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 - | + | 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 - CD^2}{2CMDM} = \boxed{\textbf{(B)} \, \frac13}.</cmath> |
~ jamesl123456 | ~ jamesl123456 | ||
Revision as of 13:45, 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, ![\[\cos(\angle CMD) = \frac{CM^2 + DM^2 - CD^2}{2CMDM} = \boxed{\textbf{(B)} \, \frac13}.\]](http://latex.artofproblemsolving.com/e/5/8/e58062990f1d3dcd349fc0c79e675e6bf637a48c.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. 
