# Congruent division of a shape in euclidean plane

Any triangle can be divided into 4 congruent shapes:

http://www.math.missouri.edu/~evanslc/Polymath/WebpageFigure2.png

An equilateral triangle can be divided into 3 congruent shapes.

Questions:

1) a triangle can be divided into 3 congruent shapes. Is it equilateral?

2) a shape in the plane can be divided into n congruent shapes for any positive integer n. What can it be? (e.g. it can be the interior of a circle or a rectangle)

let shapes be connected, open and convex. in fact I cannot define a shape exactly but its definition is intuitively clear.

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Lots of shapes satisfy (2). Parallelograms and "wedges" of circles come to mind. –  mjqxxxx Feb 1 '13 at 4:09
you're right. is there any other shape. I'll edit 2, –  user59671 Feb 1 '13 at 4:14
I think if a shape is divided into convex shapes, the only possible divisions are by straight lines. So convexity limits the problem. –  user59671 Feb 1 '13 at 20:46

Here's a large class of valid shapes. It's easiest to explain with a picture of a representative example:

$\hskip{2.5in}$

Formally, let $\gamma:[a,b]\to\mathbb R^2$ be a curve (the blue one). Let $T:[0,1]\to(\mathbb R^2\to\mathbb R^2)$ be a constant-speed rotation or translation of the plane. That is, either there is a fixed vector $x$ such that $T(\alpha)$ is a translation by $\alpha x$, or there is a fixed point $x$ and a fixed scalar $\theta$ such that $T(\alpha)$ is a rotation by $\alpha\theta$ about $x$ (the red arcs). Let us also require that the map $$s:[a,b]\times[0,1]\to\mathbb R^2,\\s(u,v)=T(v)\big(\gamma(u)\big)$$ is injective, so the transformed copies of $\gamma$ never overlap. Then the range of $s$, i.e. $s([a,b],[0,1])$, is a shape that can be divided into any number $n$ of congruent pieces, namely $$s([a,b],[0,\tfrac1n]),\quad s([a,b],[\tfrac1n,\tfrac2n]),\quad \ldots,\quad s([a,b],[\tfrac{n-1}n,1]).$$

This violates a few of your conditions (openness, convexity), but those pretty seem arbitrary to me. I don't know whether this is a complete solution, i.e. whether this includes all the possible shapes that are divisible into arbitrarily many congruent parts.

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good idea. if convex it looks be disk wedges or parallelograms. –  user59671 Feb 1 '13 at 13:32
my idea is using isometric homeomorphisms of a shape (eg triangle) and considering the automorphism group of these isometric homeomorphisms. –  user59671 Feb 1 '13 at 14:41
Not sure what you mean by that last comment. If you have a different answer, you should post it as an answer. –  Rahul Feb 1 '13 at 21:25
needs more work. –  user59671 Feb 1 '13 at 21:26
+1 A good example. I find it hard to define what parts of this space have every piece convex. Certainly if the red and blue curves are both straight you are there. You can't have the red curves arcs of circles as one will be concave inward or the pieces won't be congruent. –  Ross Millikan Feb 2 '13 at 6:09

The answer to 1) is, not necessarily. A 30-60-90 triangle can be divided into three congruent 30-60-90 triangles. If you can't see how to do this, see Figure 1 in this paper. Theorem 2 in that paper proves that this and the equilateral are the only triangles that can be divided into three congruent triangular shapes.

EDIT: If you don't insist on convexity, you can divide any triangle into three congruent shapes. First divide it into four congruent shapes by drawing line segments joining the midpoints of the sides. Then divide the one in the middle into four congruent shapes the same way, and then divide the middle one of those four into four congruent shapes the same way, and so on, ad infinitum. Now make a shape by taking one triangle of each size in the final diagram --- you can easily choose them in such a way as to make the shape connected. The remaining triangles form two shapes congruent to the first one, if you are careful when you decide which triangle goes to which of the two.

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great paper. thanks. –  user59671 Feb 1 '13 at 13:35