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Difference between revisions of "2022 AMC 12A Problems/Problem 12"

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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 1==
+
==Solution 1 (Right Triangles)==
−
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}{2(CM)(DM)} = \boxed{\textbf{(B)} \, \frac13}.</cmath>
+
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{MD}.</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>
 +
 
 +
==Solution 2 (Trigonometry)==
 +
Without loss of generality, let the edge-length of <math>ABCD</math> be <math>2.</math> It follows that <math>CM = DM = \sqrt{3}</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
 
~jamesl123456
  
−
==Solution 2==
+
==Solution 3 (Trigonometry)==
−
As done above, let the side 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</math>-<math>60</math>-<math>90</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>\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</math>-<math>60</math>-<math>90</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

Revision as of 11:29, 28 December 2022

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 $\overline{CO}\perp\overline{MD}.$ Note that $MO=\frac13 MD=\frac{\sqrt{3}}{3}.$

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

Solution 2 (Trigonometry)

Without loss of generality, let the edge-length of $ABCD$ be $2.$ It follows that $CM = DM = \sqrt{3}$.

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 (Trigonometry)

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 $30$-$60$-$90$ 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

Video Solution 1 (Quick and Simple)

https://youtu.be/wKfL1hYJCaE

~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

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