activity 2-16: triominoes
DESCRIPTION
www.carom-maths.co.uk. Activity 2-16: Triominoes. Take a blank chessboard. What happens if we try to cover this with triominoes ?. 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!. - PowerPoint PPT PresentationTRANSCRIPT
![Page 1: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/1.jpg)
Activity 2-16: Triominoes
www.carom-maths.co.uk
![Page 2: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/2.jpg)
Take a blank chessboard.
![Page 3: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/3.jpg)
What happens if we try to cover this with triominoes?
![Page 4: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/4.jpg)
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!
![Page 5: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/5.jpg)
Can we fit 21 triominoes onto the board, with a one square gap?
This is possible.
![Page 6: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/6.jpg)
But...
try as we might, the gap always appears in the same place, or in one of its rotations.
![Page 7: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/7.jpg)
Can we prove that the blue position is the only position possible for the gap?
![Page 8: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/8.jpg)
Proof 1: Colouring
So the gap must be white.
![Page 9: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/9.jpg)
![Page 10: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/10.jpg)
![Page 11: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/11.jpg)
![Page 12: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/12.jpg)
So the only possible solution has the gap as it is below:
Clearly this solution IS possible!
![Page 13: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/13.jpg)
Proof 2: PolynomialsLet us put an algebraic termlogically into each square.
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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: ?
![Page 15: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/15.jpg)
What happens if we put down a triomino horizontally and add the squares?
We get (1+x+x2)f(x, y).What happens if we put down a
triomino verticallyand add the squares?
We get (1+y+y2)g(x, y).
![Page 16: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/16.jpg)
So adding all the triominoesplus 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
![Page 17: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/17.jpg)
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
![Page 18: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/18.jpg)
What do we get?
(1+w)2 = wa+b
Sow = wa+b
So 1 = wa+b-1
So a+b-1 is divisible by 3.
![Page 19: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/19.jpg)
So possible values for a + b are
4, 7, 10, 13.
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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 = w2 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
![Page 21: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/21.jpg)
What do we get?
(1+w)(1+w2) = wa+2b
So1= wa+2b
So a+2b is divisible by 3.
![Page 22: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/22.jpg)
So possible values for a + 2b are
Possible values for a + b are
So possible values for b < 8 are
2, 5.
4, 7, 10, 13.
0, 3, 6, 9, 12, 15, 18, 21.
So possible values for a are
2, 5 (QED!)
![Page 23: Activity 2-16: Triominoes](https://reader036.vdocuments.site/reader036/viewer/2022070401/5681360a550346895d9d8153/html5/thumbnails/23.jpg)
With thanks to:
Bernard Murphy, and Nick MacKinnon.
Carom is written by Jonny Griffiths, [email protected]