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?
The two glued triangular faces immediately disappear, so the answer would seem to be 5+4-2 = 7 of the original faces showing. However, if the dihedral angles along matching edges add up to 180 degrees (in other words, are supplementary), the contiguous faces from each of the two original shapes will fuse to form one face: two equilateral triangles would produce one rhombus (a diamond shape), or a triangle and a square to make a pentagon(irregular). There are two edges where a triangle on the original tetrahedron abuts a triangle on the original pentahedron (square pyramid). There's one edge where a triangle from the tetrahedron abuts the square on the original pentahedron. As the dihedral angles around the square on the pentahedron differ from those between triangles, not all three of these pairs of faces can merge. But either the triangle-triangle face pairs can merge into diamonds or the triangle-square face pair can merge into a pentagon (or neither, depending on the dihedral angles).
The way I like to calculate dihedral angles is by erecting a sphere about one vertex and using spherical trigonometry on the triangle that is the intersection of the faces with the surface of the sphere (assuming 3 faces come together at a vertex--otherwise partition into triangles becomes necessary). The planar angles of the faces become sides of the spherical triangle, and the angles on the spherical triangle represent the dihedral angles.
Erecting a sphere about one of the vertices at the base of the square pyramid, the sides of the spherical triangle are 60, 60 and 90 degrees, from the two triangles and the square that meet there. Call the measure of its angles s, s and t (the triangle-square dihedral is s; the triangle-triangle dihedral angle is t).
By the law of cosines, for the triangle-triangle dihedral:
cos(90)=(cos(60))^2 + (sin(60))^2 * cos(t)
cos(t) = (0 - 1/4) / (3/4) = -1/3
and for the triangle-square dihedral:
cos(60)=cos(60)*cos(90) + sin(60)*sin(90)*cos(s)
cos(s) = (1/2 - 0) / (sqrt(3)/2) = 1/sqrt(3)
While for the tetrahedron, the dihedral angles are all the same, as each of the sides of any erected spherical triangle is 60 degrees. Letting d represent its dihedral angles, the law of cosines gives:
cos(60) = (cos(60))^2 + (sin(60))^2 * cos(d)
cos(d) = (1/2 - 1/4) / (3/4) = 1/3
We see that t and d are in fact supplementary as their cosines are the negative of each other. (They come out to be approx. 109.4712206344907 deg and 70.5287793655093 deg.) That is not the case with t and s, as in fact both those dihedrals are acute (with positive cosines), with the base dihedral on the square pyramid being 54.7356... deg).
So two faces of the tetrahedron merge respectively with two triangular faces of the square pyramid to form two rhombuses (diamonds). The resulting figure has a square base with equilateral triangles attached to one pair of opposing edges, with one tilted in and one tilted out. The two rhombuses meet at the top in a crest line parallel to two sides of the base. That crest line is one of the edges of the original tetrahedron. The fact that it is horizontal shows that the tetrahedron is exactly balanced on its edge which is lying on the ground coincident with the base of the square pyramid. That fact is further verified by seeing that half the dihedral of the tetrahedron is the complement of the triangle-square (base) dihedral of the original pentahedron, so the midplane of the original tetrahedron is vertical.
In fact you could add another identical square pyramid to the opposite side of the tetrahedron, and just extend the figure and still retain just 5 faces, with the base now being a rectangle of two squares put together, the two ends being two triangles from the end square pyramids, and the remaining faces being trapezoids, each formed from two triangles of the end square pyramids and an upside-down triangle of the tetrahedron wedged in between. You could go on alternating tetrahedra and square pyramids.
You could even put two or more such rows side by side, and then overlay with an upside down version to tesselate a space between two planes. Then adding more layers of these, you could tesellate space. In fact this could be done also with regular tetrahedra and octahedra, but then, a regular octahedron is just two square-to-square joined square pyramids.
I'm sure I've seen such one-layer configurations used (edges only, as struts) as something like an awning or marquee, with the triangular formations providing what Buckminster Fuller called tensegrity, and used (curved around by adjusting edge lengths) in geodesic domes.
Another way of calculating the dihedral angles:
For the square pyramid, draw a diagonal of the square base and then from each end, connect a line to the midpoint of one of the edges at which two triangles meet that extends to one of the other two points on the square. These other lines are altitudes of their triangles, and thus perpendicular to the edge, and so their angle represents the dihedral angle also. If the edges of the square pyramid have length 1, these edges of the newly formed triangle have length sqrt(3)/2, while the third side of the triangle (the diagonal of the square base) has length sqrt(2). Using the law of cosines for plane triangles,
a^2 = b^2 + c^2 - 2*b*c*cos(d)
2 = 3/4 + 3/4 - 2 * (3/4) * cos d
cos d = - (1/2) / (2 * 3/4) = -1/3
For the tetrahedron, we again use two triangle altitudes that meet at an edge, but the third side is an edge of the tetrahedron, of length 1, so we get:
1 = 3/4 + 3/4 - 2 * (3/4) * cos d
cos d = (1/2) / (2 * 3/4) = 1/3
which agree with the results from spherical trigonometry.
Posted by Charlie
on 2004-10-05 13:52:03