## Rep-tiles Revisited

*Department of Mathematics, Anderson Hall 331a, West Chester University, West Chester,
PA 19383*

The goal of this note is to take a new look at some of the most amazing objects discovered
in recreational mathematics. These objects, having the curious property of making
larger copies of themselves, were introduced in 1962 by Solomon W. Golomb [2], and soon afterwards were popularized by Martin Gardner [3] in Scientific American. In Golomb's terminology a plane figure (or tile) is defined
to be replicating of order * k* (or rep-

*) if it can be dissected into*

**k***replicas, each congruent to the other and similar to the original one. If*

**k***, an equivalent definition is that four identical figures are to be assembled into a scale model, twice as long and twice as high. All triangles and parallelograms are rep-*

**k=4***tiles. Besides these obvious examples, a great number of rep- tiles are known. We will review in this paper all known rep-*

**4***convex polygons and a couple of rep-*

**4***connected non-convex polygons. Our list originates in [2], where examples of rep-*

**4***tiles can be found for other values of*

**k***, as well as disconnected or fractal type examples. In what follows we are interested in the following question:*

**k****Question 1 ** Which polygonal rep-* 4* tiles are rep-

*tiles for any*

**k**^{2}*?*

**k ≥ 4**An equivalent formulation of the question is to find for which rep-* 4* tiles

*identical figures can be assembled into a scale model,*

**k**^{2}*times as high. If*

**k***is a given tile, we will call the scale model the*

**T***-th multiple of*

**k***, and for the rest of the paper will denote it by*

**T***. Question 1 can be viewed as a particular case of the most general question:*

**kT****Question 2** (Finite Tiling Problem) Given a finite plane region * T*, and a finite set ∑ of tiles, find if

*can be tiled by translations of tiles in ∑.*

**T**Let us only note here that methods do exist, coloring arguments among others, which
can be used to prove *non-existence* of tilings. We refer the reader to [1],[4], and [5] for other techniques, some of them stronger then coloring arguments. An open question
about non-existence of certain tilings related to reptiles will be mentioned during
our exposition. The answer to Question 1 will be positive most of the time. The main
technique used in the proof of our results is mathematical induction. Our proofs are
not necessarily the shortest, or the most elegant, possible. We encourage the reader
to try to find his own proofs before reading ours, and hope that this exercise will
get him hooked in this fascinating field.

As shown in Fig. 1, there are three trapezoids that are rep-* 4* tiles. We will denote them by

*. The first one consist of a square and a half, the second is made out of three equilateral triangles, and the third consists of an equilateral triangle and a half. We assume for the rest of the paper that the sides of the square and of the equilateral triangle are of length one.*

**T1, T2, T3****Proposition 3** All three trapezoids * T1, T2, T3*, are rep-

*tiles for any*

**k**^{2}*.*

**k ≥ 1***Proof*. For all three cases the proof is done by induction. We check first our hypothesis
for ** k = 2, 3**, and then show that rep-

**implies rep-(**

*k*^{2}**)**

*k*+3**.**

^{2}*Proof for*The cases

**T1**.**are dealt with in Fig. 2. To show that rep-**

*k*= 2, 3**implies rep-**

*k*^{2}**(**, we take a

*k*+3)^{2}**(**region and partially cover it by a

*k*+3)*T1***region as shown in Fig. 3. The**

*kT1**region can be covered by*

**kT1****tiles because of the induction hypothesis. The uncovered region can be decomposed in three parts: a rectangle (I) of dimensions (**

*T1***2**), a rectangle (II) of dimensions (

*k*, 3**), and a region of type**

*k*, 3**3**. The rectangles can be covered by rectangles of dimensions (

*T1***3, 1**), which in turn can be covered by two

**tiles.**

*T1***Figure 2:**

**Figure 3:**

*Proof for T2.* The cases k = 2, 3 are dealt with Fig. 4.

