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Unformatted text preview: truncated dodecahedron and truncated icosahedron, are not duplicated by coincidence; this is yet another consequence of duality. 3. Three hollowed, triangular prisms are used to make a torus. Carefully count the num ber of vertices, faces, and edges for this prismatic torus. Compute the Euler characteristic for this torus. (B+S 5.3.29) Solution. This torus has nine vertices, six on the “outside” of the hole and three on the “inside”. It has nine faces, three arising from each of the three original triangular prisms. It has 18 edges – these can be decomposed into three triangles going around the torus the “long” way and three triangles going around the short way. So the Euler characteristic is V E + F = 9 18 + 9 = 0. 4. Carefully count the number of vertices, faces, and edges for a twoholed torus. One way to view this twoholed torus is as two copies of the torus in problem 3, with one side removed from each and then the open edges glued together. This operation is called the connected sum . Compute the Euler characteristic for this twoholed torus. (B+S 5.3.30) Solution. Glue two copies of the torus together along a foursided face. The number of vertices of the resulting torus is twice the number of vertices of the original torus, minus the number which disappear in the gluing operation. In the gluing operation four vertices disappear – four pairs of corresponding vertices on the two copies become single vertices. So there are 2 · 9 4 = 14 vertices. Similarly, four edges disappear in the gluing operation, so there are 2 · 18 4 = 32 edges. Finally, two faces are removed in the course of the operation, so there are 2 · 9 2 = 16 faces. The Euler characteristic is thus 14 32 + 16 = 2, as it is for any twoholed torus. 5. Recall that in class we constructed the Koch snowflake, a shape which had finite area but infinite perimeter, by starting with an equilateral triangle, replacing each edge with four edges of onethird the length in a certain way, and repeating this process infinitely many times. (See p. 435 of the text for a picture of what happens to one edge when we do this.) We can construct a similar “square Koch snowflake” by beginning with a square and replacing each edge with five edges of onethird the length, such that the three middle edges form a square. (Note that the boundary of this snowflake actually touches itself at various points.) The first few iterations of this appear as follows. Assume that the original square has side length 1. 3 (a) What is the length of the n th curve in this series? (Note that the n th curve in this series is made up of a large number of short segments of the same length; you should find the length of each short segment and the number of them.) Solution. The first curve consists of four segments; in each iteration after that there are five times as many curves, so there are a total of 4...
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 Summer '09
 Lugo
 Math, Angles, triangle, vertices, Cantor, Truncation, Archimedean solid

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