Problem:
Let M be the midpoint of AB in regular tetrahedron ABCD. What is cos(∠CMD) ?
Answer Choices:
A. 41
B. 31
C. 52
D. 21
E. 23 Solution:
Without loss of generality, assume that the edges of the tetrahedron have length 2. Let O be the centroid of equilateral triangle △ABC. Then DO is perpendicular to the plane containing the △ABC. Applying the Pythagorean Theorem to △BMC gives CM=22−12=3. Similarly DM=3. Furthermore OM=31CM=313. Then
cos(∠CMD)=cos(∠OMD)=DMOM=3313=(B)31
OR
Without loss of generality, assume that CD=2. Then as above, CM=DM=3. Applying the Law of Cosines to △CMD gives
cos(∠CMD)=2⋅3⋅332+32−22=(B)31
Note: This angle measures the acute dihedral angle between plane ABC and plane ABD. Because the measure of this angle is not an integer divisor of 180∘ (it is about 70.53∘ ), it is impossible for regular tetrahedra to pack 3 -space, the way that cubes do or the way that equilateral triangles, squares, or regular hexagons pack 2 -space.