**Figure 4:**

**Figure 5:**

To show that rep-** k^{2}** implies rep-(

**)**

*k*+3**, we take a (**

^{2}**)**

*k*+3*region and partially cover it by a*

**T2***region as shown in*

**kT2****Fig. 5**. The

*region can be covered by*

**kT2***tiles because of the induction hypothesis. The region not covered by the*

**T2***region is decomposed in four parts: two parallelograms (I) and (II) of dimensions (*

**kT2****), a parallelogram (IV) of dimensions (**

*k*, 3**), and an equilateral triangle (III). All parallelograms can be covered by parallelograms of dimensions (**

*k*+3, 3**3, 1**), which consist of two

*tiles, and the equilateral triangle can be covered by*

**T2***tiles as in*

**T2****Fig. 5**.

*Proof for*The cases

**T3**.**are dealt with in**

*k*=2, 3**Fig. 6**.

**Figure 6:**

**Figure 7:**

**Figure 8:**

To show that rep-** k^{2}** implies rep-

**(**, we take a

*k*+3)^{2}**(**region and partially cover it by a

*k*+3)*T3**region as shown in the*

**kT3****Fig. 7**and

**Fig. 8**. The

*region can be covered by*

**kT3***tiles because of the induction hypothesis. The region not covered by the*

**T3***region is decomposed either in three parts, or four parts, according to*

**kT3***being even or odd. The decomposition for*

**k***even is shown in*

**k****Fig. 7**. The first two parts are parallelograms, (I) of dimensions

**(**, and (II) of dimensions (

*k*+3, 3)**). Both of them can be covered by parallelograms of dimensions**

^{k}/_{2}-3, 3**(3,1)**, which consists of four

*tiles (see*

**T3****Fig. 7**). The third part is of type

**3**

*, hence can be covered by*

**T3***tiles as well. The decomposition for*

**T3***odd is shown in*

**k****Fig. 8**. The first two parts are parallelograms, (I) of dimensions (

**), and (II) of sizes (**

*k*, 3**). Both can be covered by parallelograms of dimensions**

^{k}/_{2}-^{3}/_{2}, 3**(3,1)**. Part (III) is a equilateral triangle, with sides of length

**3**, and can be covered as in

**Fig. 8**. Finally, the fourth part is of type

**3**.

*T3*

**Figure 9:**

There are three hexagons that can be dissected into four replicas (see **Fig. 9**). The first one is made out of three squares, the second one out of four squares,
and the third one out of five squares. We will denote them by * H1*,

*, and*

**H2***, respectively.*

**H3****Proposition 4: ** All three hexagons * H1, H2, H3*, are rep-

**tiles for any**

*k*^{2}**.**

*k**≥*1Rep-tiles Revisited *Proof*. The proof is again done by induction, but here the length of the induction step
is different for each type of tile.

