Can you cover the planar with regular pentagons and decagons(all of their sides are 1 unit long), without any holes or overlaps?
I think that the answer is no, but can't really prove it.
Any ideas? Thanks!
2026-03-27 11:46:36.1774611996
How to prove if this is false or not?
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Think about what has to happen at the vertices of the tiling, comparing to Platonic or Archimedean tilings. Four squares meeting at a vertex works because it just so happens that $4\cdot 90^\circ = 360^\circ$. Two octagons and a square works because $2\cdot 135^\circ + 90^\circ = 360^\circ$.
What are the possible sums of angles meeting at a vertex if you're only allowed to use pentagons and decagons?
This won't be the complete end of the story, but it gives you a starting point for what configurations are allowable at each vertex, and you'll be able to see what those configurations imply.