NONCONGRUENT EQUIDISSECTIONS OF THE PLANE

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1 NONCONGRUENT EQUIDISSECTIONS OF THE PLANE D. FRETTLÖH Abstract. Nandakumar asked whether there is a tiling of the plane b pairwise non-congruent triangles of equal area and equal perimeter. Here a weaker result is obtained: there is a tiling of the plane b pairwise non-congruent triangles of equal area such that their perimeter is bounded b some common constant. Several variants of the problem are stated, some of them are answered.. Introduction There are several problems in Discrete Geometr, old and new, that can be stated easil but are hard to solve. Tilings and dissections provide a large number of such problems, see for instance [CFG, Chapter C]. On his blog [N], R. Nandakumar asked in 04: Question. Can the plane be filled b triangles of same area and perimeter with no two triangles congruent to each other? His webpage [N] contains several further interesting problems of this flavour. The main result of this paper, Theorem, answers a weaker form of the question above affirmativel. Section 4 is dedicated to the statement and the proof of this result, together with its analogues for quadrangles, pentagons and heagons. Section formulates several variants of this problem and gives a sstematic overview. Section contains some basic observations and a first result on a similar result for quadrangles... Notation. A tiling of R is a collection {T, T,...} of compact sets T i (the tiles) that is a packing of R (i.e., the interiors of distinct tiles are disjoint) as well as a covering of R (i.e. the union of the tiles equals R ). In general, tile shapes ma be prett complicated, but for the purpose of this paper tiles are alwas simple conve polgons. A tiling is called verte-to-verte, if the intersection of an two tiles is either an entire edge of both tiles, or a point, or empt. A tiling T is locall finite if an compact set in R intersects onl finitel man tiles of T. A tiling T is normal if there are R > r > 0 such that () each tile in T contains some ball of radius r, and () each tile is contained in some ball of radius R. B [GS,..] we have that each normal tiling is locall finite.. Basic observations In Question above, filled is to be understood in the sense that the plane is covered without overlaps. In other words: is there a tiling of the plane b pairwise noncongruent triangles all having the same area and the same perimeter? If one tries to find a solution one realises that the problem seems to be highl overdetermined. One possibilit to rela the problem is to drop the requirement on the perimeter. So one ma ask Is there a tiling of the plane b pairwise noncongruent triangles all having the same area? It is not too hard to find eamples. One possibilit is to partition the plane into half-strips and divide these half-strips into triangles, as shown in Figure. The image indicates how to fill the right half-plane b half-strips made of triangles. If all halfstrips have different widths then all triangles are distinct. The left half-plane can be filled in an analogous manner. Anwa, this tiling is not locall finite: the upper verte of an half-strip is contained in infinitel man triangles. So one ma ask Is there a locall finite tiling of the plane b pairwise noncongruent triangles of unit area? Even in this stronger form the question was alread Date: March 9, 06.

2 D. FRETTLÖH Figure. Tiling of the plane b pairwise non-congruent triangles of unit area. The perimeters of the triangles are unbounded. Moreover, the tiling is not locall finite. answered b R. Nandakumar. The image in Figure shows a solution, see also [W]. The idea is to dissect the upper right quadrant into triangles of area b zigzagging between the horizontal ais and the vertical ais. The triangles become ver long and thin soon. Nevertheless the are filling the quadrant. For the remaining three quadrants one uses an analogous construction, perturbing the coordinates slightl. (For instance, stretch a cop of the first quadrant b some irrational factor q > in the horizontal direction and shrink it b q in the vertical direction. See [W] for an alternative, more detailed eplanation.) This tiling is locall finite. Nevertheless, this eample is not reall satisfing. More precisel, this tiling is not normal, since in this solution the perimeters of the triangles become arbitrar large. (Hence the inradii become arbitrar small). So it seems natural to ask: Question. Is there a normal tiling of the plane b pairwise noncongruent triangles of unit area? This question was alread asked b Nandakumar, in the form whether there is a tiling of the plane b pairwise noncongruent triangles all having unit area such that the perimeter of the triangles is bounded b some common constant. Theorem below answers this question affirmativel. One possible approach to find a solution is the following. If one can partition a set S R into triangles of unit area such that () S tiles the plane, and () the triangles can be distorted Figure. Tiling of the plane b pairwise non-congruent triangles of area.