**Figure 10:**

*Proof for **H1**.* We show cases ** k =2, 3** in

**Fig. 10**. Then we prove that rep-

**implies rep-**

*k*^{2}**(**. The proof depends on the congruence class modulo

*k*+2)^{2}**3**of

*. If*

**k****(mod**

*k*≅ 0**3**) the proof is done according to

**Fig. 11**. We take a

**(**region and partially cover it by a

*k*+2)*H1**region. The*

**kH1***region can be covered by*

**kH1***tiles because of the induction hypothesis. The region not covered by the*

**H1***region is decomposed in seven parts: two rectangles (I) and (VII) of dimensions*

**kH1****(2,**, two rectangles (II) and (VI) of dimensions

*k*)**(2,3)**, two rectangles (III) and (V) of dimensions

**(2, 2**, and the region (IV) of type

*k*-3)**2**. All rectangles can be covered by rectangles of dimensions

*H2***(2,3)**, which in turn can be covered by two

*tiles.*

**H2**

**Figure 11:**

If ** k ≅ 1** (mod

**3**) the proof is done according to

**Fig. 12**. We take a

**(**region and partially cover it by a

*k*+2)*H1**region. The*

**kH1***region can be covered by*

**kH1***tiles because of the induction hypothesis. The region not covered by the*

**H1***region is decomposed in seven parts: two rectangles (I) and (VII) of dimensions*

**kH1****(2,**, two rectangles (II) and (VI) of dimensions

*k*-1)**(2,3)**, two rectangles (III) and (V) of dimensions

**2(2,2**, and the region (IV) of type

*k*-2)**2**. All rectangles can be covered by rectangles of dimensions

*H2***(2,3)**, which in turn can be covered by two

*tiles.*

**H2**

**Figure 12:**

**Figure 13:**

If ** k ≅ 2** (mod

**3**) the proof is done according to

**Fig. 13**. We take a

**(**region and partially cover it by a

*k*+2)*H1**region. The*

**kH1***region can be covered by*

**kH1***tiles because of the induction hypothesis. The region not covered by the*

**H1***region is decomposed in seven parts: two rectangles (I) and (VII) of dimensions*

**kH1****(2,**, two rectangles (III) and (V) of dimensions

*k*-2)**(2,2**, and the regions (II), (IV), and (VI), of type

*k*-4)*. All rectangles can be covered by rectangles of dimensions*

**2H2****(2,3)**, which in turn can be covered by two

*tiles.*

**H2**

**Figure 14:**

**Figure 15:**

**Figure 16:**

*Proof for H2.* Observe that the proof is immediate if

*is even. Indeed, if*

**k***is even, we can decompose*

**k***in two rectangles of dimensions*

**kH2****(2**. Any such rectangle can be covered by rectangles of dimensions

*k*,4*k*)**(2,4)**, which in turn can be covered by two

*tiles (see*

**H2****Fig. 14**). If

**is odd, we proceed by induction. We already know cases**

*k***, and the case**

*k*=2,4**is dealt with in**

*k*=3**Fig. 14**. We show that rep-

**implies rep-**

*k*^{2}**(**. Take a

*k*+4)^{2}**(**region and partially cover it by a

*k*+4)*H2**region as shown in the*

**kH2****Fig. 15**. The

*region can be covered by*

**kH2***tiles because of the induction hypothesis. The region not covered by the*

**H2***region is decomposed in seven parts: a rectangle (I) of dimensions*

**kH2****(**, a rectangle (III) of dimensions

*k*-1,4)**(2**, a rectangle (V) of dimensions

*k*-2, 4)**(3**, a rectangle (VI) of dimensions

*k*-1, 4)**(**, a rectangle (VII) of dimensions

*k*-1, 8)**(5,8)**, and two squares, (II) and (IV), of dimensions

**(5,5)**with a unit square missing in one of the corners. The rectangles (I),(III),(V), and (VI), can be covered by rectangles of dimensions

**(2,4)**, which consist of two

*tiles, and the regions (II) and (IV), and the rectangle (VII), can be covered by*

**H2***tiles as in*

**H2****Fig. 16**.

**Figure 17:**

**Figure 18:**

**Figure 19:**

*Proof for H3.* We show cases

**in**

*k*=2,3,5**Fig. 17**. The case

**follows immediately from the case**

*k*=4**. Then we prove that rep-**

*k*=2**implies rep-**

*k*^{2}**(**. The proof depends on on the congruence class modulo

*k*+5)^{2}**2**of

*. If*

**k****(mod**

*k*≅ 0**2**), the proof is done according to

**Fig. 18**. We take a

**(**region and partially cover it by a

*k*+5)*H3**region. The*

**kH3***region can be covered by*

**kH3***tiles because of the induction hypothesis. The region not covered by the*

**H3***region is decomposed in five parts: a rectangle (I) of dimensions*

**kH3****(5,**, a rectangle (II) of dimensions

*k*)**(5,3**, a rectangle (IV) of dimensions

*k*)**(10,2**, a rectangle (V) of dimensions

*k*)**(5,2**, and the region (III) of type

*k*)**5**. The rectangles (I),(II),(IV), and (V) can be covered by rectangles of dimensions

*H3***(2,5)**, which consist of two

*tiles (see*

**H3****Fig. 17**). If

**(mod**

*k*≅ 1**2**), the proof is done according to

**Fig. 19**. We take a

**(**region and partially cover it by a

*k*+5)*H3**region. The*

**kH3***region can be covered by*

**kH3***tiles because of the induction hypothesis. The region not covered by the*

**H3***region is decomposed in five parts: a rectangle (I) of dimensions*

**kH3****(5,**, a rectangle (II) of dimensions

*k*+5)**(5, 3**, a rectangle (IV) of dimensions

*k*-5)**(10, 2**, a rectangle (V) of dimensions

*k*)**(5, 2**, and the region (III) of type

*k*)**5**. The rectangles (I),(II),(IV), and (V) can be covered by rectangles of dimensions

*H3***(2,5)**, which consist of two

*tiles.*

**H3**

**Figure 20:**

There is a pentagon that can be dissected into four replicas (see **Fig. 20**). This figure is known as the "sphinx". It consists of five equilateral triangles.
We will denote it by * P*.

