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2022 AMC 12A Problems/Problem 12: Difference between revisions

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==Problem==
==Problem==


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


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


==Solution 1==
==Diagram==
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>
<asy>
/* Made by MRENTHUSIASM */
size(200);
import graph3;
import solids;


~ jamesl123456
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);


==Solution 2==
currentprojection=orthographic((-2,0,1));
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 <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>\frac{2}{3} - \frac13 = \boxed{\textbf{(B)} \, \frac13}.</cmath>


~ Misclicked
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 <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{MOD}.</math> Note that <math>MO=\frac13 MD=\frac{\sqrt{3}}{3}.</math>
 
In right <math>\triangle CMO,</math> we have <cmath>\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.</cmath>
~MRENTHUSIASM
 
==Solution 2 (Law of Cosines)==
Without loss of generality, let the edge-length of <math>ABCD</math> be <math>2.</math> It follows that <math>CM=DM=\sqrt3.</math>
 
By the Law of Cosines, <cmath>\cos(\angle CMD) = \frac{CM^2 + DM^2 - CD^2}{2(CM)(DM)} = \boxed{\textbf{(B) } \frac13}.</cmath>
~jamesl123456
 
==Solution 3 (Double Angle Identities)==
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
 
==Solution 4 (Vector Methods)==
Without loss of generality, let tetrahedron <math>ABCD</math> lie in three-dimensional space with vertices
<math>A(-1, 1, 1)</math>, <math>B(1, 1, -1)</math>, <math>C(1, -1, 1)</math>, and <math>D(-1, -1, -1)</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{MC} = \langle 1, -2, 1 \rangle</cmath>
 
We know that the dot product of two vectors equals the product of their magnitudes multiplied by the cosine of the angle between them:
 
<cmath>\vec{MD} \cdot \vec{MC} = (-1)(1) + (-2)(-2) + (-1)(1) = -1 + 4 - 1 = 2</cmath>
 
Compute the magnitudes:
 
<cmath>||\vec{MD}|| = \sqrt{(-1)^2 + (-2)^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}</cmath> 
<cmath>||\vec{MC}|| = \sqrt{(1)^2 + (-2)^2 + (1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}</cmath>
 
Then:
 
<cmath>
\vec{MD} \cdot \vec{MC} = ||\vec{MD}|| \cdot ||\vec{MC}|| \cdot \cos(\angle CMD)
</cmath>
 
<cmath>
2 = \sqrt{6} \cdot \sqrt{6} \cdot \cos(\angle CMD) = 6 \cos(\angle CMD)
</cmath>
 
Finally: <cmath>\cos(\angle CMD) = \frac{2}{6} = \boxed{\textbf{(B) } \frac13}.</cmath>
~TylerTrikowsky


==Video Solution 1 (Quick and Simple)==
==Video Solution 1 (Quick and Simple)==
Line 19: Line 87:


~Education, the Study of Everything
~Education, the Study of Everything
==Video Solution 1 (Smart and Simple)==
https://youtu.be/7yAh4MtJ8a8?si=9uWHOngb2PTMRpg8&t=2423
~Math-X


== 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 23:27, 28 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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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