# 2022 AMC 12A Problems/Problem 12

## Problem

Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD$. What is $\cos(\angle CMD)$? $\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}$

## Diagram $[asy] /* Made by MRENTHUSIASM */ size(200); import graph3; import solids; triple A, B, C, D, M; A = (2/3*sqrt(3)*Cos(90),2/3*sqrt(3)*Sin(90),0); B = (2/3*sqrt(3)*Cos(210),2/3*sqrt(3)*Sin(210),0); D = (2/3*sqrt(3)*Cos(330),2/3*sqrt(3)*Sin(330),0); C = (0,0,2/3*sqrt(6)); M = midpoint(A--B); currentprojection=orthographic((-2,0,1)); draw(A--B--D); draw(A--D,dashed); draw(C--A^^C--B^^C--D); draw(C--M,red); draw(M--D,red+dashed); dot("A",A,A-D,linewidth(5)); dot("B",B,B-A,linewidth(5)); dot("C",C,C-M,linewidth(5)); dot("D",D,D-A,linewidth(5)); dot("M",M,M-C,linewidth(5)); [/asy]$ ~MRENTHUSIASM

## Solution 1 (Right Triangles)

Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $MC=MD=\sqrt3.$

Let $O$ be the center of $\triangle ABD,$ so $\overline{CO}\perp\overline{MOD}.$ Note that $MO=\frac13 MD=\frac{\sqrt{3}}{3}.$

In right $\triangle CMO,$ we have $$\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.$$ ~MRENTHUSIASM

## Solution 2 (Law of Cosines)

Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $CM=DM=\sqrt3.$

By the Law of Cosines, $$\cos(\angle CMD) = \frac{CM^2 + DM^2 - CD^2}{2(CM)(DM)} = \boxed{\textbf{(B) } \frac13}.$$ ~jamesl123456

## Solution 3 (Double Angle Identities)

As done above, let the edge-length equal $2$ (usually better than $1$ because we can avoid fractions when dropping altitudes). Notice that the triangle stated in the question has two side-lengths that are the altitudes of two equilateral triangles. By dropping the equilateral triangles' altitude and using $30^{\circ}$- $60^{\circ}$- $90^{\circ}$ properties, we find that the other two sides are equal to $\sqrt{3}$. Now by dropping the main triangle's altitude, we see it equals $\sqrt{2}$ from the Pythagorean Theorem. we can use the Double Angle Identities for Cosine. Doing this, we obtain $$\cos(\angle CMD) = \frac{2}{3} - \frac13 = \boxed{\textbf{(B) } \frac13}.$$ ~Misclicked

## Video Solution 1 (Quick and Simple)

~Education, the Study of Everything

## Video Solution 1 (Smart and Simple)

~Math-X

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 