We say that a figure 'tiles the plane' if the whole Euclidean plane may be covered by non-overlapping congruent copies of that figure.

What is the necessary and sufficient condition for a polygon in general to tile the plane?

Do provide a rigorous combinatorial proof for your claim.

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## Comments

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TopNewestWhat I'm sure is that any form of triangle and quadrilateral is always able to tile a plane. Do someone has any proof that hexagon also always tile a plane AND none of the n-gon for n>=7 are able to tile a plane?

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If you are referring to regular polygons, then it is easy to show that tiling can only occur for \( n = 3, 4, 6 \).

If you are referring to not-necessarily regular polygons, then there exist many counter-examples for \( n \geq 7 \). For example, MC Escher art. The T-tetramino and S-tetramino are simple shapes with 8 edges, that easily tile the plane.

I would likely disagree with your claim that "any quadrilateral / hexagon will always tile the plane". The easiest examples I can think of are V-shaped (with distinct slopes). I agree that "any triangle will always tile the plane", with the parallelogram proof that you stated.

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By the way, I re-read an article and it actually says 3 hexagons. So sorry for that. Also, I forgot to refer it as convex polygons, and sorry too for that. Well, sure, I agree with you, just that I have a proof for n = 3 and 4 but it's in a photo and I don't know how can I send it to here directly...

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The concept cof tetranimoes from murderous maths

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Proof for your "sure" claim, that any form of triangle and quadrilateral tile the plane ?

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No. I can prove both myself, but the problem is (nevermind the hexagon... I have re-read an article that says there's 3 of it) I need a proof that it's impossible for any n-gon to tile an Euclidean plane for every n>=7. Triangles are easy solution, quadrilateral: a little bit harder

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2 triangles in a right positioning obviously make a parralelogram, which obviously tile a plane thanks to each pair of opposite side being parallel. OR do I need to prove that 2 congruent triangle are always able to form a parallelogram???

For quadrilateral, I guess I need to draw it... but the point is to use one of it in an oriented angle but use the other one by rotating it 180 degrees.

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Parallelogram

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@Daniel Liu @Calvin Lin @Trevor Arashiro @Kishlaya Jaiswal @Agnishom Chattopadhyay

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Planar tiling isn't easy to deal with. There isn't a simple classification of such polygons. You can look at the wallpaper group, and try to determine certain conditions.

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