Difference between revisions of "2022 AMC 12A Problems/Problem 12"
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== See Also == | == See Also == | ||
{{AMC12 box|year=2022|ab=A|num-b=11|num-a=13}} | {{AMC12 box|year=2022|ab=A|num-b=11|num-a=13}} | ||
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| + | [[Category:Introductory Geometry Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:07, 15 November 2022
Problem
Let 
be the midpoint of 
in regular tetrahedron 
. What is 
?
Solution 1
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
Solution 2
As done above, let the side length equal 2 (usually better than one 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 30-60-90 properties, we find that the other two sides are equal to 
. Now by dropping the main triangle's altitude, we see it equals 
from the Pythagorean Theorem. we can use the Double Angle Identities for Cosine. Doing this, we obtain ![\[\frac{2}{3} - \frac13 = \boxed{\textbf{(B)} \, \frac13}.\]](http://latex.artofproblemsolving.com/e/8/c/e8cfc3ebec7820c7e3127e3051faeac3c8e43e80.png)
~ Misclicked
Video Solution 1 (Quick and Simple)
~Education, the Study of Everything
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. 
