2022 AMC 12A Problems/Problem 12
Contents
Problem
Let 
be the midpoint of 
in regular tetrahedron 
. What is 
?
Diagram
![[asy] /* Made by MRENTHUSIASM */ size(200); import graph3; import solids; triple A, B, C, D, M; A = (2/3*sqrt(3)*Cos(90),2/3*sqrt(3)*Sin(90),0); B = (2/3*sqrt(3)*Cos(210),2/3*sqrt(3)*Sin(210),0); D = (2/3*sqrt(3)*Cos(330),2/3*sqrt(3)*Sin(330),0); C = (0,0,2/3*sqrt(6)); M = midpoint(A--B); currentprojection=orthographic((-2,0,1)); draw(A--B--D); draw(A--D,dashed); draw(C--A^^C--B^^C--D); draw(C--M,red); draw(M--D,red+dashed); dot("$A$",A,A-D,linewidth(5)); dot("$B$",B,B-A,linewidth(5)); dot("$C$",C,C-M,linewidth(5)); dot("$D$",D,D-A,linewidth(5)); dot("$M$",M,M-C,linewidth(5)); [/asy]](http://latex.artofproblemsolving.com/c/e/9/ce9c6f555508412bc11945246e1c2b33dd08bd80.png)
~MRENTHUSIASM
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 (usually better than 
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 
-
-
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. 
