TESSELLATIONS. This is a sample (draft) chapter from: MATHEMATICAL OUTPOURINGS. Newsletters and Musings from the St. Mark s Institute of Mathematics

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1 TESSELLATIONS This is a sample (daft) chapte fom: MATHEMATICAL OUTPOURINGS Newslettes and Musings fom the St. Mak s Institute of Mathematics James Tanton This mateial was and can still be used as the basis of a successful MATH CIRCLE activity.

TESSELLATIONS: Newslette Septembe 010 SEPTEMBER 010 THIS MONTH S PUZZLER: TESSELLATIONS AND TILINGS A polygon is said to tessellate the plane if it is possible to cove the entie plane with conguent

2 TESSELLATIONS: Newslette Septembe 010 SEPTEMBER 010 THIS MONTH S PUZZLER: TESSELLATIONS AND TILINGS A polygon is said to tessellate the plane if it is possible to cove the entie plane with conguent copies of the figue without ovelap (except along the edges of the figues). The tessellation is called a tiling if each edge of one polygon matches an entie edge of an adjacent polygon. 1. The tiling with tiangles shown is peiodic, meaning that it possesses tanslational symmeties in (at least) two non-paallel diections. That is, it is possible to shift the entie tiling along some diection of tanslation and etun to pecisely the same coveing of the plane, and again along some second, non-paallel diection. Is it possible to use a tiangula tile to ceate a non-peiodic tiling of the plane? [Comment: The definition of peiodic equies a tiling to possess at least two independent tanslational symmeties. Is it possible fo a tiling to possess tanslational symmety in just one diection? If so, constuct an example of such a tiling.]. Squaes, ectangles, and paallelogams all cetainly tile the plane. Does evey quadilateal tile the plane? (Even concave quadilateals?) Paallelogams tile the plane. As two copies of the same tiangle can be placed side-byside to fom a paallelogam we see that: Evey tiangle tiles the plane. 3. Is thee a five-sided polygon that tessellates the plane? How about one that tiles the plane? Think about these puzzles befoe eading on. DOES IT TILE? Regula hexagons tile the plane:

In fact, any hexagon with opposite sides paallel will tile the plane: TESSELLATIONS: Newslette Septembe 010 In summay, we have: Evey tiangle tiles the plane. Evey quadilateal tiles the plane.

) Daw an example of a hexagon that does not tile the plane. Notice that two copies of any quadilateal (convex o concave) placed side-by-side poduces a hexagon with paallel opposite sides.

This shows that all quadilateals do in fact tile the plane: Thee is a pentagon that tessellates the plane: Do thee exist convex n -gons, with n 7, that tile the plane? Mathematicians know the answe.

3 In fact, any hexagon with opposite sides paallel will tile the plane: TESSELLATIONS: Newslette Septembe 010 In summay, we have: Evey tiangle tiles the plane. Evey quadilateal tiles the plane. Thee exist pentagons that tile the plane. Thee exist hexagons that tile the plane. QUESTION: Why doesn t a egula pentagon tile the plane? (Which egula polygons do?) Daw an example of a hexagon that does not tile the plane. Notice that two copies of any quadilateal (convex o concave) placed side-by-side poduces a hexagon with paallel opposite sides. Fo each n 7 it is easy to ceate a concave n -gon that tiles the plane. This shows that all quadilateals do in fact tile the plane: Thee is a pentagon that tessellates the plane: Do thee exist convex n -gons, with n 7, that tile the plane? Mathematicians know the answe. It s to the negative. Theoem: No convex n -gon with n 7 tiles the plane. Even wose, no convex n - gon with n 7 tessellates the plane. A NON-PERIODIC TILING: An isosceles tiangle of the appopiate dimensions can be used to make a non-peiodic tiling of the plane. The otational symmety does not tanslate. Thee is even one that tiles the plane! Excise an equilateal tiangle fom a squae to poduce a five-sided shape. It tiles! Split the design to make a spial! RESEARCH CORNER: It emains an unsolved poblem to this day as to which pentagons tessellate the plane. Cuently only 14 diffeent types of pentagons ae known to wok. Cae to finish solving this classification poblem? [High-schooles in New South Wales, Austalia, have contibuted to this poject!] 010 James Tanton St. Mak s Institute of Mathematics. mathinstitute@stmaksscnhool.og

