(b) These are each eightsided regions; they can thus each be guarded with [8
/
3] = 2
guards. In the case of the top region, we have the triangulation and coloring
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
0
0
0
0
0
0
0
0
0
2
2
2
2
2
2
2
2
p
p
p
p
p
p
p
p
p
p
p
p
p
p
p
R
Y
B
B
R
Y
B
R
and we can guard this region by placing guards at the two points labeled
Y
. Reflecting this
over a horizontal line gives a triangulation and coloring of the bottom region.
(c) We can combine these two results to give a means of guarding the original region
using four guards. However, two of those guards will be in the same place! In fact, for the
original hexagonwithahole region we can position the guards as illustrated below:
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Y
Y
Y
3.
Draw a picture illustrating the fact that, although we can place 2 regular pentagons
(5sided polygons) and a regular decagon (10sided polygon) around a single vertex, we
cannot extend this to a tessellation of the place which has this arrangement at every vertex.
4.
Consider the truncated cube, which is obtained from the ordinary cube by cutting off
each corner. (See the picture on p. 370 of the text, or on Wikipedia.)
(a) How many faces does it have? Edges? Vertices? Your answer should refer back to
the (ordinary) cube to make the counting easier; recall that the ordinary cube has 6 faces,
12 edges, and 8 vertices.
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 Summer '09
 Lugo
 Math, Polyhedron, 120 degrees, Art Gallery Theorem, 72 degrees, Angle sum

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