A regular tetrahedron has four equilateral triangles as faces. A regular square pyramid has four equilateral triangles and a square as faces. The faces of the tetrahedron are congruent to the triangular faces of the square pyramid.
A new polyhedron is created by gluing the tetrahedron and the square pyramid together at a triangular face so that the vertices of the triangles coincide. How many faces does this polyhedron have?
This problem appeared on the SAT's once, and the correct solution of 5 did
not appear among the choices. Only one student in the nation realized
this, and he wrote to the SAT board to tell them about their error.
To show that the new polyhedron has 5 faces, we need to show that two
triangular faces merge to form a single rhombus-shaped face. That
is, if the square pyramid is given by ABCDP and the tetrahedron is
given by ABPE, we need to show that A, D, P, and E are coplanar (B, C,
P, and E will be coplanar by symmetry).
Assume all sides lengths to be 1. We need to show that the
distance DE is equal to the length of the long diagonal of a
unit-length 60-120-60-120 rhombus, namely, sqrt(3). Let M and N
be the midpoints of AB and CD, respectively. Clearly, P, E, M,
and N are coplanar. In fact, computing distances, we have that
both NPM and PME are sqrt(3)/2 - sqrt(3)/2 - 1 triangles.
Thus, PEMN is a parallelogram, and the distance NE is given by twice
the length of the median from N in triangle NPM. This can be
found by the formula a(p^2+mn)=b^2m+c^2n for a cevian of length p
dividing the side of length a into segments of length m and n. We
find NE = 2sqrt(11/16) = sqrt(11)/2. We get the distance DE then
from the Pythagorean Theorem on right triangle DNE: magically, DE
= sqrt(NE^2+DN^2) = sqrt(11/4+1/4) = sqrt(3) as desired.
Solution
