HMMT February

Geometry Difficulty 5.1 AIME, harder Find the answer

Two vertices of a cube are given in space. The locus of points that could be a third vertex of the cube is the union of $n$ circles. Find $n$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Let the distance between the two given vertices be 1. If the two given vertices are adjacent, then the other vertices lie on four circles, two of radius 1 and two of radius $\sqrt{2}$ . If the two vertices are separated by a diagonal of a face of the cube, then the locus of possible vertices adjacent to both of them is a circle of radius $\frac{1}{2}$ , the locus of possible vertices adjacent to exactly one of them is two circles of radius $\frac{\sqrt{2}}{2}$ , and the locus of possible vertices adjacent to neither of them is a circle of radius $\frac{\sqrt{3}}{2}$ . If the two given vertices are separated by a long diagonal, then each of the other vertices lie on one of two circles of radius $\frac{\sqrt{2}}{3}$ , for a total of 10 circles.

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Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.