Magic Hexagons in Regular Tilings - Recent Developments

Transcription

1 Magic Hexagons in Regular Tilings - Recent Developments by Hans F. Bauch from Stralsund revised October 010 Hans Friedrich.Bauchfh stralsund.de In this article we shall investigate simple figures on regular tilings. The task is to find magic allocations for these figures. The figures are compounded from triangles or hexagons with 3 times n straight lines. The number f(n) is the number of cells of the figure - like we know it from the well known squares, where f(n) = n. An allocation with the numbers from c+l to c+ f n, where c is any integer, f.i. zero, is called magic if the same (magic) sum m = m(n) in each straight line occurs. Which conditions and examples exist? We will give an overview. Contents 1. Regular tilings. Squares and their counterparts.1 Polygon-centred figures. Vertex-centred figures 3. Classical magic allocations and sums - necessary conditions 3.1 Polygon centred figures (n=k-1) Hexagon tiling (p=3) 3.1. Triangle tiling (p=6) 3. Vertex centred figures (n=k) 3..1 Hexagon tiling (p=3) 3.. Triangle tiling (p=6). Generalized magic allocations and sums - necessary conditions.1 Polygon centred figures (n=k-1).1.1 Hexagon tiling (p=3).1. Triangle tiling (p=6). Vertex centred figures (n=k)..1 Hexagon tiling (p=3).. Triangle tiling (p=6) 5. Classical magic hexagons - solutions 5.1 The original magic hexagon HC(5) of Ernst von Haselberg 5. Vertex-centred T-hexagons 6. Generalized magic hexagons - solutions 6.1 Vertex-centred H-hexagons 6. Triangle-centred T-hexagons 7. Resume References 1. Literature. Web sites

2 010magic1110.nb 1. Regular tilings By a regular tiling, we mean that the Euclidean plane is covered by regular polygons with common sides and vertices. There are three regular tilings, triangular (T), square (S) and hexagonal (H). At each vertex, p different polygons meet; p = 6 triangles meet on the triangular tiling, p = squares meet on the square tiling and p = 3 hexagons meet on the hexagonal tiling. We want to look at series of simple figures on the three tilings. The figure F(l) is an individual polygon. We then add all neighbouring polygons to enclose the polygon. Since the new figure consists of three parallel rows, we call it F(3). From each figure F(k- l) with k-1 parallel rows we can develop the figure F(k+l) with k+1 parallel rows by this ringing process. But squares with even numbers of little squares are centred not around a square but around a vertex. So we get two possibilities in each of the three tilings. We will call them in general T-hexagons and H-hexagons following [BA/KI 006] ( and squares)..1. Polygon-centred figures If n = k-1 (k ), the described figures are triangle centred TC-hexagons in the T-tiling and hexagon centered HC-hexagons in the H-tiling ( and square centered SC-squares in the S-tiling). The number of cells is [BAU 1991]: f n p n p It is clear that each f(n) is odd. p k 1 p k for n k 1

3 010magic1110.nb n f n p p 6.. Vertex-centred figures If n = k (k ), the described figures are vertex centred TV-hexagons in the T-tiling and vertex centred HV-hexagons in the H-tiling (and SV-squares in the S-tiling). The number of cells is [BAU 1991] f n p n p k p k for n k It is clear that for p = 6 each f(n) is even and for p = 3 f(n) is odd and even alternating with n. n f n p p 6 3. Classical magic allocations and sums - necessary conditions If we want to write the natural numbers from 1 to f(n) into the cells - triangles or hexagons - of the T-hexagon or the H- hexagon with f(n) cells such that in each row results the same sum m(n) we have to divide the sum of all numbers from 1 to f(n) by the number n of parallel rows to get m(n). This is the classical case. Later we will investigate the possibilities which arise if we use c+1, c+,..., c+f(n) with an integer c which is not zero. m n 1... f n n f n f n 1 n

4 010magic1110.nb A necessary condition for a magic allocation is: m has to be integer! 3.1 Polygon centred figures (n=k-1) n=k-1: m n p n p p n p n 1, that is the same as m k 1 p k 1 p p k p k Hexagon tiling (p=3) m n 3 n 1 3 n 5 3 n m n 9 n 3 18 n 5 n 3 This number is integer if and only n = 1 (trivial) or n = 5 with m = 38. That's why only HC(5) is under further consideration for magic allocation with 1,,..., 19. Only HC(5) is to be investigated. HC Triangle tiling (p=6) m n 6 n 6 n 3 n m n 9 n3 1 n 8

