# On convex planar regions that can be cut into only a specified number of mutually congruent and connected pieces

References:

Question 1: Given a number $$N$$, can we construct a convex planar region that can be cut into $$N$$ mutually congruent, connected, convex pieces but not into any other number of connected, mutually congruent convex pieces?

Partial Answer (guess): For prime $$N$$, there seems to be a simple way. Take a regular $$N$$-gon and mark from it $$N$$ mutually congruent quadrilaterals by drawing lines from center to mid points of the N faces. Now in each quadrilateral, replace the two 'outward' edges by copies of a polyline with say $$p$$ edges and with angles that are irrational fractions of $$\pi$$ (see ref 3 for some justification for 'irrational') in such a way that the $$N$$-gon becomes a convex $$Np$$-gon. This $$Np$$-gon seems to allow partition into $$N$$ and only $$N$$ pieces that are mutually congruent, convex and connected.

Remark: As per the answers below, one can upgrade above attempt to work for all values of $$N$$, not only primes.

Question 2: Are there convex planar regions that allow partition into mutually congruent and connected pieces only when the number of pieces is one of exactly $$2$$ specified values — for example, is there a convex region that can only be cut into $$3$$ connected congruent pieces or $$5$$ congruent pieces but not into any other number of congruent pieces?

Remark: Answer to question 1 can be slightly modified to yield planar regions that seem to allow partition into only $$N$$ mutually congruent pieces or $$kN$$ mutually congruent pieces where $$N$$ and $$k$$ are primes.

Note: One can widen question 2 and ask if given a set S of numbers relatively prime to one another, one can construct a planar region that allows partition only into sets of congruent pieces with cardinalities equal to each element in set S and no other number. One can also consider less constrained versions - eg. allow the mutually congruent pieces and the input region to be non-convex.

• I think it's inappropriate to widen a question after two answers have already been given -- instead, it would be appropriate to roll back to v4 of the question, accept one of the answers, and then possibly ask another question on the same theme.
– user44143
Commented Jan 13, 2022 at 16:54
• Please note that you can accept an answer by clicking on the check mark below the vote score (accepting satisfactory answers is part of the site etiquette). Commented Jan 13, 2022 at 18:14
• thanks stefan kohl. and thanks matt f. i just accepted the answers - i didn't know how to do it earlier - and i shall go back and accept some earlier answers! reg. the other point, question 2, which remains, was there from the beginning of the post and i didn't really widen the question. Commented Jan 14, 2022 at 12:15

This construction (the picture has the case $$N=6$$) seems to work. It is obtained in 2 steps:

-Begin with a convex set formed by $$N$$ equal pieces, each having a boundary formed by two segments and a piece of circumference.

-Change the orientation of one piece.

• I agree. This widens the answer for question 1 from prime N - assuming that mirror images are congruent. If we treat mirror images as non-congruent, what could one say? Shall add a remark in the question. Thanks. Commented Jan 12, 2022 at 9:47

What if you take $$N$$ thin slices of pizza (as in your example, but thinner) and arrange them like this?

• This could be divided into $kN$ pieces for any $k$. But if you change the slices by strange slices like the ones from my answer it seems to work without orientation changes Commented Jan 12, 2022 at 10:05
• @SaúlRodríguezMartín That’s what I meant by ‘as in your example’ —- meaning the OP’s example, with a polyline of $p$ segments. You know, a pizza is not perfectly round;) Commented Jan 12, 2022 at 10:07
• Guess this rearrangement with one piece going the other way - but with all pieces being congruent - could work for any N, including all composites. IOW, this polygon appears to be partitionable only into N congruent pieces. And it seems to have no problem with mirrors. thnx! Commented Jan 12, 2022 at 10:57