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Is it possible to tessellate a planar surface from triangles but with the following constraints:

  • density (average number of triangles) can be varied.
  • a finite set of unique triangles are used for the tessellation. Say 5-15 unique triangles.

Optionally also with the ability to increase density for sub areas.

If it is possible, can you point me to some relevant material.


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This question needs a lot of work to become comprehensible. To begin with, what's a triangle? I know what a triangle is in the plane, but you want to tessellate a surface, and surfaces are generally curved, and don't have lots of straight lines in them, so - what's a triangle? – Gerry Myerson Feb 25 '12 at 11:26
Yes, a planar surface. – quayd Feb 25 '12 at 12:08
Well, it is possible to tessellate a triangle, but impossible to tessellate a circle, so it would be nice to be more specific about the kind of planar surfaces you wish to allow. Or is that exactly what you want, a characterization of such surfaces? Or perhaps by "planar surface" you mean "polygon"? – Dejan Govc Feb 25 '12 at 15:37
I do not have a requirement for the surface shape. It can be pretty much anything. A square, triangle, rectangle, etc. or even something like:… – quayd Feb 25 '12 at 19:46
Well, then, it seems that you have your answer. How do you propose to tessellate a disk? – Gerry Myerson Feb 25 '12 at 22:38
up vote 3 down vote accepted

I'm going to make an interpretation of the question, and then answer it. If my interpretation is wrong, OP can let us know.

We are given triangles $T_1,T_2,\dots,T_n$, and we want to know whether it is possible to tessellate an arbitrary polygonal region $P$ with a finite number of triangles, each triangle similar to one of those given.

I claim it's not possible. Let $P$ have an angle that is smaller than any of the angles in the triangles. Then there is no way to get to that angle.

Now, what if we are allowed to pick the triangles $T_1,T_2,\dots,T_n$ after we have seen the region $P$? If $n$ is fixed, we're still out of luck. The angles we can get lie in an extension field of transcendence degree at most $2n+1$ over the rationals, so if we are faced with a region with more than $2n+1$ algebraically independent angles, we can't tessellate it.

In short, under the sort of assumptions I've been making, the class of tessellatable polygonal regions is very restricted.

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I appreciate your taking the plunge. I am a layman in math, so I could not have predicted the challenges in interpreting my question. By the statement, "P have an angle", does that mean the sum of all its internal angles? – quayd Feb 26 '12 at 6:43
Actually, it seems you interpret the question accurately. To clarify, my intent is that "n" is 5-15, and the total number of triangles used for the tessellation is ">n" selecting at least once (preferably at least 4 times) each triangle in the set "T". However, what assumptions (in layman's terms preferably) are being made that do not lead to the following, which I would suspect my description allows for as a successful candidate (assuming one divides the polygons into respective triangles): – quayd Feb 26 '12 at 7:10
$P$ is a polygon. It has a lot of angles, one at each vertex. "Let P have an angle..." means suppose that one of those angles is smaller than the smallest angle in any of the triangles. – Gerry Myerson Feb 26 '12 at 7:44
The Penrose picture shows that one particular polygon can be tessellated with a small number of basic shapes of triangle. Notice that as you go around the perimeter of that polygon, you never come to any really, really sharp angles. If you choose your triangles in advance, and then someone confronts you with a polygon with an angle sharper than any of the ones in your triangles, you are out of luck. – Gerry Myerson Feb 26 '12 at 7:48

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