Focus on Problem Solving 11

FOCUS ON PROBLEM SOLVING 11 One important problem-solving technique is the strategy of establishing subgoals. We use this strategy in the classification of the regular polyhedra. These are called Platonic solids because they were mentioned in the writings of Plato. The Regular Polyhedra A regular polygon is one in which all sides and all angles are equal. There are infinitely many regular polygons, as indicated in Figure 1. Triangle Square Pentagon Hexagon Heptagon Octagon Figure 1 For three-dimensional shapes, the analogous concept is that of regular polyhedra. A regular polyhedron is a solid in which all faces are congruent regular polygons, and the same number of polygons meet at each corner, or vertex. We would like to find all possible regular polyhedra. It might seem at first that there should be infinitely many, just as for regular polygons. But we will see that then are just finitely many regular polyhedra. Tetrahedron Octahedron Cube Dodecahedron Icosahedron Figure 2 We now prove that there are exactly five regular polyhedra (see Figure 2). To show this, we first establish the subgoal of finding all polyhedra with faces which are equilateral triangles. If we can find these, maybe we can find the polyhedra whose faces are other regular polygons. Regular Polyhedra with Equilateral Triangle Faces Suppose the faces of a regular polyhedron are equilateral triangles. How many such Three, four, or five equilateral triangles can meet at a corner? To make a corner there must be at least triangles can be folded up to three triangles. We can also have four or five. But six equilateral make a corner, but six such triangles cannot meet at a point to make a corner. (Why?) If three triangles lie flat. 1 triangles meet at each vertex, we can complete the polyhedron by adding one more triangle to make a tetrahedron. If four triangles meet at each vertex, we have an octahedron. If five triangles meet at each vertex, the resulting regular polyhedron is an icosahedron. Thus we have found all the regular polyhedra with triangular faces. Regular Polyhedra with Square Faces Suppose the faces of a regular polygon are squares. If three squares meet at each point, then the polyhedron is a cube. It’s impossible for four or more squares to meet at a point to make a corner. (Why?) Thus, the only regular polyhedron with square faces is the cube. Regular Polyhedra with Regular Pentagon Faces Suppose the faces of a regular polygon are pentagons. If three pentagons meet at each vertex, the resulting polyhedron is a dodecahedron. Since the angles of a regular pentagon are 108°, it’s impossible for more than three regular pentagons to meet at a vertex. Thus, the only regular polyhedron with pentagonal faces is the dodecahedron. Regular Polyhedra with Other Regular Polygons for Faces Is it possible for the faces of a regular polygon to be regular hexagons? Since the angle of a regular hexagon is 120°, when three such hexagons meet at a point they do not form a corner, so it’s impossible for a regular polyhedron to have hexagonal faces. The same reasoning shows that no other regular polygon can be the face of a regular polyhedron. Euler’s Formula How many faces, edges, and vertices does a regular polyhedron have? In the 18th century, Euler observed that FEV− +=2 Euler’s Formula where F is the number of faces, E is the number of edges, and V is the number of vertices. We can use Euler’s Formula to answer the question. For example, the icosahedron is assembled from F equilateral triangles: In these triangles the total number of sides is 3F, and the total number of angles is also 3F. In an icosahedrons, five angles of these triangles meet to form a vertex, so the total number of vertices must be 2 3F V = 5 Since two sides of adjacent triangles meet to form one edge of a polyhedron, the number of edges must be 3F E = 2 Substituting into Euler’s Formula gives 33FF F − +=2 . 25 Solving gives F = 20 . Substituting this value of F into the formulas for edges and faces gives E = 30 andV =12. Thus for the icosahedron, F = 20 , E = 30 , and V =12. Using similar reasoning we can find the number of faces, edges, and vertices for the other regular polygons. PROBLEMS 1. Find the number of faces, edges, and vertices for each of the regular polyhedra, not by counting them, but by using Euler’s Formula. 2. Describe the polyhedron whose edges are the line segments joining the centers of the faces of an octahedron, as shown in the figure at the left. Do the same for the other Platonic solids. 3 3. As the following figures indicate, it’s possible to tile the plane (that is, completely cover it) with equilateral triangles and with squares. Find all other regular polygons that tile the plane. Prove your answer. 4. A group of pulleys, all of radius 1, is fixed so that their centers form a convex n-gon of perimeter P. (The figure show the case n = 4 .) Find the lengths of the belt that fits around the pulleys. [Hint: Try fitting together the sectors of the pulleys that touch the belts.] 5. A pair of pulleys is connected by a betlt, as shown in the fgure. Find the length of the belt. (a) 4 (b) 6. Two circles of radius 1 are placed so that their centers are one unit apart. Find the area of the region common to both circles. 7. Suppose that the lengths of sides of a triangle are rational numbers. Prove that the cosine of each angle is a rational number. [Hint: Use the Law of Cosines.] 8. Verify the identity ()()()()a2222++=+− b c d ac bd 2 ad bc 2 (a) Write 13 as a sum of two squares. (b) Write 41 as a sum of two squares. (c) Express 533 as the sum of two squares in two different ways. [Hint: Factor 533 and use parts (a) and (b) together with the above identity.] 5.

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