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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 Spring '08
 Mitchell,J
 edges, Delaunay triangulation, Voronoi diagram, Joe Mitchell, Delaunay diagram

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