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 is the way I calculated whether the triangle faces would merge.
AB This is a picture of the two
/\ / pyramids seen from above the
/  \ /  square pyramid.
/  \ /  AEDP is the tetrahedron,
E+P  ABCDP is the square pyramid.
\  / \  Since the faces merge, AP and DP
\  / \  aren't really needed, but are shown
\/ \ here for clarity.
DC
To show that the faces merge, I have to prove that E, D, C, and P are planar. What better way to do this, then to show that EP and DC are parallel?
Now, without even calculating lengths. Imagine there are two congruent square pyramids side by side. One has the point P at its tip, and the other has point E. It is simple to prove that EP, ED, and EA are all the same lengths as the sides of the square pyramids. Therefore, AEDP is the tetrahedron that we want. It is also simple to prove that EP is parallel to DC. So E, D, C, and P are planar. QED
There are 5 faces in this new solid, rather than the more intuitve answer of 7.

Posted by Tristan
on 20041005 18:56:19 