4 TESSELLATIONS: Commentay TESSELLATIONS Septembe 010 COMMENTARY, SOLUTIONS and THOUGHTS In the newslette we pesented a esult due to Ivan Niven: Theoem: No convex n -gon with n 7 tessellates the plane. Hee is an oveview of the poof. The full details can be found in [NIVEN1] (and also [NIVEN] fo the case of a tiling). Poof: Suppose we have a tessellation of the plane with a convex n -gon, n 7. Suppose the plane is equipped with a set of coodinate axes scaled so that the peimete of each tile is one unit in length. (Each tile consequently has width no moe than half a unit.) Let A be the aea of each tile. Fo each eal numbe let S be the set of tiles in the tessellation that each cove some point in the disc of adius centeed about the oigin, and let S denote the numbe of tiles in this sub-tessellation. As no tile has width geate than 1, the tiles of S cove the disc of adius but do not extend beyond a disc of adius + 1. Consequently we have: π S A ( ) < π + 1 Let v be the numbe of vetices that appea in the sub-tessellation given by S. Of the π adians of tuning that suounds each vetex, some potion of this coesponds to the measue of inteio angles of tiles. (In a tiling, all inteio vetices of the sub-tiling ae fully suounded by inteio angles of tiles, and bounday vetices only patially so. Fo a tessellation, some inteio vetices lie on the side of anothe tile and ae not fully suounded by tile inteio angles.) The quantity π v is an ove-estimate of the sum of the inteio angles of the tiles of S. As the sum of inteio angles of one tile is ( n ) π we have: ( ) π v n π S On the othe hand, since each tile is convex, at least thee tiles suound each vetex in the full tessellation. As all the vetices that appea in S ae inteio vetices of the subtessellation S + 1, and as each tile of S + 1 has n vetices we have: n S 3v James Tanton

5 TESSELLATIONS: Commentay 5 Now we just play with these thee sets of inequalities to find a contadiction. The latte two inequalities give: n n S+ 1 S 3 Multiplying by A and invoking the fist inequality gives: n n n n π ( ) + > S A S A π Consequently: n 3n 6 + A quick execise in algeba shows that fo n 7 we have 14 < + 15 n 14, and so: 3n 6 15 fo all values of. This is a contadiction as the left hand side appoaches the value 1 as inceases.

NON-PERIODIC TESSELLATIONS AND TILINGS TESSELLATIONS: Commentay 6 Tile the plane with squaes and divide each tile along its diagonal to obtain a tiling with isosceles ight tiangles.

6 NON-PERIODIC TESSELLATIONS AND TILINGS TESSELLATIONS: Commentay 6 Tile the plane with squaes and divide each tile along its diagonal to obtain a tiling with isosceles ight tiangles. If we do this in a haphazad manne we can ensue that no tanslational symmety holds and theeby poduce a non-peiodic tiling of the plane. COMMENT: In some eal sense this set of diections is unsatisfactoy: How can we, as humans, specify the diection of each diagonal fo an infinite aay of cells and also be sue that no tanslational symmety holds? The bette thing to do would be to specify some cleve algoithm fo assigning choices, pehaps based on the distibution of pime numbes, fo instance one that we can pove is sue to disupt peiodicity. Cae to ty? We can cetainly use this appoach to poduce a tiling of the plane with isosceles ight tiangles possessing only one diection of tanslational symmety: declae one column of squaes to be column 1 and assign one type of squae diagonal just to the pimenumbeed columns to its ight. CHALLENGE: Pove that if p is pime, thee is no numbe k such that p, p+ k, p+ k, p+ 3k, ae all pime.

7 TESSELLATIONS: Commentay 7 A tile is said to self-eplicate if a finite numbe of conguent copies of itself fit togethe to make a lage scaled copy of the tile. Self-eplicating tiles povide an inteesting souce of tessellations fo the plane. Fo example, fou copies of a squae stack togethe to make a lage squae. Repeating this pocess ad infinitum poduces the egula squae tiling of the plane: All tiangles self-eplicate, as does the L-shape and an isosceles tapezoid (each in moe than one way!) CHALLENGE: Find a figue with the popety that just two copies of itself fit togethe to poduce a scaled vesion of that figue. CHALLENGE: The L-shape shown above is subdivided, if you like, into fou and into 5 copies of itself. Show that one can also subdivide the L-shape into 9 and into 16 copies of itself. Can it be subdivided into a non-squae numbe of copies?

8 TESSELLATIONS: Commentay 8 I am pesonally enamoed with the 1 5 ight tiangle: five copies of this figue stack togethe to fom anothe 1 5 ight tiangle. And if we iteate this self-eplication an infinite numbe of times we obtain a tessellation of the entie plane. (It is called the pinwheel tessellation. See [RADIN].) And the esulting tessellation is non-peiodic! COMMENT: Students at St. Mak s School have coveed a coido wall with six iteations of this self-eplication stating with 1-inch wide constuction-pape tiangles. The esultant, and enomous, poduct is beautifully mesmeizing! This tiling was also used to decoate the Fedeation Squae Buildings in Melboune, Austalia. (See, fo example, [WEISSTEIN].) The non-peiodicity of the tessellation follows swiftly fom the following lemma. (Unfotunately, poving the lemma takes some wok! We ll leave this to the end of ou discussion.) Lemma: The smallest angle in a 1 5 ight tiangle, namel,y iational multiple of π. 1 x= actan, is an