5 010magic1110.nb this number is an integer if and only if n = 1. That's why no TC(n) is under further consideration for magic allocation with 1,, 3,,.... No TC(n) is to be investigated. TC 3 3. Vertex centred figures (n=k) n=k: m n p n p n 1 n, that is the same as m k p k p k Hexagon tiling (p=3) m k 3 k 3 k 1 3 k 3 3 k 1 k k 1 This number is integer if and only if k is not of the form k = m+ (m ), that is, n is not of the form n = k = 8m+ (m ). That's why HV(), HV(1), HV(0),... are out of further consideration for magic allocations with 1,, 3,,..., f(n).

6 6 010magic1110.nb HV(6), HV(8), HV(10), HV(1), HV(16), HV(18), HV(),... have to be investigated. HV 6 but: Theorem [BAU 1991] No HV(n) has a classical magic allocation. No HV(n) has to be investigated. 3.. Triangle tiling (p=6) m k 6 k 6 k 1 9 k 3 3 k This number is integer if and only if k is even, that is n is of the form n = k = m (m ). That's why TV(), TV(6), TV(10),..are out of further consideration for magic allocation with 1,, 3,,..., f(n). TV(), TV(8), TV(1), TV(1), TV(16),... has to be investigated. TV. Generalized magic allocations and sums - necessary conditions If we try to choose an uninterruped sequence of integers starting not only with 1 but with an arbitrary integer c+1 to have the same sum m(n,c) in each row of a figure F(n) from our series, we define this as a generalized magic allocation. We will give now necessary conditions for the generalized magic sum m(n,c):

7 010magic1110.nb m n, c c 1 c... c f n n m n, c c f n n f n f n 1 n m n, c f n f n c 1 n m(n,c) = 0 We can ask especially for the possibility of m(n,c) = 0: m n, c 0f n c 1 0 f n 1 f n is an odd integer and c Then we have to use exactly the integers from c+1 = - (f(n)+1)/ + 1 = - (f(n)-1)/ to (f(n)-1)/ for a possible magic allocation with magic sum 0.. If f(n) is odd, then it always possible to fulfil the necessary condition for m=0..1 Polygon centred figures (n=k-1) LEMMA: Only the following integers c fulfill the necessary condition for a generalized magic allocation, that is, m is integer. f n 1 c n j j Proof: If we look at the formula for c, we see immediately that the generalized magic sum m(n,c) is integer because each f(n) is odd: m f n f n f n1 n that is n j 1, m f n j If we add to that c an integer r, cc=c+r, with 0 < r < n, we will not get mm=m(cc,n) as integer: Let fn=f(n), then It follows fn 1 cc n j r mm fn fn fn1 n 0 r n n j r 1

8 8 010magic1110.nb cc 1 1 fn j n r mm fn 1 fn 1 1 fn j n r n fn j fn r n and this is not integer because: fn r n p n p r n 1 r n 1 r n 3 n r for p 3 3 n r for p Hexagon tiling (p=3) For the hexagon tiling and j=0 we get: n f n c m HC 5 Different ranges of integers in HC(5) (following Hoelbling, Zahray, Sillke) The impossibility of c > 0 for HC(5): We use the sequence c+1,..., c+19 : See [BAU 1990] or [SIL] for the notations: Eliminating M = m(5,c) from B = M + d and C = M - d gives B = C + 6d. Now B 19+c+18c+17+c+16+c+15+c+1+c = 99+6c but C+6d 6(1+c)+(+c+3+c++c+5+c+6+c+7+c) = 60+18c. As 100+6c > B = C + 6d > 59+18c we have > 1c. Therefore c > 0 is impossible because c has to be divisible by 5. For c = -10 we have M = m(5,-10) = 0, for c = -5 we use the sequence -,...,1, for c = -15 the sequence -1,...,. That is why a solution for c = -5 by changing the signs gives a solution for c = -15 and the original solution gives the unique solution for c = -0 where in every row M = -38. Smaller c are, following the proof above, impossible. Only for c=0, -5, -10, -15 the necessary condition for HC(5) is fulfilled..1. Triangle tiling (p=6) For the triangle tiling and j = 0 we get:

