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Unformatted text preview: AMS 345/CSE 355 (Fall, 2010) Joe Mitchell COMPUTATIONAL GEOMETRY Homework Set # 5 – Solution Notes (1). [10 points] O’Rourke, problem 4, section 4.1.6, page 108. The cuboctahedron has F = 8 + 6 = 14 faces (one equilateral triangle face per corner of the original cube, plus one diamond face per original face of the cube). It has E = 6 · 4 = 24 edges. It has V = 12 vertices. Thus, we can easily verify Euler’s formula: F E + V = 14 24 + 12 = 2. (2). [21 points] (a). Assume there exists a 7edge polyhedron. We will arrive at a contradiction. Let V be the number of vertices, E = 7 the number of edges, and F the number of faces of the polyhedron. We know that each vertex of any polyhedron has degree at least 3 (this follows from the definition, since a vertex must lie at the intersection of 3 or more faces (it takes 3 planes to determine a point)). Thus, the sum of the vertex degrees, which equals 2 E = 14, is at least 3 V : 14 ≥ 3 V (which implies that the integer V must be at most 4). But the inequality E ≤ 3 V 6 (which holds for planar graphs) implies that 7 ≤ 3 V 6, so 13 ≤ 3 V , which contradicts the fact that 14 ≥ 3 V . Thus, a polyhedron with E = 7 cannot exist....
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This note was uploaded on 03/13/2011 for the course AMS 345 taught by Professor Mitchell,j during the Spring '08 term at SUNY Stony Brook.
 Spring '08
 Mitchell,J

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