Difference between revisions of "2022 AMC 12A Problems/Problem 12"
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==Solution== | ==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, <cmath>\cos(\angle CMD) = \frac{CM^2 + DM^2 - BC^2}{2CMDM} = \boxed{\textbf{(B)} \, \frac13}.</cmath> | 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) = \frac{CM^2 + DM^2 - BC^2}{2CMDM} = \boxed{\textbf{(B)} \, \frac13}.</cmath> | ||
| + | |||
| + | ~ jamesl123456 | ||
| + | |||
| + | == See Also == | ||
| + | {{AMC12 box|year=2022|ab=A|num-b=11|num-a=13}} | ||
| + | {{MAA Notice}} | ||
Revision as of 12:44, 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, ![\[\cos(\angle CMD) = \frac{CM^2 + DM^2 - BC^2}{2CMDM} = \boxed{\textbf{(B)} \, \frac13}.\]](http://latex.artofproblemsolving.com/1/4/f/14f715efe242680e4cee8904a690f319cb98666c.png)
~ jamesl123456
See Also
| 2022 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 11 |
Followed by Problem 13 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
