Difference between revisions of "2022 AMC 12A Problems/Problem 12"
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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 = \sqrt{3}</math>. | 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> | By the Law of Cosines, <cmath>\cos(\angle CMD) = \frac{CM^2 + DM^2 - CD^2}{2(CM)(DM)} = \boxed{\textbf{(B) } \frac13}.</cmath> | ||
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~jamesl123456 | ~jamesl123456 | ||
Revision as of 11:30, 28 December 2022
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 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 (Trigonometry)
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
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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