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Unformatted text preview: So the number of 2dimensional faces of the ntriangle is the number of 2faces of the ( n 1)triangle, plus the number of edges. That is, letting F 2 ( n ) denote the number of 2faces of the ntriangle, F 2 ( n ) = F 2 ( n 1) + E ( n 1) and working backwards, F 2 ( n ) = 1 + 3 + 6 + ... + ( n 1) n/ 2 . This is in fact n ( n + 1)( n 1) / 6, but this is hard to guess. Similarly, 3faces of the ntriangle come from 3faces of the ( n 1)triangle and 2faces of the ( n 1)triangle. Letting F 3 ( n ) denote their number, F 3 ( n ) = F 3 ( n 1) + F 2 ( n 1) . In fact, all of these numbers are binomial coefficients, the numbers from Pascal’s triangle! This occurs since there is a face of an ntriangle corresponding to any set of its vertices we might pick. 2. Consider the five regular solids with their vertices cut off. These objects are called truncated solids . For each truncated solid, count the number of vertices, edges, and faces, and verify that the Euler characteristic is correct. (See p. 370 of the text for pictures. Also, note that the things you should be “counting” here include those vertices, edges, and faces which are hidden in the drawings.) (B+S 5.3.22) Solution. For the truncated tetrahedron, there are three vertices where each vertex of the original tetrahedron used to be. Thus there are 4 · 3 = 12 vertices. The six edges of the original tetrahedron remain, and in addition there are three “new” edges surrounding each vertex of the original tetrahedron, giving 6 + 4 · 3 = 18 edges. Finally, the number of faces is the number of faces of the original tetrahedron, plus one new face for each vertex, giving 4 + 4 = 8 faces. The Euler characteristic is therefore 12 18 + 8 = 2. This pattern holds in general. The number of vertices of a truncated solid is the number of vertices of the original solid, times the number of faces meeting at each vertex of the original solid. The number of edges of a truncated solid is the number of edges of the original solid, plus the number of vertices of the truncated solid which we have already computed. Finally, the number of faces of the truncated solid is the number of faces of the original solid, plus the number of vertices of the original solid. This gives the values solid vertices edges faces trunc. tetrahedron 4 · 3 = 12 6 + 4 · 3 = 18 4 + 4 = 8 trunc. cube 8 · 3 = 24 12 + 8 · 3 = 36 6 + 8 = 14 trunc. octahedron 6 · 4 = 24 12 + 6 · 4 = 36 8 + 6 = 14 trunc. dodecahedron 20 · 3 = 60 30 + 20 · 3 = 90 12 + 20 = 32 trunc. icosahedron 12 · 5 = 60 30 + 12 · 5 = 90 20 + 12 = 32 We can easily check that V E + F = 2 for each solid. Note that the numbers which are duplicated between the truncated cube and truncated octahedron, and between the 2 truncated dodecahedron and truncated icosahedron, are not duplicated by coincidence; this...
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 Summer '09
 Lugo
 Math, Angles, triangle, vertices, Cantor, Truncation, Archimedean solid

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