SECTION IV REGULAR POLYHEDRONS 537. DEFINITION. A regular polyhedron is one whose faces are equal regular polygons and whose polyhedral angles are all equal. Since the polyhedral angles are all equal, the same number of faces and edges must meet at each vertex; and since the faces are all equal, the same number of vertices and edges must lie on each face. All the edges of a regular polyhedron are equal. The faces of a convex regular polyhedron are all convex polygons, and the polyhedral angles are all convex. 538. At each vertex there must meet at least three faces, and in each face there must lie at least three vertices. Moreover, the sum of the face angles at any vertex of a convex regular polyhedron is less than four right angles, or 360 degrees. (Art. 460.) From these two properties we are able to show that the greatest possible number of convex regular polyhedrons is five. The faces of a regular convex polyhedron must be regular convex polygons, equilateral triangles, squares, etc. 1. Since the interior angle of an equilateral triangle is 60 degrees, if the faces of a regular convex polyhedron are equilateral triangles, at each vertex there may meet three, four, or five faces; but not six. 2. Since the interior angle of a square is 90 degrees, if the faces of a regular convex polyhedron are squares, at each vertex there may meet three faces, but not four. 3. Since the interior angle of a regular pentagon is 108 degrees, if the faces of a regular convex polyhedron are pentagons, at each vertex there may meet three faces, but not four. 4. The interior angle of a regular hexagon is 120 degrees; hence it is impossible for three or more regular hexagons to meet at a vertex so as to form a convex polyhedral angle. Similarly no regular polygon of more than six sides can be a face of a regular convex polyhedron. Therefore the only polygons which can enter into the construction of regular convex polyhedrons are equilateral triangles, squares, and regular pentagons and the ways in which these can be combined are: (1) triangular faces meeting three at a vertex. (5) pentagonal faces meeting three at a vertex. What is here shown is that no other regular convex polyhedrons than the five answering the above conditions can exist; that these five varieties do exist can most easily be demonstrated by actually making models of them. Tetrahedron Hexahedron Octahedron Dodecahedron Icosahedron 539. The five regular polyhedrons are: 1. The regular tetrahedron, having four triangular faces meet ing three in a vertex. It has four vertices and six edges. 2. The regular octahedron, having eight triangular faces meeting four in a vertex. It has six vertices and twelve edges. 3. The regular icosahedron, having twenty triangular faces meeting five in a vertex. It has twelve vertices and thirty edges. 4. The regular hexahedron or cube, having six square faces meeting three in a vertex. It has eight vertices and twelve edges. 5. The regular dodecahedron, having twelve pentagonal faces meeting three in a vertex. It has twenty vertices and thirty edges. 540. Models of the five regular convex polyhedrons can easily be constructed by cutting pieces of cardboard in the shape of the diagrams below, and folding them along the dotted lines till the edges come together. The edges should then be glued or pasted over with strips of cloth or paper. To make the folding easy it would be well to cut the cardboard halfway through along the dotted lines. SECTION V POLYHEDRONS IN GENERAL PROPOSITION XX 541. In any polyhedron the number of edges increased by two is equal to the number of faces together with the number of vertices. [Euler's Theorem.] Let E be the number of edges in any given polyhedron, F the number of faces, and V the number of vertices. It is required to prove that E + 2 = V+ F. Proof. Imagine the polyhedron to be built up by taking one face and to it attaching another, and another, and so on, till the whole figure is completed. Let f be the number of faces of the incomplete figure at any stage, e the number of edges, and v the number of vertices. In any face taken alone, the number of vertices equals the number of edges. Now to the first face add a second. One edge of the second is made to coincide with an edge of the first, while two vertices of the second coincide with two vertices of the first. That is, in the new face by itself the number of edges equals the number of vertices, but when this face is attached to the one already taken, two vertices are lost, while only one edge is lost. Therefore, when ƒ = 2, e = v + 1. To these attach a third face, and again the number of vertices lost will exceed the number of edges lost by one. Therefore, when ƒ = 3, e = v + 2. Continue this process till all the faces are added but the last one. Then, when f= F− 1, e = v + (F-2), since the number of vertices lost for each of the F 2 faces added has exceeded the number of edges lost by one. When the last face is added, all of its edges and vertices are made to coincide with edges and vertices already counted. So that adding the last face merely completes the figure without changing the relation between the numbers of edges and vertices. Therefore, for the complete figure, 542. The sum of the face angles of any polyhedron, together with eight right angles, is equal to four times as many right angles as the polyhedron has vertices. Let F denote the number of faces, E the number of edges, V the number of vertices, and S the sum of the face angles of any polyhedron. It is required to prove that S +8 right angles = 4 V right angles. Proof. In any face, the sum of the face angles + 4 right angles twice as many right angles as the face has edges. = (Art. 116.) Therefore, adding for all the faces, remembering that each edge is counted twice, or S+4 F right angles 2 E x 2 right angles, = S=4(E-F) right angles |