Source: Original.
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 read them!).
[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
3,7,42 3,8,24 3,9,18 3,10,15 4,5,20 4,8,8 5,5,10If 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.