3 NONCONGRUENT EQUIDISSECTIONS 4 Figure. Partitioning a square into 4 distinct quadrangles of equal area. There are uncountabl man possibilities for such a dissection. continuousl, in a wa such that all triangles in S are distinct (but still having unit area), this solves the problem. We will illustrate this concept (where triangle is replaced b quadrangle ) in the proof of the following result. Theorem. There is a normal tiling of the plane b pairwise non-congruent conve quadrangles of unit area. Proof. Consider a square Q of edge length. Let Q be a point such that () for the distance d of to the centre of Q holds 0 < d < 0, and () is neither contained in the diagonals of Q nor in the line segments connecting mid-points of opposite edges of Q. Let be a point on the boundar of Q such that for the distance d of to the mid-point of the edge containing holds 0 < d < 0. The choice of and determines three further unique points,, 4 on the boundar of Q such that the line segments i ( i 4) partition Q into four quadrangles of unit area. B the choice of, avoiding the mirror aes of Q, it can alwas be achieved that all quadrangles in a single partition of Q are distinct. (In fact the author believes that no two congruent quadrangles can occur in a partition where is not contained in the mirror aes of Q, but this might be tedious to prove. Here we prefer rather to use the free parameters to achieve that all quadrangles are distinct.) Figure indicates such a partition. The two coordinates determining can be changed continuousl within a small range independentl, ielding two free parameters. One coordinate of can be changed continuousl, too, within some small range. Hence we obtain the desired tiling as follows: Tile the plane R with copies of the square Q. Dissect each cop of Q into four quadrangles of area, in some order. In each dissection, choose and such that the resulting quadrangles have not shown up earlier in the construction. This is alwas possible since, in each step, there are onl finitel man quadrangles constructed alread, whereas there are uncountabl man choices for and. At this point it becomes obvious that Questions and lead to several variants. The eample in the proof above ields a normal tiling, but in general not one that is verte-to-verte. One ma ask the questions for triangles, for quadrangles, for pentagons, and in each case with or without requiring equal perimeter, or normal, or verte-to-verte. The net section aims to give a sstematic overview of the questions.. Variants of the problem The general propert we will require throughout the paper is that a tiling consists of conve tiles of unit area such that all tiles are pairwise distinct. The tiles can be triangles (as in the original question), but also quadrangles, rectangles, pentagons or heagons. We ma or ma not require additionall that all tiles have equal perimeter, or that the perimeter is bounded b some common constant, or that the tilings are normal, or just locall finite. Furthermore, it ma be possible to construct a tiling analogous to the proof of Theorem, that is, b tiling a tile S in infinitel man was, where S in turn can tile the plane. The connections between these properties is shown in the following diagram. } equal perimeter () perimeter is bounded normal locall finite tiling a tile

