2022 AMC 12A Problems/Problem 12

Problem

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

$\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}$

Solution

Let the side length of $ABCD$ be $2$ . Then, $CM = DM = \sqrt{3}$ . By the Law of Cosines, $\cos(\angle CMD) = (CM^2 + DM^2 - BC^2)/(2CMDM) = \boxed{\textbf{(B)} \, \frac13}.$

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2022 AMC 12A Problems/Problem 12

Problem

Solution