3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ what do you mean by "regular"? It's easy to partition a square into two non-convex pentagons. $\endgroup$ – Robert Israel Nov 28 '12 at 19:40
  • $\begingroup$ Something which looks nice enough, for example piecewise $C^1$. $\endgroup$ – Beni Bogosel Nov 28 '12 at 19:42
  • 1
    $\begingroup$ 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! $\endgroup$ – TonyK Nov 28 '12 at 20:31
  • $\begingroup$ @TonyK: for the infinite case, see my answer. $\endgroup$ – Robert Israel Nov 28 '12 at 21:45
  • $\begingroup$ @Robert: That's why I said "bounded"! $\endgroup$ – TonyK Nov 28 '12 at 22:40
4
$\begingroup$

I copy'n'paste the solution from problem 5 at http://www.imomath.com/index.php?options=543&lmm=0 ---

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.

$\endgroup$
  • $\begingroup$ So simple!${}$${}$ $\endgroup$ – TonyK Nov 28 '12 at 22:47
  • $\begingroup$ I also found this solution last night. :) $\endgroup$ – Beni Bogosel Nov 29 '12 at 12:47
4
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.