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

2022 AMC 12A Problem 12

Attempt Problem 12 of the 2022 AMC 12A below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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}

CM

and

DM

are medians of equilateral faces with edge length

1

CM=DM=\dfrac{\sqrt3}{2}

and

CD=1;

apply the Law of Cosines in

\triangle CMD

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,

\begin{aligned} \cos(\angle CMD) &=\frac{\frac34+\frac34-1}{2\cdot\frac34} \\ &=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac13. \end{aligned}

Thus, the correct answer is B.