**Proposition 5** The pentagon * P* is rep-

**tiles for any**

*k*^{2}*.*

**k ≥ 1**

**Figure 21:**

*Proof*. We show cases ** k=2,3** in

**Fig. 21**. Then we prove that rep-

**implies rep-**

*k*^{2}**(**. The proof depends on the congruence class modulo

*k*+3)^{2}**2**of

*.*

**k**

**Figure 22:**

If ** k ≅ 0** (mod

**2**), we take a

**(**region and partially cover it by a

*k*+3)*P**region as shown in the*

**kP****Fig. 22**. The

*region can be covered by*

**kP***tiles because of the induction hypothesis. The region not covered by the*

**P***region is decomposed in four parts: a parallelogram (I) of dimensions*

**kP****(2**, a parallelogram (III) of dimensions

*k*, 3)**(3**, a parallelogram (IV) of dimensions

*k*,3)**(3,**, and the region (II) which is of type

*k*)**3**. Parallelograms (I),(III) and (IV) can be covered by parallelograms of dimensions

*P***(2,3)**, which according to

**Fig. 21**can be covered by two

*tiles.*

**P**

**Figure 23:**

If ** k ≅ 1** (mod

**2**), we take a

**(**

*k*+3)*region and partially cover it by a*

**P***region as shown in the*

**kP****Fig. 23**. The

*region can be covered by*

**kP***tiles because of the induction hypothesis. The region not covered by the*

**P***region is decomposed in four parts: a parallelogram (I) of dimensions*

**kP****(2**, a parallelogram (III) of dimensions

*k*, 3)**(3**, a parallelogram (IV) of dimensions

*k*-3, 3)**(3,**, and the region (II) which is of type

*k*+3)**3**. Parallelograms (I),(III) and (IV) can be covered by parallelograms of dimensions

*P***(2,3)**, which can be covered by two

*tiles.*

**P**

**Figure 24:**

The last two convex polygonal rep-**4** tiles we consider are obtained by changing slightly two of the tiles investigated
so far. Note that * H1* is being made out of three-fourths of a square. One can actually use here three-fourths
of any rectangle (see

**Fig. 24**). The new tile is also rep-

**for any**

*k*^{2}*. The proof of this fact is similar with the proof for the hexagon*

**k ≥ 1***, and we leave the details to the interested reader.*

**H1**

**Figure 25:**

**Figure 26:**

A more interesting modification is to take the second hexagon * H2* and to skew it as in

**Fig. 25**. The angle

*is between*

**A****0**and

**Π**. We will call the new tile

*(skewed hexagon). It is easy to see that*

**SH***is rep-*

**SH****for any**

*k*^{2}*even. Nevertheless, we believe that*

**k***is*

**SH***not*rep-

**for any**

*k*^{2}*odd. We hope to solve this question in a future article. Let us only mention here that this question is equivalent to the question of tiling*

**k***using only translations (and no symmetries about the vertical or horizontal axis!) of the tiles shown in*

**kH2****Fig. 26**.

**Figure 27:**

There are two stellated polygons in the list, both shown in **Fig. 27**. We will denote them by * S1* and

*. The proof of the following proposition is left as an exercise to the reader.*

**S2****Proposition 6**

is rep-**S1**for any*k*^{2};**k ≥ 1**is rep-**S2**for any k even;*k*^{2}is not rep-**S2**for any*k*^{2}odd.**k**

### Bibliography

- J. H. Conway and J. C. Lagarias: Tiling with polyominoes and combinatorial group theory J. Combin. Theory Ser. A 53 1990 183-208
- S. W. Golomb: Replicating figures in the plane Mathematical Gazette 48 1964 403-412
- M. Gardner: On "rep-tiles", polygons that can make larger and smaller copies of themselves Scientific American 208 1963 154-157
- I. Pak: Ribbon tile invariants Trans. AMS 352 2000 5525-5561
- W. P. Thurston: Conway's tiling groups Amer. Math. Monthly 97 1990 757-773

### About this Document

**Rep-tiles Revisited**

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