Let us define a perfect congruent partition of a planar region $R$ as a partition of it with no portion left over into some finite number n of pieces that are all mutually congruent (ie any piece can be transformed into another piece by an isometry. We consider only cases where each piece is connected and is bounded by a simple curve).

Note: It is known there are convex planar regions - indeed quadrilaterals - which do not allow perfect congruent partition for any n ([1] proves a stronger result).

Claim: If a convex polygonal $R$ allows a perfect congruent partition of itself into $N$ non-convex pieces each with finitely many sides, then $R$ also allows a perfect congruent partition into $N$ convex pieces with finitely many sides. In other words, allowing the pieces to be non-convex polygons does not improve the chances of a convex planar region achieving a perfect congruent partition into $N$ pieces.

I know no proof, no counter example. One can consider replacing 'congruent' with 'similar' in the above question. Some more related thoughts are in [2].


1.https://www.research.ibm.com/haifa/ponderthis/challenges/December2003.html 2.https://arxiv.org/abs/1002.0122

A bit added on March 15th, 2024: Are there convex polygonal regions with even number of edges that has this property of being perfect congruent partitionable only into 2 non-convex pieces?

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    $\begingroup$ If you do not demand the regions to be simple polygons and congruence includes rotations, the counterexample is simple:take the regular hexagon and build something on its sides in the pattern aabbcc. Then you can split it into two congruent pieces by connecting the center to the "first" a,b,c and then to the "second" but no convex partition (i.e, splitting with a single line) into two congruent pieces will be possible. If you mean something more restricted, just say what exactly. $\endgroup$
    – fedja
    Aug 30, 2020 at 13:00
  • $\begingroup$ Thanks. The partition is to be into simple polygonal regions. Made a couple of edits to the question. $\endgroup$ Aug 31, 2020 at 17:18

1 Answer 1


This picture looks like a counterexample with $N=2$ and $R$ a convex pentagon:

This should work more generally starting from an $n \times (n+1)$ rectangles for any integer $n>1$, removing two congruent right triangles that are not isosceles (the picture shows $n=3$ with $2:5$ triangles). Replacing the heavy lines with more complicated polygonal convex paths yields convex polygons with more sides.

  • $\begingroup$ Beautiful! Nice to use a pentagon that cannot be partitioned into two congruent convex pieces, as that would require a chord to create identical angles. $\endgroup$ Sep 1, 2020 at 15:19
  • $\begingroup$ Really elegant example; thanks!! Let me add a small extension to the question, basically to bound this property: to find that convex shape with such a perfect congruent partition into 2 non-convex pieces that 'wastes' the highest fraction of its area when given the best partition into 2 convex pieces. $\endgroup$ Sep 1, 2020 at 17:11

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