## More than Three Polygons

1. In a previous problem, Three Polygons, I asked: Three regular polygons, all with unit sides, share a common vertex. Each polygon has a different number of sides, and each polygon shares a side with the other two; there are no gaps or overlaps. Find the number of sides for each polygon.
2. I now expand on that idea: If the restriction that each polygon has a different number of sides is removed, how many more solutions are there? There can be more than three polygons (up to six triangles). What are all the possible combinations?
3. In a plane, one set of polygons may be found at one vertex, while different sets could be found at others. There could be all triangles (1 unique polygon), or a combination of triangles and hexagons (2 unique polygons), and others. What is the largest number of unique polygons which can tile a plane?

Source: Original.

Solutions were received from Nick Baxter (nick@visigenic.com) and R. Barger (rbarger@erols.com):
1. 1/2 = 1/a + 1/b + 1/c
```        solutions are:
3,7,42
3,8,24
3,9,18
3,10,15
4,5,20
4,6,12
Technique is similar to latter 2 solutions given for previous problem
(actually, those solutions pretty much give away these, now that I've
```
2. for n regular polygons,
```[I insert some math to explain this solution:
The internal angles must sum to 360:
180(a-2)/a + 180(b-2)/b + ... = 360
n - 2(1/a + 1/b + ...) = 2     -KD]

n/2 - 1 = 1/a + 1/b + ... (n terms)
solutions are:
3,7,42
3,8,24
3,9,18
3,10,15
3,12,12
4,5,20
4,6,12
4,8,8
5,5,10
6,6,6
3,3,4,12
3,3,6,6
3,4,4,6
4,4,4,4
3,3,3,3,6
3,3,3,4,4
3,3,3,3,3,3
```
3. 4 unique polygons is the maximum. Each vertex in the plane must be one of the above. Consider a single vertex. Each neighboring vertex must share two of the same polygons. In the list, there are several which have unique pairs of polygons (no other set has any similar pair):
```3,7,42          3,8,24          3,9,18          3,10,15
4,5,20          4,8,8           5,5,10
```
If a tiling exists for any of the above sets, it can only include polygons from that set, for a maximum of three polygons in the plane. If there is a tiling that uses more, it must use the other sets of vertices, which only include polygons of sides 3, 4, 6, and 12.

Simply start with the tiling containing only vertices of the kind 4,6,12 and then replace one hexagon with 6 triangles. This gives a tiling containing each of the 4 kinds: 3,4,6,12.

Mail to Ken