Difference between revisions of "2022 AMC 12A Problems/Problem 12"
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Without loss of generality, let the edge-length of <math>ABCD</math> be <math>2.</math> It follows that <math>MC=MD=\sqrt3.</math> | Without loss of generality, let the edge-length of <math>ABCD</math> be <math>2.</math> It follows that <math>MC=MD=\sqrt3.</math> | ||
| − | Let <math>O</math> be the center of <math>\triangle ABD,</math> so <math>\overline{CO}\perp\overline{ | + | Let <math>O</math> be the center of <math>\triangle ABD,</math> so <math>\overline{CO}\perp\overline{MOD}.</math> Note that <math>MO=\frac13 MD=\frac{\sqrt{3}}{3}.</math> |
| − | In <math>\triangle CMO,</math> we have <cmath>\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.</cmath> | + | In <math>\triangle CMO,</math> we have <cmath>\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.</cmath> |
| + | ~MRENTHUSIASM | ||
==Solution 2 (Trigonometry)== | ==Solution 2 (Trigonometry)== | ||
Revision as of 11:29, 28 December 2022
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 (Right Triangles)
Without loss of generality, let the edge-length of be 
It follows that 
Let 
be the center of 
so 
Note that 
In 
we have ![\[\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.\]](http://latex.artofproblemsolving.com/f/f/3/ff3b30b41c028481a7cafc83de73717d09b6abd7.png)
~MRENTHUSIASM
Solution 2 (Trigonometry)
Without loss of generality, let the edge-length of be It follows that 
.
By the Law of Cosines, ![\[\cos(\angle CMD) = \frac{CM^2 + DM^2 - CD^2}{2(CM)(DM)} = \boxed{\textbf{(B) } \frac13}.\]](http://latex.artofproblemsolving.com/1/e/0/1e04e444d3c7c7226d3fc5ef965d6f4ab3e3f4fe.png)
~jamesl123456
Solution 3 (Trigonometry)
As done above, let the edge-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 ![\[\cos(\angle CMD) = \frac{2}{3} - \frac13 = \boxed{\textbf{(B) } \frac13}.\]](http://latex.artofproblemsolving.com/3/9/c/39c30864d4fc8d75d5c06e69e0da3d2dcbc692b0.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. 
