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A reptile is a shape that can be dissected into smaller copies of the same shape. A reptile is labelled rep-n if the dissection uses n copies.

Given any positive n, any parallelogram with sides in ratio 1:√n is easily seen to be a rep-n reptile.

  1. given any n, is it possible to construct some convex polygon other than parallelograms that is a rep-n reptile?

  2. Or can some values of n be proved not to support non parallelogram rep-n reptiles ( for example, I can’t think of any convex, non parallelogram rep-7 reptile)?

Note: If question 2 has a “yes” answer, we could ask: if convexity is relaxed can we form non parallelogram rep-n reptiles for any n?

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  • $\begingroup$ By 'non-rectangular' presumably you mean that either some or all of the angles should be other than right angles? $\endgroup$ Commented Jun 26 at 4:43
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    $\begingroup$ Don’t parallelograms allow for the same trivial reptile construction as rectangles? $\endgroup$ Commented Jun 26 at 6:03
  • $\begingroup$ Sure. Edited the question to reflect that. Thanks. Whatever applies to rectangles goes for parallelograms too. The question statement needed to be more precise. $\endgroup$ Commented Jun 26 at 6:19
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    $\begingroup$ Yes; triangles work for many values of n but it seems no triangle can be a rep-7 reptile. $\endgroup$ Commented Jun 26 at 11:34
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    $\begingroup$ For triangoes, the only $n$ values with evident solutions are a square, a sum of two squares (right triangle with integer legs), or three times a square (right triangle with 30° and 60° angles). $\endgroup$ Commented Jun 26 at 15:51

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Jin Heo in April 2024* gave an extensive treatment, based on previous work whose review implies that all convex rep-tiles are either triangles or quadrilaterals, and among the quadrilaterals only parallelograms and trapezoids may be rep-tiles.

The authors examine the cases of right trapezoids and isosceles trapezoids, and they identify only the following as rep-tiles:

Right trapezoids

  • The well-known case of a right trapezoid with bases $1,2$ and altitude $1$, which is a rep-$4$-tile.

  • A second previously found rep-tile, again with bases $2,4$ but this time the altitude $\sqrt3$. This is again a rep-$4$-tile.

  • The last possible case, which the authors discover in their analysis: the bases are $1,5$ and the altitude is $2\sqrt3$. This one is a rep-$25$-tile.

Isosceles trapezoids

  • The previously known isosceles trapezoid with bases $2,4$ and altitude $\sqrt3$, doubling the second case of right trapezoids noted above. This is both a rep-$4$-tile and a rep-$9$-tile.

  • Another previously known case, with bases $1,3$ and again the altitude us $\sqrt3$. This is a rep-$9$-tile.

  • No additional isosceles trapezoids are found in this work.

Thus, barring nonobvious divisions, the only $n$ values that can be reached with right or isosceles trapezoids are either $(2^p3^q)^2$ for whole numbers $p,q$ using the first isosceles case noted above, or $5^{2p}$ using the newly discovered right trapezoid.

Below are illustrations of all the identified cases. For the isosceles trapezoid that is identified above as both a rep-$4$- and rep-$9$-tile (R3 in the first figure), only the $4$-division is shown; the reader is invited to identify the (nonunique) $9$-division by first dividing this trapezoid into three equilateral triangles.

enter image description here

enter image description here

*Note: Only Jin Heo's name is seen as an author, but the abstract refers to "we". The manuscript may not be properly displaying all authors.

A rep-7-tile ... with fractals

Among all $n$ values from $2$ through $10$, $n=7$ uniquely lacks an obvious solution in terms of convex polygons or the the L-shaped polygons presented by Per Alexandersson. Here a solution is rendered ... with fractals.

The procedure is essentially ancadapted version of the Sierpiński carpet. Begin with a square block. Divide it into 3×3 smaller square blocks using cut lines parallel to the sides of th original block, then remove two smaller blocks situated along opposing edges making an H shape:

enter image description here

The H shape consists of the seven renaining blocks, each of which is divided and decremented similarly to the original and the process is iterated infinitely to produce the fractal.

Because of its self-similarity, the fractal automatically contains seven smaller copies of itself arranged in the H shape, and so qualifies as a rep-7-tile. Note that because the H shape is connected the fractal will be too.

Note the connection with fractal dimension. The Hausdorff dimension is easily rendered as $\ln(7)/\ln(3)\approx 1.7712$ instead of $2$, which goes along with the $3:1$ scaling producing seven replicates instead of nine.

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    $\begingroup$ It's not uncommon to use 'we' in single author papers (I like to think of it as including the reader in the activity being described). $\endgroup$ Commented Jun 26 at 16:17
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For any even $n \geq 4$, there is a non-convex reptile. You can take an L-shape, and put two of these together to form a rectangle. One such rectangle can then form the base of the L, and then you can take arbitrary many more copies of this rectangle to form the tall part of the L. It is straightforward to work out the proportions needed.

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