MATH 5210/7210 Exam 1 Solutions
February 25, 2008
1. Volumes. Suppose that a regular tetrahedron has volume 1. What is the volume of the truncated tetrahedron whose faces are regular hexagons and equilateral triangles? Hint: Compute the volume of one of the little tetrahedra that is cut off of a corner of the original tetrahedron.
The faces of the truncated tetrahedron are equilateral triangles and regular hexagons. To obtain a hexagonal face, three equilateral triangles are removed from a triangular face of the tetrahedron:
Thus the truncated tetrahedron is obtained by removing four little regular tetrahedra from the big tetrahedron, and the edge length of the little tetrahedra is 1/3 the edge length of the big tetrahedron. Thus the volume of a little tetrahedron is (1/3)3 times the volume of the big tetrahedron. So the volume of each little tetrahedron is 1/27, and the volume of the truncated tetrahedron is 1 - 4(1/27) = 23/27.
2. Dihedral angles. A pyramid has base a regular hexagon with edge length 1 and its sides are isosceles triangles with edge lengths 1, 2, 2. Compute the dihedral angle between the base and a side of the pyramid.
Let A be the apex of the pyramid, let B be the midpoint of an edge of the hexagon, and let C be the center of the hexagon. Then the plane ABC is perpendicular to the edge containing B, and so the dihedral angle of the pyramid at this edge is the angle ABC. Let D be a vertex of the edge containing B. Then ABD is a right triangle with hypotenuse AD = 2 and leg BD = 1/2, so the leg AB = √(22 - (1/2)2) = (√15)/2. Now ABC is a right triangle with hypotenuse AB = (√15)/2 and leg BC = (√3)/2. (BC is the altitude of the equilateral triangle obtained by joining the center C of the hexagon to the two endpoints of the edge containing B.) Thus cos(ABC) = [(√3)/2]/[(√15)/2] = √3/√15 = √5/5. So ABC = arccos(√5/5) = 63.4 degrees.
3. Semiregular polyhedra.
(a) What is the definition of a semiregular polyhedron?
A semiregular polyhedron is a convex polyhedron P such that all of the faces are regular and all of the vertices are the same. More precisely, (1) every face of P is a regular polygon, and not all faces of P are congruent. (Regular polyhedra are not semiregular.) And (2) if v is a vertex of P, let P(v) be the union of the faces of P that have v as a vertex. If v and w are vertices of P, then P(v) is congruent to P(w). Another way to express condition (2) is to say that the vertex symbols of v and w are equivalent.
(b) Define the vertex symbol of a semiregular polyhedron.
The vertex symbol of a semiregular polyhedron P gives the configuration of faces in order around a vertex of P. Let f1, f2, . . . , fq be the faces that have v as a vertex, and suppose that each of the pairs (f1,f2), (f2,f3), . . . , (fq,f1) has an edge in common. If the face fi has p
(c) What is the vertex symbol of a traditional soccer ball?
A soccer ball is a spherical truncated icosahedron. Each truncated vertex of the icosahedron becomes a pentagon, and each truncated face of the icosahedron becomes a hexagon. Thus the vertex symbol of a truncated icosahedron is 5.6.6.
(d) Which of the following symbols are vertex symbols of a semiregular polyhedron? Give reasons for your answers.
(i) 3.4.5.5 - No.
The sum of the face angles around the common vertex is 60 + 90 + 108 + 108 = 366 degrees. Since this sum is greater than 360 degrees, this symbol is not the vertex symbol of a convex polyhedron.
(ii) 3.4.3.4 - Yes.
The sum of the face angles is 60 + 90 + 60 + 90 = 300 degrees, and this is the vertex symbol of the cuboctahedron.
(iii) 3.3.4.4 - No.
The sum of the face angles is 60 + 60 + 90 + 90 = 300 degrees, but this configuration cannot be extended to a semiregular polyhedron. Suppose the polygons around a vertex v, in order, are triangle-triangle-square-square:
Consider the yellow triangle and the adjacent green square sharing vertex v. Let w be the other vertex shared by these two polygons. If the vertex symbol at w is also 3.3.4.4, there must be a triangle (dashed) adjacent to the yellow triangle at the vertex w. Now consider the third vertex x of the yellow triangle. There are three triangles at the vertex x, so the vertex symbol at x cannot be 3.3.4.4, and the vertex x cannot be the same as the vertex v.
