Doodling over lunch, I can say that the convexity constraint isn't a big deal. Starting with the regular tiling of the plane with hexagons, it's easy to convert the four hexagons touching a single edge into two pentagons and two heptagons, by "rotating the edge 90 degrees". Hmm, I'll try an ascii art rendition; apologies if this comes through muddled... _______ / \ / \ / \ _______/ \_______ / \ / \ / \ / \ / \ / \ _______/ \_______/ \_______ / \ / \ / \ / \ / \ / \ / \ / \ / \ / \_______/ \_______/ \ \ / \__ __/ \ / \ / \__ __/ \ / \ / \ / \ / \_______/ | \_______/ / \ | / \ / \ __/ \__ / \ / \ __/ \__ / \ / \_______/ \_______/ \ \ / \ / \ / \ / \ / \ / \ / \ / \ / \_______/ \_______/ \_______/ \ / \ / \ / \ / \ / \ / \_______/ \_______/ \ / \ / \ / \_______/ Anyway, that doesn't help us with the question of what your friend really wants. --Michael Kleber On Wed, Jun 18, 2008 at 10:04 AM, James Buddenhagen <jbuddenh@gmail.com> wrote:
This problem surely needs to be made more specific. There are many ways to tile the plane with identical convex pentagons. Similarly for convex hexagons. The plane can be tiled with identical non-convex heptagons. It would be surprising if there were not infinitely many possibilities if all 3 must occur, but I don't know either if they could all be convex.
I would suggest your friend have a look at the book "Tilings and Patterns" by Grünbaum and Shephard.
Jim Buddenhagen
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