2022 AMC 12A Problem 12

Below is the professionally curated solution for Problem 12 of the 2022 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 12A solutions, or check the answer key.

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Concepts:3D geometry median (geometry)law of cosines

Difficulty rating: 1630

12.

Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD.$ What is $\cos(\angle CMD)?$

$\dfrac14$

$\dfrac13$

$\dfrac25$

$\dfrac12$

$\dfrac{\sqrt3}{2}$

Solution:

Take edge length $1.$ Since $M$ is the midpoint of $\overline{AB},$ segments $CM$ and $DM$ are altitudes of the equilateral faces, each of length $\dfrac{\sqrt3}{2}.$ Also $CD=1.$

By the Law of Cosines in $\triangle CMD,$ $\cos(\angle CMD)=\frac{\frac34+\frac34-1}{2\cdot\frac34}=\frac{1/2}{3/2}=\frac13.$

Thus, the correct answer is B.

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