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Triominoes Puzzle: Find the One-Square Gap on a Chessboard

Explore the challenge of fitting 21 triominoes onto an 8x8 chessboard with a one-square gap. Discover the unique solution and the mathematical proofs behind it. Can you prove that the gap's position is fixed? Delve into colorings and polynomials to unveil the secrets of this intriguing puzzle.

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Triominoes Puzzle: Find the One-Square Gap on a Chessboard

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  1. www.carom-maths.co.uk Activity 2-16: Triominoes

  2. Take a blank chessboard.

  3. What happens if we try to cover this with triominoes?

  4. 64 = 3 x 21 + 1 So if we can fit 21 triominoes onto the board, then there will be a one square gap. Is this possible? Draw an 8 by 8 grid and try it!

  5. Can we fit 21 triominoes onto the board, with a one square gap? This is possible.

  6. But... try as we might, the gap always appears in the same place, or in one of its rotations.

  7. Can we prove that the blue position is the only position possible for the gap?

  8. Proof 1: Colouring So the gap must be white.

  9. So the only possible solution has the gap as it is below: Clearly this solution IS possible!

  10. Proof 2: Polynomials Let us put an algebraic term logically into each square.

  11. Now we will add all the terms in two different ways. Way 1: (1+x+x2+x3+x4+x5+x6+x7) (1+y+y2+y3+y4+y5+y6+y7) Way 2: ?

  12. What happens if we put down a triominohorizontally and add the squares? We get (1+x+x2)f(x, y). What happens if we put down a triominovertically and add the squares? We get (1+y+y2)g(x, y).

  13. So adding all thetriominoesplus the gap we get: F(x, y)(1+x+x2) + G(x, y)(1+y+y2) +xayb and so (1+x+x2+x3+x4+x5+x6+x7)(1+y+y2+y3+y4+y5+y6+y7) = F(x, y)(1+x+x2) + G(x, y)(1+y+y2) + xayb

  14. Now we know that if w is one of the complex cube roots of 1, then 1 + w + w2 = 0. So let’s put x = w, y = w in (1+x+x2+x3+x4+x5+x6+x7)(1+y+y2+y3+y4+y5+y6+y7) = F(x, y)(1+x+x2) + G(x, y)(1+y+y2)+xayb

  15. What do we get? (1+w)2 = wa+b So w= wa+b So 1= wa+b-1 So a+b-1is divisible by 3.

  16. So possible values for a + b are 4, 7, 10, 13.

  17. Now we know that if w is one of the complex cube roots of 1, then 1 + w2 + w4= 0. So let’s put x = w, y = w2in (1+x+x2+x3+x4+x5+x6+x7)(1+y+y2+y3+y4+y5+y6+y7) = F(x, y)(1+x+x2) + G(x, y)(1+y+y2)+xayb

  18. What do we get? (1+w)(1+w2)= wa+2b So 1= wa+2b So a+2bis divisible by 3.

  19. So possible values for a + 2b are 0, 3, 6, 9, 12, 15, 18, 21. Possible values for a + b are 4, 7, 10, 13. So possible values for b < 8 are 2, 5. So possible values for a are 2, 5 (QED!)

  20. With thanks to: Bernard Murphy, and Nick MacKinnon. Carom is written by Jonny Griffiths, mail@jonny.griffiths.net

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