Difference between revisions of "2022 AMC 12A Problems/Problem 12"
MRENTHUSIASM (talk | contribs) (→Problem) |
MRENTHUSIASM (talk | contribs) (→Solution 1) |
||
Line 36: | Line 36: | ||
==Solution 1== | ==Solution 1== | ||
− | 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}{ | + | 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> |
~jamesl123456 | ~jamesl123456 |
Revision as of 15:57, 14 December 2022
Contents
[hide]Problem
Let be the midpoint of in regular tetrahedron . What is ?
Diagram
~MRENTHUSIASM
Solution 1
Let the side length of be . Then, . By the Law of Cosines,
~jamesl123456
Solution 2
As done above, let the side 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.