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

(→‎Solution 2 (Trigonometry))
(→‎Video Solution 1 (Smart and Simple))
 
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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>
 
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 right <math>\triangle CMO,</math> we have <cmath>\cos(\angle CMD)=\frac{MO}{MC}=\boxed{\textbf{(B) } \frac13}.</cmath>
 
~MRENTHUSIASM
 
~MRENTHUSIASM
  
 
==Solution 2 (Law of Cosines)==
 
==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 = \sqrt{3}</math>.
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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>
 
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 3 (Trigonometry)==
+
==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</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>
+
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
  
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~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 ==

Latest revision as of 22:42, 25 October 2023

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

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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