9 010magic1110.nb n f n c m TC 3 Different ranges of integers in TC (3) (following Rudolph-oral communication) There are the following equations for TC(3): b 6 a 1 b 1 m b a b 3 m b a 3 b 5 m a b 3 m a 1 b 1 c 1 b a m a b 3 c b a 3 m a 3 b 5 c 3 b 6 a 1 m a b c 3 m b 1 c 1 d c b m b 3 c d c 3 b 6 m b 5 c 3 d c 1 b m b c 3 d 3 m

10 10 010magic1110.nb a d b 3 m d c d Because c+d=0, in {c 1, c, c 3, d} are integers with different signs. So following.1 generalised allocations could be possible only for 13 integers -6 to 6 with m=0 or -3 to 9 with m=13 or -9 to 3 with m=-13 but not 0 to 1 with m=6 or -1 to 0 with m=-6. We would like to investigate the special case m = 0 for all HC(k-1) and TC(k-1). Vertex centred figures (n=k)..1 Hexagon tiling (p=3) m n, c 3 k 3 k 1 c k 3 k 3 3 k 1 k k 1 3 k c this number is not integer if and only if k is of the form k = m+ (m ), that is n is of the form n = k = 8m+ (m ). That's why HV(), HV(1), HV(0),..are out of further consideration for generalized magic allocations. If k is odd and c is even then m(n,c) is integer as well as if k is divisible by and c is integer. So there is a big amount of possibilities. We prefer to choose, if possible, m = 0, that means c = -(f(n)+1)/. See a table for the first n: n f n c m HV(6), HV(10), HV(1), HV(18), HV(),... have to be investigated for m = 0. HV 6

11 010magic1110.nb HV(), HV(1), HV(0),... are out of further consideration, because no integer m is possible. HV For the last case, if n is divisible by 8, we have no possibility for magic allocation with m = 0, but with other m's it could be possible. See a table with minimal positive generalized magic sums : n f n c m HV(8), HV(16), HV(),... have to be investigated for m = 3n/8, the smallest positive possible integer for m. HV 8.. Triangle tiling (p=6) m k, c 6 k 6 k 1 c k 9 k 3 3 k 3 k c This number m is integer if and only if k is even, that is, n is of the form n = k = m (m ). Then f(n) is even as we have seen before and m = 0 is impossible but other m's could be possible. if n is not divisible by four then m is not an integer, that's why the cases TV(), TV(6), TV(10),... are out of further consideration for generalized magic allocations. See a table with minimal positive m if n is divisible by : n f n c m

12 1 010magic1110.nb TV(), TV(8), TV(1), TV(16),... have to be investigated for m=3n/, the smallest positive possible integer. TV TV(), TV(6), TV(10), TV(1),... are out of further consideration for generalized magic allocation, because no integer m is possible. TV 5. The traditional magic hexagon and new classical magic hexagons 5.1. The original magic hexagon HC(5) of Ernst von Haselberg Look for a lot of informations about this mathematical jewel the literature [Bau 1990], [Bau 1991], [HEM 1988] and of course [GAR 1963]. The latest paper gave the problem to the mathematical public in 1963 after some other attempts like [HAS 1888], [HAS 1889] and [VIC 1958] which were in vain in finding publicity.

13 010magic1110.nb Vertex-centred T-hexagons This problem for TV () was as far as we know first posed and published and examples of solutions were given in the paper [BAU 1991]. See two examples for classical magic allocations with the smallest sum in the inner ring and the maximal possible sum in the inner ring for TV () which were done by computer programs by our colleagues J. G. Lehmann and H.- D. Gronau. There are listed some properties of these allocations.

