hw6-solutions

# B these are each eight sided regions they can thus

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(b) These are each eight-sided 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 hexagon-with-a-hole 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 (5-sided polygons) and a regular decagon (10-sided 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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