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Difference between revisions of "2022 AMC 12A Problems/Problem 12"
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As done above, let the edge-length equal <math>2</math> (usually better than <math>1</math> 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 <math>30^{\circ}</math>-<math>60^{\circ}</math>-<math>90^{\circ}</math> properties, we find that the other two sides are equal to <math>\sqrt{3}</math>. Now by dropping the main triangle's altitude, we see it equals <math>\sqrt{2}</math> from the Pythagorean Theorem. we can use the Double Angle Identities for Cosine. Doing this, we obtain <cmath>\cos(\angle CMD) = \frac{2}{3} - \frac13 = \boxed{\textbf{(B) } \frac13}.</cmath>
 
As done above, let the edge-length equal <math>2</math> (usually better than <math>1</math> 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 <math>30^{\circ}</math>-<math>60^{\circ}</math>-<math>90^{\circ}</math> properties, we find that the other two sides are equal to <math>\sqrt{3}</math>. Now by dropping the main triangle's altitude, we see it equals <math>\sqrt{2}</math> from the Pythagorean Theorem. we can use the Double Angle Identities for Cosine. Doing this, we obtain <cmath>\cos(\angle CMD) = \frac{2}{3} - \frac13 = \boxed{\textbf{(B) } \frac13}.</cmath>
 
~Misclicked
 
~Misclicked
−
 
  
 
==Solution 4 (Vector Methods)==
 
==Solution 4 (Vector Methods)==
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Let point <math>M</math> be located at <math>(0, 1, 0)</math>.
 
Let point <math>M</math> be located at <math>(0, 1, 0)</math>.
  
−
<cmath>\vec{MD} = \langle -1, -2, -1 \rangle</cmath>,  
+
<cmath>\vec{MD} = \langle -1, -2, -1 \rangle</cmath>   
 
<cmath>\vec{MC} = \langle 1, -2, 1 \rangle</cmath>
 
<cmath>\vec{MC} = \langle 1, -2, 1 \rangle</cmath>
  
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<cmath>\vec{MD} \cdot \vec{MC} = (-1)(1) + (-2)(-2) + (-1)(1) = -1 + 4 - 1 = 2</cmath>
 
<cmath>\vec{MD} \cdot \vec{MC} = (-1)(1) + (-2)(-2) + (-1)(1) = -1 + 4 - 1 = 2</cmath>
  
−
Now compute the magnitudes:
+
Compute the magnitudes:
  
 
<cmath>||\vec{MD}|| = \sqrt{(-1)^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}</cmath>   
 
<cmath>||\vec{MD}|| = \sqrt{(-1)^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}</cmath>   
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</cmath>
 
</cmath>
  
−
So we result with <cmath>\cos(\angle CMD) = \frac{2}{6} = \boxed{\textbf{(B) } \frac13}.</cmath>
+
Finally: <cmath>\cos(\angle CMD) = \frac{2}{6} = \boxed{\textbf{(B) } \frac13}.</cmath>
 
~TylerTrikowsky
 
~TylerTrikowsky
  
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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}}
−
 
−
[[Category:Introductory Geometry Problems]]
 
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category:3D Geometry Problems]]

Latest revision as of 00:27, 29 July 2025

Problem

Let $M$ be the midpoint of $\overline{AB}$ in regular tetrahedron $ABCD$. What is $\cos(\angle CMD)$?

$\textbf{(A) } \frac14 \qquad \textbf{(B) } \frac13 \qquad \textbf{(C) } \frac25 \qquad \textbf{(D) } \frac12 \qquad \textbf{(E) } \frac{\sqrt{3}}{2}$

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] ~MRENTHUSIASM

Solution 1 (Right Triangles)

Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $MC=MD=\sqrt3.$

Let $O$ be the center of $\triangle ABD,$ so $\overline{CO}\perp\overline{MOD}.$ Note that $MO=\frac13 MD=\frac{\sqrt{3}}{3}.$

In right $\triangle CMO,$ we have \[\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.\] ~MRENTHUSIASM

Solution 2 (Law of Cosines)

Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $CM=DM=\sqrt3.$

By the Law of Cosines, \[\cos(\angle CMD) = \frac{CM^2 + DM^2 - CD^2}{2(CM)(DM)} = \boxed{\textbf{(B) } \frac13}.\] ~jamesl123456

Solution 3 (Double Angle Identities)

As done above, let the edge-length equal $2$ (usually better than $1$ 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 $30^{\circ}$-$60^{\circ}$-$90^{\circ}$ properties, we find that the other two sides are equal to $\sqrt{3}$. Now by dropping the main triangle's altitude, we see it equals $\sqrt{2}$ from the Pythagorean Theorem. we can use the Double Angle Identities for Cosine. Doing this, we obtain \[\cos(\angle CMD) = \frac{2}{3} - \frac13 = \boxed{\textbf{(B) } \frac13}.\] ~Misclicked

Solution 4 (Vector Methods)

Without loss of generality, let tetrahedron $ABCD$ lie in three-dimensional space with vertices $A(-1, 1, 1)$, $B(1, 1, -1)$, $C(1, -1, 1)$, and $D(-1, -1, -1)$. Let point $M$ be located at $(0, 1, 0)$.

\[\vec{MD} = \langle -1, -2, -1 \rangle\] \[\vec{MC} = \langle 1, -2, 1 \rangle\]

We know that the dot product of two vectors equals the product of their magnitudes multiplied by the cosine of the angle between them:

\[\vec{MD} \cdot \vec{MC} = (-1)(1) + (-2)(-2) + (-1)(1) = -1 + 4 - 1 = 2\]

Compute the magnitudes:

\[||\vec{MD}|| = \sqrt{(-1)^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\] \[||\vec{MC}|| = \sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}\]

Then:

\[\vec{MD} \cdot \vec{MC} = ||\vec{MD}|| \cdot ||\vec{MC}|| \cdot \cos(\angle CMD)\]

\[2 = \sqrt{6} \cdot \sqrt{6} \cdot \cos(\angle CMD) = 6 \cos(\angle CMD)\]

Finally: \[\cos(\angle CMD) = \frac{2}{6} = \boxed{\textbf{(B) } \frac13}.\] ~TylerTrikowsky

Video Solution 1 (Quick and Simple)

https://youtu.be/wKfL1hYJCaE

~Education, the Study of Everything

Video Solution 1 (Smart and Simple)

https://youtu.be/7yAh4MtJ8a8?si=9uWHOngb2PTMRpg8&t=2423

~Math-X

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

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