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The property I'm talking about is:

There is some partition of the plane figure P into $n$ congruent figures for any $n$.

Is it true that only discs, sectors of discs, annuli, sectors of annuli and parallelograms have this property?

user132181
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  • Sectors of discs and sectors of annuli work too - both are sort of an extension of the disc example, so perhaps include them into your question? – curious Nov 01 '14 at 21:57
  • Do you mean $n+1$ congruent plane figures? – Matt Samuel Nov 01 '14 at 21:58
  • @curious thanks for pointing it out. – user132181 Nov 01 '14 at 22:03
  • @MattS no, that's why I've edited the question. – user132181 Nov 01 '14 at 22:03
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    Would "there is some partition of the figure into $n$ congruent figures for any $n$" be an accurate statement of the property you want? – Milo Brandt Nov 01 '14 at 22:06
  • "Plane figure" is a very general concept. $\mathbb R^2\setminus{,(\cos \alpha, \sin \alpha)\mid \alpha\in\mathbb Q\pi,}$ has the property as well – Hagen von Eitzen Nov 01 '14 at 22:06
  • @Meelo yes, that would be the proper way to state the property, indeed. I've edited the question appropriately. – user132181 Nov 01 '14 at 22:09
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    Any set of the form ${,(x,f(x)+y),\mid x\in X, 0\le y\le a}$ has the property as well. (And this includes some polygons that would even fit the strictest definitions of "plane figure", except possibly this which restricts the term to "lines only" stuff) – Hagen von Eitzen Nov 01 '14 at 22:12
  • @HagenvonEitzen is the necessity true as well? – user132181 Nov 01 '14 at 22:14
  • I think one may proceed as follows: First show that for infinitely many $n$, the movements witnessing the congruences of parts generate a cyclic group. Then show that for infinitely many $n$, this cyclic group is in fact a subgroup of a group isomorphic to $\mathbb Q$ (translations) or $\mathbb Q/\mathbb Z$ (rotations). Then somehow conclude that the figure is obtained by "sweeping" an infinitesimal figure with a connected set of translations or rotations. - However I'm not sure how to show the two steps (nor if they are even true). – Hagen von Eitzen Nov 01 '14 at 22:21
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    Conjecture. If $P$ is compact and has the user132181 property, then there exists a compact subset $A\subset P$ (often $A\subset \partial P$?) such that $P=\bigcup_{t\in[a,b]}\phi_t(A)$ where the union is disjoint up to possibly one point and $t\mapsto \phi_t$ is a homomorphism from $\mathbb R$ to the topological group of plane movements. – Hagen von Eitzen Nov 01 '14 at 22:33
  • possible duplicate of Congruent division of a shape in euclidean plane –  Nov 04 '14 at 10:15

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To expand on Hagen von Eitzen's remarks: the answer to your question is "no".

Let $a, b$ be positive real numbers with $b>a$, and $f$ be a continuous function mapping [a,b] to $\mathbb R$. Let $X = \{(x,f(x)) : x \in [a,b]\}$.

Now sweep out a planar area by rotating $X$ around the origin through some angle $\theta$ in $(0, 2 \pi]$.

When $f(x) = 0$, this yields your (sectors of, when $\theta<2 \pi$) annuli (or disks, when $a=0$).

We can think of your case of parallelograms as being the limit when $a$ is very, very large compared to $b-a$; and where $f(x) = (x-a)c$; we're (very loosely speaking!) rotating the figure around "a point at infinity". (This corresponds to Hagen's set of translations).

But we aren't limited to a linear $f$ in either case; the only requirement is that there be no "self-intersections" as we sweep out the figure. And that can be satisfied if $d(x, f(x))$ (the distance from the point $(x, f(x))$ to the origin) is injective; i.e., for all $x_0, x_1 \in [a,b]$, $d(x_0, f(x_0)) = d(x_1, f(x_1)) \leftrightarrow x_0 = x_1$.

So for example: Take a triangle shaped wedge out of one end of your sector-of-an-annulus, and slap it onto the other end. Assuming the triangle isn't too acute, the resulting figure will have the property you describe.

Lots of much more complicated possibilities can be constructed, as long as we obey the constraint of "distance from the origin is injective".

(Edit: replace "strictly increasing" with "injective".)

Chas Brown
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