'BEERS' TILES and 'BREEN'S' STAGGERING DOTS' |
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HISTORY - 'BEERS' TILESMany years ago when I was a member of the ACAVA artists studios at Faroe Road, an artist friend/colleague, one Tony Beers, told me about an intriguing discovery he had made, a mathematical/artistic curiosity concerning rectilinear square-tile patterns.This is a typical example of such a pattern of tiles with a central set (red) surrounded on its perimeter by another set (green). In this case a 'nominal' pattern of 6 by 5 tiles made up of 18 turquoise tiles surrounding 12 red tiles. In his wanderings, adventures and research Tony discovered a number examples of such arrangements but only a seemingly rare few where...
the number of perimeter tiles IS EQUAL TO the number of the central tiles.
He found these rare examples in primitive and vernacular decorations, in contemporary street furniture and architecture, etc, seemingly intentionally created or designed and usually aesthetically embellished and produced. 'BEERS' TILESBut of these rare examples he found only two variations limited to 24 tiles (48 in total) and 30 tiles (60 in total) respectively. And there were ONLY in the ratios of 6 x 8 overall (see blue/yellow below); and/or 12 x 5 overall (see purple/turquoise below).AND REFERENCE:Beers, Tony. 'Two Static States', in Julian Brown,'What's in a Number', Channel 4 Television. 1996. p.16.Tony subsequently developed his findings into a amazing portfolio of applications, variations and studies in various media. He can be contacted through ACAVA London. |
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BUT ARE THERE ONLY TWO SOLUTIONS ?The proof that there are only TWO UNIQUE SOLUTIONS starts with a little maths.The number of squares in the OUTER tile (perimeter) set MUST EQUAL the number of tiles on the INNER set. Take a 'nominal' tile set:
The OUTER (perimeter) tiles (blue squares) total is: 2(a -1 + b - 1) or, 2a + 2b - 4)
The INNER (centre) tiles (red squares) total must be: (a - 2) x (b - 2) Solving algebraically for positive integer solutions:
(a - 2)(b - 2) = 2(a -1 + b - 1)
ab - 2a - 2b + 4 = 2a + 2b - 4 ab - 4a = 4b - 8 a(b - 4) = 4b - 8 a = (4b - 8) / (b - 4) For this to have positive (integer) values the denominator (b - 4) must have b > 4 and hence, a > 4. Now simply check/substitute positive integar values:
To view/plot this graphically, running this equation/function through Graphsketch by Andy Schmitz: This is a hyperbola tending to 4.0 at infinity. This confirms that the only positive integer solutions are the couplets - 5,12 (12,5) and 6,8 (8,6). CONCLUSIONThere are only two cases of square tile rectilinear arrangements 30 + 30 and 24 + 24 where the sum of the perimeter tiles EQUALS the sum of the tiles forming the rectangle it surrounds. |
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QED |
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FOR FURTHER CONSIDERATIONTony Beers also found this example: This research revealed another possibility: |
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BREEN'S STAGGERING DOTSConsider the case of staggered dots (or even staggered rectilinear tiles, OR even BEER MATS). Typically: We need to determine if there are any solutions where the number of dots in the INNER tile set is EQUAL TO the number of dots on the OUTER (perimeter) tile set . Exploring for solutions starts again with some maths. The number of dots in the INNER tile set MUST EQUAL the number of tiles on the OUTER (perimeter) tile set:
The INNER tiles total is: (a - 1)(b - 1) + (a - 2)(b - 2) (dark blue dots + light blue dots), where a and b are:
VARIATION:
The OUTER (perimeter) tiles total must be: 2 x (a + (b - 2)) (white dots) Solving algebraically for positive integer solutions: (a -1)(b - 1) + (a - 2)(b - 2) = 2(a + (b - 2) (ab - a - b + 1) + (ab - 2a - 2b + 4) = 2a + 2b - 4 2ab - 3a - 3b + 5 = 2a + 2b - 4 2ab - 5a = 5b - 9 a( 2b - 5) = 5b - 9 a = (5b - 9) / (2b- 5) For positive integer values in this function, then for 2b - 5, b > 2.5 a = 6 and b = 3 and a = 3 and b = 6: the couplet 3,6 (6,3) as the ONLY SOLUTION. And running this equation/function (again through Graphsketch by Andy Schmitz): This is a hyperbola tending to +/- 2.5 at infinity. With the key values at 3,6 and 6,3 This confirms that there is only ONE POSITIVE INTEGER SOLUTION - the couplet 3,6 (6,3) resulting in 14 INNER dots and 14 OUTER dots. FURTHER RESEARCH UNDERWAYPlease send me corrections/improvements to the above; ideas or advice very welcome. April 17 2018 |