How many faces does the resulting polyhedron have? Take a regular tetrahedron of edge one.
Also take a square-based pyramid, whose edges are all one (therefore the side faces are equilateral triangles of same size as the faces of the tetrahedron).
Glue a face of the tetrahedron to a triangular face of the pyramid so that their edges match up.
Considering the volume taken up by the two pieces as a single polyhedron, how many faces does it have?
 A: Recall that the regular tetrahedron is self-dual:  it is its own dual polyhedron, thus for a regular tetrahedron of edge length 2, consider its compound with its own dual such that both tetrahedra share the same circumradius.  The resulting compound is known as the stella octangula.  The intersection of the two tetrahedra (i.e., the region of space common to both) is a regular octahedron of edge length 1, and half of this octahedron as bisected by a plane perpendicular to a fourfold axis, forms the aforementioned square pyramid.  This pyramid, upon which a smaller regular tetrahedron of edge length 1 is attached, is the figure of interest.  But from this description, it becomes immediately obvious that two of the three faces of the small tetrahedron are coplanar with two of the triangular faces of the square pyramid, thus there are only 5 distinct faces to this polyhedron.
Explanatory figure taken from the MathWorld link above:

Here we see that the coplanarity is evident.
A: Let $ABCD$ be the base square of the pyramid, $S$ its tip, and $M$ the midpoint of $BS$. The segment $BS$ is a hinge connecting two equilateral triangles;  therefore the plane of the triangle $\triangle:=AMC$ intersects $BS$ orthogonally. It follows that the angle $\alpha:=\angle(AMC)$ is the angle between two adjacent walls of the pyramid. Using the cosine theorem one obtains
$$\cos\alpha={{3\over4}+{3\over4}-2\over 2\cdot {\sqrt{3}\over2}\cdot{\sqrt{3}\over2}}=-{1\over3}\ .$$
The angle $\beta$ between two faces of the tetrahedron is the angle at the tip of an isosceles triangle with sides ${\sqrt{3}\over2}$, ${\sqrt{3}\over2}$, and $1$; so
$$\cos\beta={{3\over4}+{3\over4}-1\over 2\cdot {\sqrt{3}\over2}\cdot{\sqrt{3}\over2}}={1\over3}\ .$$
It follows that $\alpha+\beta=\pi$. Therefore the resulting solid does not have $5+4-2=7$ faces, as expected, but only $5$ of them: the base square and one triangular side wall of the pyramid, two rhombi composed of a side wall of the pyramid and a facet of the tetrahedron, and one facet of the tetrahedron.
A: One way to think about this problem is as follows:  


*

*In your mind, take 2 pyramids, and place them together on a flat surface, square side down, with the edges touching.  Note that each of their 2 adjacent sides are parallel to the equivalent sides on the other pyramid, and are in the same plane.

*Draw a line between the tips.  

*Think of the length of that line.  It covers half of the square base of each pyramid, thus it is equal to the square base.  

*The shape formed from this line and the closest side of each pyramid is the tetrahedron described (the other 3 dimensions are defined by the triangular sides of the pyramids, which are also equal to the square base). The two added sides are in the same plane as those of both pyramids, forming a single shape with 6 sides.

*Now take off one of the pyramids.  It's two sides are now replaced with the 1 side where it was connected to the tetrahedron.  Thus, 5 sides.
A: I believe that it would be the number of faces of the tetrahedron ($4$), plus the number of faces of the pyramid ($5$), minus the two faces that got glued together since they will not be on the outside surface of the resulting polyhedron, so $7$ faces total.
A: I visualized the solution Nick provided:


(I found it because it had used this arrangement to build boats models with a magnet game)
A: I tried to do this with vectors and angles at first, but it got a bit hand-wavy and I couldn't nail down the details. Any alternative answer which does this more elegantly would be very welcome.
Let's work out the Cartesian coordinates of all the vertices.
For the sake of convenience and symmetry let's double the edge length to 2 and place the base of the pyramid at vertices: 
$A=(-1, -1, 0)$, $B=(-1, 1, 0)$, $C=(1, 1, 0)$, $D=(1, -1, 0)$
Then the apex of the pyramid will be at $E=(0, 0, h)$ for some value of $h$ which we can find with 3d Pythagorus:
$2^2 = DE^2 = 1^2 + 1^2 + h^2$
$4 = 2+h^2$
$h = \sqrt2$
Now place the tetrahedron against face
CDE, and call its fourth vertex F, located at (x, y, z).
By symmetry (equidistant from C and D) we can see that y must be 0; let's calculate x and z.
3d Pythagorus again gives:
$2^2 = EF^2 = x^2 + 0^2+ (z-\sqrt2)^2$ (1)
$2^2 = CF^2 = (x-1)^2 + 1^2 + z^2$ (2)
Rearranging (1):
$x = \sqrt{4 - (z-\sqrt2)^2}$
Substituting in (2) and rearranging (a messy exercise left to the reader!), it turns out the only sensible solution is $z = \sqrt2 = h, x = 2$
So EF is parallel to BC and B, C, E, F are coplanar. Instead of two triangular faces, these vertices form a single parallelogram face. Similarly for A, D, E, F.
So the resulting solid is a pentahedron comprising one square face, two triangular faces (one from the pyramid, one from the tetrahedron) and two parallelograms.
A: Here's a variation on your solution to the problem.
If you truncate the regular tetrahedron at the midpoints (instead of the 1/3 points) this gives a regular octahedron. The above problem looks at 1/2 of the octahedron (a square pyramid) and puts back one of the tetrahedra that were truncated and removed in the first step. The two faces were coplanar when they were together originally.
To see that the above truncation gives an octahedron notice that each face on the original tetrahedron loses corner triangles but keeps an "old" equilateral triangle. The removal of the corner tetrahedron gives a new equilateral triangular face. 4 old equilateral faces + 4 new equilateral faces yield an 8 faced solid which has 4 faces at each vertex--a regular octahedron.
