Difference between revisions of "2022 AMC 12A Problems/Problem 12"
Oxymoronic15 (talk | contribs) (Created page with "==Problem== Let <math>M</math> be the midpoint of <math>AB</math> in regular tetrahedron <math>ABCD</math>. What is <math>\cos(\angle CMD)</math>? <math>\textbf{(A) } \frac1...") |
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<math>\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}</math> | <math>\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}</math> | ||
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| + | ==Solution== | ||
| + | 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}. | ||
Revision as of 12:43, 12 November 2022
Problem
Let 
be the midpoint of 
in regular tetrahedron 
. What is 
?
Solution
Let the side length of be 
. Then, 
. By the Law of Cosines, $$ (Error compiling LaTeX. Unknown error_msg)\cos(\angle CMD) = (CM^2 + DM^2 - BC^2)/(2CMDM) = \boxed{\textbf{(B)} \, \frac13}.
