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Difference between revisions of "2022 AMC 12A Problems/Problem 12"
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==Solution==
 
==Solution==
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Let the side length of <math>ABCD</math> be <math>2</math>. Then, <math>CM = DM = \sqrt{3}</math>. By the Law of Cosines, <math></math>\cos(\angle CMD) = (CM^2 + DM^2 - BC^2)/(2CMDM) = \boxed{\textbf{(B)} \, \frac13}.
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Let the side length of <math>ABCD</math> be <math>2</math>. Then, <math>CM = DM = \sqrt{3}</math>. By the Law of Cosines, <cmath>\cos(\angle CMD) = (CM^2 + DM^2 - BC^2)/(2CMDM) = \boxed{\textbf{(B)} \, \frac13}.</cmath>

Revision as of 13:43, 12 November 2022

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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