Difference between revisions of "2022 AMC 12A Problems/Problem 12"
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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== | + | ==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 <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 | ||
+ | |||
+ | ==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 == | ||
+ | {{AMC12 box|year=2022|ab=A|num-b=11|num-a=13}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 22:42, 25 October 2023
Contents
[hide]Problem
Let be the midpoint of in regular tetrahedron . What is ?
Diagram
~MRENTHUSIASM
Solution 1 (Right Triangles)
Without loss of generality, let the edge-length of be It follows that
Let be the center of so Note that
In right we have ~MRENTHUSIASM
Solution 2 (Law of Cosines)
Without loss of generality, let the edge-length of be It follows that
By the Law of Cosines, ~jamesl123456
Solution 3 (Double Angle Identities)
As done above, let the edge-length equal (usually better than 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 -- properties, we find that the other two sides are equal to . Now by dropping the main triangle's altitude, we see it equals from the Pythagorean Theorem. we can use the Double Angle Identities for Cosine. Doing this, we obtain ~Misclicked
Video Solution 1 (Quick and Simple)
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.