4 4 D. FRETTLÖH For instance, Equation () tells that if there is tiling obtained b tiling a tile S in infinitel man was, then the perimeters of the tiles in this tiling are bounded b some common constant. In turn, the latter is equivalent to the tiling being normal (since all tiles are conve and have unit area), which in turn implies (b [GS,..]) that the tiling is locall finite. These implications help to give an overview of the several variants of the questions. The following tables list, for each of the cases of triangles, conve quadrangles, conve pentagons, and conve heagons, whether there is some tiling known fulfilling the properties in Equation (), and whether there is such a tiling that is even verte-to-verte. Because of the implications in Equation (), if there is es in some column, then the entries above in the same column contain also a es. Note that not vtv is usuall a weaker condition than vtv, but a tiling b conve heagons that is not vtv is much harder to find than one that is vtv. Triangles vtv not vtv locall finite? es bounded perimeter? es tiling a tile?? Pentagons vtv not vtv locall finite? es bounded perimeter? es tiling a tile? es Quadrangles vtv not vtv locall finite es es bounded perimeter es es tiling a tile es es Heagons vtv not vtv locall finite es? bounded perimeter es? tiling a tile es? Theorem above proves the case quadrangles: tiling a tile, not vtv (thus also the two cases above it in the same column in the corresponding table). Theorem below proves triangles: bounded perimeter, not vtv, Theorem proves quadrangles: tiling a tile, vtv, Theorem 4 proves heagons: tiling a tile, vtv, and Theorem 5 proves heagons: tiling a tile, vtv. 4. Main results Theorem. There is a normal tiling of the plane b pairwise non-congruent triangles of unit area. Proof. The idea of the proof is a refinement of the construction in Figure. Basicall we add additional fault lines in each quadrant. Moreover, we make use of some free parameter in some range, allowing for uncountabl man choices, where in each step of the construction onl finitel man triangle shapes must be avoided. Choose some constant c big enough. This serves as the upper bound on the perimeter of the triangles. For our purposes c = 00 will do. Consider the upper right quadrant Q. Pick a point 0 on the positive horizontal ais with 0 < c. Let T be the unique triangle in Q with vertices 0, 0 and area. (For this and what follows compare Figure 4.) Denote the third verte of T b. Choose on the horizontal ais such that the triangle T with vertices 0, and has area. Continue zigzagging in this wa between horizontal and vertical ais. I.e., choose i+ on the ais not containing i such that the triangle T i+ with vertices i, i, i+ has area. Repeat this until the net triangle T i+ would have perimeter larger than c. Omit T i+. Pick such that the triangle i, i+, has area. There are uncountabl man choices for. For the sake of smmetr let be close to the bisector {(, ) R} of Q. Choose a half-line l proceeding from. There are uncountabl man choices for l. Again, for the sake of smmetr, let l be close to the bisector of Q. Continue b zigzagging in two regions, between the horizontal ais and l, and between the vertical ais and l. I.e., if k is the last point on the horizontal ais, pick t on the horizontal ais such that the triangle k,, t has area. Continue b choosing t on l such that the triangle, t, t has area and so on, until the perimeter of the net triangle t i, t i+, t i+ would be larger than c. Omit this triangle. Choose such that the new triangle t i, t i+, has area. Choose a half-line l proceeding from. Again there are uncountabl man choices for and l.

5 NONCONGRUENT EQUIDISSECTIONS 5 5 t 4 t l l l 0 0 t 4 t Figure 4. Tiling the upper right quadrant Q b pairwise non-congruent triangles of unit area and bounded perimeter. Do the analogous construction in the upper region between l and the vertical ais. Continue in this manner. Whenever a triangle occurs with perimeter larger than c choose a new point k and a new line l k dividing the old region into two. The uncountabilit of choices for k and l k ensures that we can alwas avoid to add a triangle that is congruent to some triangle added earlier. Indeed, whenever we are in the situation to choose k and l k there are at most countabl man triangles constructed alread. Hence k can be chosen such that no triangle with verte k is congruent to an alread constructed one, and l k can be chosen such that no triangle occurring in the two new regions defined b l k is congruent to an alread constructed one. Hence the quadrant Q can be tiled b pairwise non-congruent triangles with area and perimeter less than c. The other quadrants can be tiled accordingl. Whenever a choice of a new point and a new half-line happens there are uncountabl man possibilities, hence all (at most countabl man) alread constructed triangles can be avoided. Theorem. There is a normal vtv tiling of the plane b pairwise non-congruent quadrangles of unit area. The tiling consists of squares that are dissected into four distinct quadrangles of equal area. Proof. The idea is to use the construction in the proof of Theorem, adding (dissected) squares consecutivel, using the degrees of freedom to achieve verte-to-verte in neighbouring squares. Figure 5 indicates the order in which squares are added, and the degrees of freedom in the dissection of each square. Start with some square S, dissected as in the proof of Theorem. There are three degrees of freedom how to dissect S into four quadrangles, two for placing the centre of dissection, one for a point on the boundar. This square is indicated b a circled in Figure 5. Add four more dissected squares adjacent to S, such that the quadrangles are verte-to-verte. These squares are numbers to 5 in the figure. In each of these squares there are still two degrees of freedom for placing the centre. The third parameter is determined uniquel b the verte-to-verte condition. Still one ma use the two degrees of freedom to avoid adding a quadrangles that is congruent to one added alread. Now add four more squares (6 to 9), each one adjacent to two edges of squares to 5, respectivel. Now the position of two points of the dissection are determined for each of the squares 6 to 9. Hence, b the area condition, the centre of the square is restricted to some line. Anwa, it can