14 1 010magic1110.nb Further magic allocations and special properties are given in [BA/KI 006], [BAK], [KIN] and [KR/HA 005]. Baker and King found also examples of classical magic allocations for TV(8), TV(1) and TV(16), see [BAK]. 6. Generalised magic hexagons Following. there are a lot of figures of the series HC, TV, HV and TC which are to be investigated. For examples for generalised magic allocations of HC (5), HC (7), HC (9), HC (11), HC (13) and HC(15) see [ZAH], [HOE], [SIL] and [KOE]. We want to examine possible HV-hexagons and TC-hexagons with m equal zero or m near zero following a hint of Torsten Sillke. The TV series is not examined till now - perhaps because there is a lot of classical magic allocations. 6.1 Vertex-centred H-hexagons Problem: Find a sequence of integers c+1, c+,c+3,...,c+f(n), where f(n) is the number of cells in the hexagon HV(n) and n is the number of parallel rows such that in each row the same sum m of the numbers in the cells of the row occurs. Choosing f.i. n = 6 we have 3 times 6 parallel rows and f(6)=7 integers. THEOREM [BAU 1991]: There is no classical magic allocation for HV(k). THEOREM: If m 0, there is no generalized magic allocation for HV(k). The proof works the same way for generalised magic allocations and for all integer c. COROLLARY: No HV(m) has a magic allocation. The proof is trivial, because m=0 is impossible as we have seen in..1 the smallest generalised magic hexagon of type HV(n) The smallest HV( m + ) is HV(6). The number f (6) of tiles is 7. To reach m = 0, the start number c + 1 has to be - (7-1)/ = -13. Problem : Find a generalised magic allocation for HV(6) with the numbers - 13, -1,..., 0, 1,..., 13! A6 B3 A5 C5 D6 D5 C C6 E3 F3 E C3 A1 D1 F1 F D A B1 D E1 D3 B A C1 C A3 HV 6

15 010magic1110.nb One Solution : By direct communication after posing this problem to my students as a christmas challenge I got solutions from Christian Schäfer, Ronny Zapf and Birgit Eschen. There are = ****3*7*351 solutions. If we consider the symmetries of mirroring (3 axes), rotating (k*10 ) and changing the signs (+/-) then we get.757 strong different magic allocations (R. Zapf, 006). Allocations for other HV(m+), m=, 3,... are open. Of course HV() has no solution. 5.. Triangle-centered T-hexagons For all n is f(n) for TC (n) odd. It follows that m = 0 is always possible. The allocation numbers are the unbroken sequence integers between - (f(n)-1)/ and (f(n)-1)/ which is symmetric to zero. TC(3) is the smallest hexagon from our four series with only f(3)=13 tiles. Is this the smallest generalised magic hexagon? Problem: Find a magic allocation with the 13 integers -6,...,6 and magic sum 0! This problem I posed in 007 together with TC (5), f (5) = 37, m = 0, c = -19 and consequently integers from - 18 till 18. There are ten substantial different solutions which give by symmetries 10 solutions. They - and also solutions for TC(5) - were given to me by my student Matias Palomäki who solved the problem by a computer program in the beginning of 008. After some discussions my high school student Kathlén Kohn gave an algorithm to solve the problem without the use of a computer in November 008.

16 16 010magic1110.nb So find solutions for the smallest magic hexagon TC(3) by your own without a computer!! 7. Resume The smallest magic figures in the six series are SC(3) with 9 tiles, TC(3) with 13 tiles, SV() with 16 tiles, HC(5) with 19 tiles, TV() with tiles, SC(5) with 5 tiles and HV(6) with 7 tiles. References 1. Papers [BAU 1990] Bauch, H.F.: Das magische Sechseck von Ernst von Haselberg.Wissenschaft und Fortschritt 0:9 (1990) [BAU 1991] Bauch, H.F.: Magische Figuren in Parketten, Math. Semesterberichte 38:1(1991) [BA/KI 006] Baker, J.E. and D.R. King: Magic Hexagons - Magic Moments, Math. Gaz., Vol. 90, Num. 18 (006) [GAR 1963] GARDNER, M.: Permutations and Paradoxes in Combinatorial Mathematics, Scientific American 09 (1963) [HAS 1887] von Haselberg, E.: Das magische Sechseck (manuscript) (finished ) [HAS 1888] von Haselberg, E.: Section 795: Zeitschrift für math. und naturwiss. Unterricht 19 (1888) 9, Problem [HAS 1889] von Haselberg, E.: Section 801: Zeitschrift für math. und naturwiss. Unterricht 0 (1889) 63-6, Solution [HEM 1988] Hemme, H.: Das magische Sechseck, Bild der Wissenschaft (Oktober 1988) [KR/HA 005] Krause, S. and H. Harborth: Magic Euclidean Gameboards, Congressus Numerantium 173 (005) [VIC 1958] Vickers, T. : Magic hexagon, The mathematical gazette (December 1958) 91, note 799. Web sites [BAK] BAKER, J.E. [HOE] HOELBLING, L. in [KIN] KING, D.R. [KOE] KÖLLER, J. [SIL] SILLKE, T. [ZAH] Zahray, A. in