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I found the following problem:

Is it possible to partition every convex polygon into a finite number concave quadrilaterals?

The answer seems negative, because heuristically if we remove a concave quadrilateral the new polygon is still convex, and after a finite number of steps we arrive at a concave quadrilateral in end, and therefore a contradiction.

The problem is that it is possible to have some weird configurations, and removing a any quadrilateral from the partition may make the resulting polygon non-convex.

What is the answer to the question, and what is the proof?

Moreover, is there a more general result like:

It is impossible to partition a convex set into a finite number of regular, connected non-convex sets? (Answer: NO)

What happens if we remove the finiteness assumption?

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what do you mean by "regular"? It's easy to partition a square into two non-convex pentagons. – Robert Israel Nov 28 '12 at 19:40
Something which looks nice enough, for example piecewise $C^1$. – Beni Bogosel Nov 28 '12 at 19:42
Is it possible to construct any convex set at all from a finite number of non-overlapping concave quadrilaterals? Is it even possible to construct a bounded convex set as an infinite union of non-overlapping concave quadrilaterals? Good question! – TonyK Nov 28 '12 at 20:31
@TonyK: for the infinite case, see my answer. – Robert Israel Nov 28 '12 at 21:45
@Robert: That's why I said "bounded"! – TonyK Nov 28 '12 at 22:40
up vote 4 down vote accepted

I copy'n'paste the solution from problem 5 at ---

The answer is no. Assume that, on the contrary it is possible to partition a polygon $P$ into non-convex quadrilaterals. Let $n$ be the number of quadrilaterals. Denote by $S$ the total sum of all internal angles of all the quadrilaterals. Since the sum of internal angles of each quadrilateral is $360^\circ$ we have $S=360^\circ n$. However, each of the nonconvex angles has to be in the interior of $P$, hence the sum of angles around the vertex of that angle has to be $360^\circ$. This immediately gives $360^\circ n$ as the sum of angles around such vertices. Since those are not the only vertices (at least the vertices of $P$ will contribute to the sum $S$), we have that $S\gt360^\circ n$ and this is a contradiction.

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So simple!${}$${}$ – TonyK Nov 28 '12 at 22:47
I also found this solution last night. :) – Beni Bogosel Nov 29 '12 at 12:47

Partitioning a square into two non-convex pentagons:

enter image description here

EDIT: Partitioning the plane into concave quadrilaterals:

enter image description here

EDIT: Partitioning a triangle into infinitely many concave quadrilaterals.

enter image description here

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That solves the last part :) – Beni Bogosel Nov 28 '12 at 19:51
If I could upvote twice I would :) Thank you. – Beni Bogosel Nov 29 '12 at 13:30

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