harder Find the answer

Let AA and BB be points in space for which AB=1A B=1. Let R\mathcal{R} be the region of points PP for which AP1A P \leq 1 and BP1B P \leq 1. Compute the largest possible side length of a cube contained within R\mathcal{R}.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Let hh be the distance between the center of one sphere and the center of the opposite face of the cube. Let xx 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 h2+(2x2)2=1h^{2}+\left(\frac{\sqrt{2} x}{2}\right)^{2}=1. Because the centers of the spheres are 1 unit apart, h=12x+12h=\frac{1}{2} x+\frac{1}{2}, giving us the quadratic (12x+12)2+(2x2)2=1\left(\frac{1}{2} x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{2} x}{2}\right)^{2}=1. Solving yields x=1013x=\frac{\sqrt{10}-1}{3}.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.