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

Revision as of 19:48, 14 November 2022

Problem

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

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

Solution 1

Let the side length of $ABCD$ be $2$ . Then, $CM = DM = \sqrt{3}$ . By the Law of Cosines, $\cos(\angle CMD) = \frac{CM^2 + DM^2 - CD^2}{2CMDM} = \boxed{\textbf{(B)} \, \frac13}.$

~ jamesl123456

Solution 2

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 $\sqrt{3}$ . 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 $\frac{2}{3} - \frac13 = \boxed{\textbf{(B)} \, \frac13}.$

~ Misclicked

Video Solution 1 (Quick and Simple)

https://youtu.be/wKfL1hYJCaE

~Education, the Study of Everything

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.

@@ Line 14: / Line 14: @@
 ~ Misclicked
+==Video Solution 1 (Quick and Simple)==
+https://youtu.be/wKfL1hYJCaE
+~Education, the Study of Everything
 == See Also ==
 {{AMC12 box|year=2022|ab=A|num-b=11|num-a=13}}
 {{MAA Notice}}

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