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
Activity 2-16: Triominoes
www.carom-maths.co.uk
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!
Can we fit 21 triominoes onto the board, with a one square gap?
This is possible.
But...
try as we might, the gap always appears in the same place, or in one of its rotations.
Can we prove that the blue position is the only position possible for the gap?
Proof 1: Colouring
So the gap must be white.
So the only possible solution has the gap as it is below:
Clearly this solution IS possible!
Proof 2: PolynomialsLet us put an algebraic termlogically into each square.
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: ?
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).
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
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
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.
So possible values for a + b are
4, 7, 10, 13.
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
What do we get?
(1+w)(1+w2) = wa+2b
So1= wa+2b
So a+2b is divisible by 3.
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!)
With thanks to:
Bernard Murphy, and Nick MacKinnon.
Carom is written by Jonny Griffiths, [email protected]