HMMT February

Geometry Difficulty 5.6 AIME, harder Prove it United States

Problem:

A paper equilateral triangle of side length $2$ on a table has vertices labeled $A$ , $B$ , $C$ . Let $M$ be the point on the sheet of paper halfway between $A$ and $C$ . Over time, point $M$ is lifted upwards, folding the triangle along segment $BM$ , while $A$ , $B$ , and $C$ remain on the table. This continues until $A$ and $C$ touch. Find the maximum volume of tetrahedron $ABCM$ at any time during this process.

Solution

Solution:

View triangle $ABM$ as a base of this tetrahedron. Then relative to triangle $ABM$ , triangle $CBM$ rotates around segment $BM$ on a hinge. Therefore the volume is maximized when $C$ is farthest from triangle $ABM$ , which is when triangles $ABM$ and $CBM$ are perpendicular. The volume in this case can be calculated using the formula for the volume of a tetrahedron as
$\frac{1}{6} \cdot 1 \cdot 1 \cdot \sqrt{3} = \frac{\sqrt{3}}{6}.$

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