4. Regular tilings.
(a) What is the definition of a regular tiling of the plane?
A tiling of the plane is a decomposition of the plane into polygons that meet along edges or at vertices. More precisely, a tiling of the plane is a set T of polygons in the plane, called tiles, such that the union of all the tiles is the whole plane, and for all pairs P and Q of tiles either (1) P and Q have empty intersection, (2) the intersection of P and Q is a vertex of both P and Q, (3) the intersection of P and Q is an edge of both P and Q, or (4) P = Q.
A tiling of the plane is regular if all the tiles are regular, and all the tiles are congruent.
(b) Prove that there are exactly three regular tilings of the plane.
If a tiling is regular, then there's an integer p ≥ 3 such that all the tiles are congruent regular p-gons. If α is the interior angle of a regular p-gon, then α = 180 - 360/p degrees. The sum of the angles around a vertex must be 360 degrees, so if there are q polygons around a vertex, then qα= 360.
Start with p = 3, equilateral triangles. Then α = 60, and q = 360/60 = 6.
Next consider p = 4, squares. Then α = 90, and q = 360/90 = 4.
For p = 5, regular pentagons, we have α = 108, and q = 360/108 = 6.67, which is not an integer, so tiling by pentagons is not possible.
For p = 6, regular hexagons, we have α = 120, and q = 360/120 = 3.
For p > 6 we have we have α > 120, and q < 360/120 = 3, so p > 6 is not possible, since there must be at least three polygons around each vertex.
Therefore the only possible regular tilings of the plane are with triangles, squares, or hexagons. All three of these types of tilings exist:
5. Euler’s formula. Suppose that every face of a convex polyhedron P has at least five sides. In other words, there are no triangular or quadrilateral faces. (It is not assumed that the faces are regular polygons.) Let V be the number of vertices of P, let E be the number of edges of P, and let F be the number of faces of P.
(a) State Euler’s formula relating V, E, and F. (Do not prove it.)
V - E + F = 2
(b) The numbers of vertices and edges are related by the inequality 2E ≥ 3V. Explain why this is true.
Every edge has two vertices, and at least 3 edges come in to each vertex, since there are at least three faces around each vertex. If we count all the ends of all the edges, each edge will be counted twice (once for each of its ends), and each vertex will be counted at least three times. Thus the number of ends is 2E, and the number of ends is at least 3V, so 2E ≥ 3V.
(c) For p = 5, 6, 7, . . . , let Fp be the number of faces that are p-gons. Thus F = F5 + F6 + F7 + . . . .
This just says that the total number of faces equals the number of pentagons plus the number of hexagons plus the number of septagons, etc.
(d) The numbers of faces and edges are related by the equation 2E = 5F5 + 6F6 + 7F7 + . . . . Explain why this is true.
Suppose we count the number of edges around each face and add all these numbers up. On the one hand, each edge will be counted twice, since it belongs to exactly two faces. On the other hand, each pentagon has 5 edges, so the number of edges of pentagons is 5F5 , each hexagon has 6 edges, so the number of edges of hexagons is 6F6 , etc., and the total number of edges we count is 5F5 + 6F6 + 7F7 + . . . .
(e) Combine equations (a), (b), (c), (d) to prove that F5 ≥ 12. In other words, the number of pentagons is at least 12.
2 = V - E + F (from a)
V ≤ 2/3 E (from b)
2 ≤ 2/3 E - E + F
2 ≤ -1/3 E + F
E = 1/2(5F5 + 6F6 + 7F7 + . . . ) (from d)
F = F5 + F6 + F7 + . . . (from c)
2 ≤ -1/6 (5F5 + 6F6 + 7F7 + . . . ) + (F5 + F6 + F7 + . . .)
12 ≤ -(5F5 + 6F6 + 7F7 + . . . ) + 6(F5 + F6 + F7 + . . .)
12 ≤ F5 + 0F6 - 1F7 - 2F8 - . . .
12 + 1F7 + 2F8 + . . . ≤ F5
12 ≤ F5