2022 AMC 12A Problems/Problem 12
Contents
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
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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