# Cutting a polygon into 2 or 3 smaller, rationally-scaled copies of itself?

I've noticed that many 2D geometric figures can be tiled using four smaller copies of themselves. For example, here's how to subdivide a rectangle, equilateral triangle, and right triomino into four smaller copies: Each smaller figure here is scaled down by a factor of $$\frac{1}{2}$$ in width and height, dropping its area by a factor of four, which is why there are four smaller figures in each.

You can also tile some 2D figures with nine smaller copies, each $$\frac{1}{3}$$ of the original size, or sixteen smaller copies, each $$\frac{1}{4}$$ of the original size, as shown here: By mixing and matching sizes, we can get other numbers of figures in the subdivisions. For example, here's a $$2 \times 1$$ rectangle subdivided into five rectangles of the same aspect ratio, an equilateral triangle subdivided into eleven equilateral triangles, and a right triomino tiled by thirty-eight right triominoes: $2 \times 1$ rectangle subdivided into five rectangles of the same aspect ratio, an equilateral triangle subdivided into eleven equilateral triangles, and a right triomino tiled by thirty-eight right triominoes" />

I've been looking for a shape that can tile itself with exactly two or three smaller copies. I know this is possible if we allow the smaller copies to be scaled down by arbitrary amounts, but I haven't been able to find a shape that can tile itself with two or three copies of itself when those smaller copies are scaled down by rational amounts (e.g. by a factor of $$\frac{1}{2}$$ or $$\frac{3}{5}$$).

## My Question

My question is the following:

Is there a 2D polygon that can be tiled with two or three smaller copies of itself such that each smaller copy's dimensions are a rational multiple of the original size?

If we drop the restriction about the smaller figures having their dimensions scaled by a rational multiple, we can do this pretty easily. For example, a rectangle of aspect ratio $$\sqrt{2} : 1$$ can tile itself with two smaller copies, and a rectangle of aspect ratio $$\sqrt{3} : 1$$ can tile itself with three smaller copies: $\sqrt{2}:1$ rectangle cut into two self-similar copies, and a $$\sqrt{3}:1$$ rectangle cut into three self-similar copies" />

However, in these figures, the two smaller copies are scaled down by a factor of $$\frac{\sqrt{2}}{2}$$ and $$\frac{\sqrt{3}}{3}$$, respectively, which aren't rational numbers.

If we move away from classical polygons and allow for fractals, then we can do this with a Sierpinski triangle, which can be tiled by three smaller copies of itself. However, it's a fractal, not a polygon.

## What I've Tried

If we scale down a 2D figure by a factor of $$\frac{a}{b}$$, then its area drops to a $$\frac{a^2}{b^2}$$ fraction of its original area. This led me to explore writing $$1$$ as a sum of squares of rational numbers, such as $$1 = \frac{4}{9} + \frac{4}{9} + \frac{1}{9}$$ or $$1 = \frac{9}{25} + \frac{16}{25}$$. This gives several possible values for how to scale down the smaller copies of the polygon, but doesn't give a strategy for choosing the shapes of the reduced-size polygon to get the smaller pieces to perfectly tile it.

I've looked into other problems like squaring the square and other similar tiling problems. However, none of the figures I've found so far allow for a figure to be tiled with two or three copies of itself.

I've also tried drawing a bunch of figures on paper and seeing what happens, but none of them are panning out.

Is this even possible in the first place?

Thanks!

Take a 345 triangle, and draw the altitude from the hypotenuse to the opposite corner.
To get three pieces, cut one of the two pieces in the same way.

• Same should work for any right triangle with all sides integral. Sep 4, 2021 at 10:49

If all three pieces are the same size, then it is impossible, without invoking fractals, for the smaller pieces to have dimensions that are a rational multiple of the original object's size.

In order for the area of the new shape to be $$n$$ times the area of the original, and still be similar, we must multiply the linear dimensions by $$\sqrt{n}$$. Since $$1/3$$ is not a perfect square, $$\sqrt{1/3}$$ is not a rational number. In order to avoid this we can use fractals, as you saw with the Sierpinski triangle, which allows us to use the fractal dimension instead of the "real-space" dimension.

A trapezoid with side lengths in the ratio of 1:1:2:sqrt(2) is an irreptile. It can be cut into three similar trapezoids, two small and one large.

Consider the trapezoid bounded by 0<x<2 and 0<y<4-x. It can be cut into three similar trapezoids bounded by:

1. 0<x<1 and 0<y<2-x;
2. 1<x<2 and 0<y<x; and
3. 2-x<y<4-x, 0<x, and y>x. 