6 6 D. FRETTLÖH Figure 5. Tiling a square with distinct quadrangles of unit area (compare Figure ) can be done in a wa such that partitions of adjacent squares are verte-toverte. Circled numbers indicate the order in which squares are added consecutivel, ordinar numbers indicate the degrees of freedom in each square. A means that the centre can be wiggled within a small ball. A means that the centre can be shifted along some line b a small amount. The (approimate) directions of these lines are indicated b dashed line segments. be shifted along a small segment of this line continuousl. Hence there is still one free parameter that we can use to avoid adding a quadrangle that is congruent to some quadrangle added earlier. In this wa we continue filling the plane: add four squares along the horizontal and vertical aes (the net step would be adding squares 0 to in the figure), add more squares to the pattern to complete a square pattern. Proceeding in this wa ensures that in each step there is at least one free parameter that can be used to avoid adding a square congruent to one added earlier. Theorem 4. There is a normal tiling of the plane b pairwise non-congruent pentagons of unit area. The tiling consists of heagons that are dissected into three distinct pentagons of equal area. Proof. A regular heagon of area three can be divided into three heagons of unit area in uncountabl man was, compare the left part of Figure 6. a a Figure 6. A regular heagon can be divided into three distinct pentagons of equal area in uncountabl man was (left). A non-conve 4-gon can be divided into four distinct heagons of equal area in uncountabl man was.

7 NONCONGRUENT EQUIDISSECTIONS 7 Theorem 5. There is a normal vtv tiling of the plane b pairwise non-congruent heagons of unit area. The tiling consists of non-conve 4-gons that are dissected into four distinct heagons of equal area. Proof. Consider a non-conve 4-gon assembled from three regular heagons and a fourth heagon that is obtained from a regular heagon b stretching it slightl in the direction of one of the edges, see Figure 6 right. The longer edges are labelled with a in the figure. This 4-gon can be dissected into four heagons of equal area. There is still one parameter of freedom: one verte of the dissection can be shifted continuousl along a line segment, the other interior verte of the dissection is then determined uniquel b the area condition. The 4-gons ield a tiling of the plane: gluing 4-gons together at their edges of length a ields biinfinite strips. These strips in turn can be assembled into a tiling. During working on the problem the author tried several approaches. Based on this eperience we want to highlight the following problems for further stud. () Is there a compact conve region in the plane that can be tiled b non-congruent triangles of unit area in infinitel man (uncountabl man) was? () Is there a compact region in the plane that (a) can be tiled b non-congruent triangles of unit area in infinitel man (uncountabl man) was, and (b) tiles the plane? () Is there a verte-to-verte tiling of the plane b pairwise non-congruent triangles of unit area? (4) Is there a verte-to-verte tiling of the plane b pairwise non-congruent triangles of unit area such that the perimeter of the triangles is bounded b some common constant? (5) Is there a tiling of the plane b pairwise non-congruent rectangles of unit area such that the perimeter of the rectangles is bounded b some common constant? Acknowledgments The author epresses his gratitude to R. Nandakumar for providing several interesting problems. Special thanks to Jens Schubert for helpful discussions on this topic during a pleasant summer weekend in Bochum. References [CFG] H.T. Croft, K.J. Falconer, R.K. Gu: Unsolved Problems in Geometr, Springer, New York (99). [GS] B. Grünbaum, G.C. Shephard: Tilings and Patterns, W.H. Freeman, New York (987). [N] R. Nandakumar: post from Dec 04 and Jan 05. [W] S. Wagon: Bielefeld Universit, Postfach 00, 50 Bielefeld, German