9 TESSELLATIONS: Commentay 9 Notice that the hypotenuse of the initial ight tiangle makes an angle x with the hoizontal. We depict this as follows: The fist iteation of self eplication poduces a line segment of angle x with the hoizontal (its hypotenuse) and a cental line segment of angle x to the hoizontal (the hypotenuse of the cental copy of the oiginal): And the second iteation poduces line segments of angles x, x and 3x to the hoizontal: And so on. As x is an iational multiple of π, the infinite list of multiples of x contains no epeat angles (even modulo π ). Thus the final infinite tessellation of the plane contains line segments at infinitely many diffeent angles to the hoizontal. If the tessellation is peiodic, thee must be two diections of tanslation symmety. That is, thee is a distance a in one diection one can tanslate the tessellation to poduce an identical copy of itself, and a distance b in a second diection one can do the same. Thus thee is a paallelogam defined by these two isometies which offes the template fo the epeating design of the tessellation. But being a bounded egion, this template can only contain a finite numbe of line segment angles, focing the same fo the entie tessellation. As we have seen, this is not the case. Thus the tessellation cannot be peiodic.

10 TESSELLATIONS: Commentay 10 HARD CHALLENGE: Is the tessellation esulting fom the self-eplication of the L- shape peiodic? 1 PROOF OF LEMMA: We need to establish that actan is an iational multiple of π. The poof hee is based on the wok of J.M.H. Olmsted ([OLMSTED]). We begin by collating some facts about the tangent function. Recall that the angle addition fomula states: tan + = tan a+ tan b 1 tan a tan b ( a b) fom which it follows that if tan x t tan x= 1 t 3 t 3t tan 3x= tan 4x 3t 1 3 = 4t 4t 4 t + t 6 1 = t, then and so on. An induction agument establishes that tan nx= ( ) ( ) p t q t with p( t ) and q( t ) polynomials in t beginning with highest ode tems of the fom: 1 ( ) = nt n n + and q( t) t n 1 ( ) = t + and q( t) nt n p t p t = + if n is even = + if n is odd This shows:

11 OBSERVATION 1: If tan x TESSELLATIONS: Commentay 11 = t is ational, then tan nx is also ational (if it is finite). p OBSERVATION : If tan x= fo integes p and q, and tan nx= 0, then p q q is a ational oot of a polynomial with leading coefficient n o 1. We thus have q n. We ae inteested in the case whee tan x equals 1. By obsevation 1, each value of tan nx is also ational (when it is finite). Wite: tan x= p q n n as a faction in educed fom (when the quantity is finite). OBSERVATION 3: The denominatos qn can be abitaily lage. p To see this, suppose that tanα = is a faction witten in educed fom with one of p o q pq q even, and set so that q is positive. Then it is not difficult to see that tan α = q p is also in educed fom. Moeove, if q> p, then ( )( ) q p q p q p q p q = + > + >, and if p > q, then p q p q q > + >. pq Eithe way, when is witten as a faction with positive denominato that q p denominato is lage than q. 1 Now we have tan x=, which is a faction in educed fom with even denominato. It follows that the denominatos of tan x, tan 4x, tan 8x, thus gow without bound. We ae now set to complete the poof.

12 TESSELLATIONS: Commentay 1 1 a Suppose tan x= with x= π fo some integes a and b. Let α = nx. Then b pn tanα = tan nx= and tan( bα) = tan( na π) = 0. By Obsevation, this means that q qn n b fo all n. Thus each value q n is bounded by b. This contadicts Obsevation 3. REFERENCES: [NIVEN 1] Niven, I., Convex polygons that cannot tile the plane, Ameican Mathematical Monthly, 54 (1978), [NIVEN ] Niven, I., Maxima and Minima without Calculus, Mathematical Association of Ameica, Washington D.C., [OLMSTED] Olmsted, J.M.H., Rational values of tigonometic functions, Ameican Mathematical Monthly, 5 (1945), [RADIN] Radin, C., The pinwheel tilings of the plane, Annals of Mathematics, 139 (1994), [WEISSTEIN] Weisstein, E. W., Apeiodic tiling fom MathWold-A Wolfam Web Resouce. URL:

Also available at ISSN (printed edn.), ISSN (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 3 (2010)

Also available at ISSN (printed edn.), ISSN (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 3 (2010) Also available at http://amc.imfm.si ISSN 1855-3966 (pinted edn.), ISSN 1855-3974 (electonic edn.) ARS MATHEMATICA CONTEMPORANEA 3 (2010) 109 120 Fulleene patches I Jack E. Gave Syacuse Univesity, Depatment