activity 2-16: triominoes

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

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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?. - PowerPoint PPT Presentation

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Page 1: Activity  2-16:  Triominoes

Activity 2-16: Triominoes

www.carom-maths.co.uk

Page 2: Activity  2-16:  Triominoes

Take a blank chessboard.

Page 3: Activity  2-16:  Triominoes

What happens if we try to cover this with triominoes?

Page 4: Activity  2-16:  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!

Page 5: Activity  2-16:  Triominoes

If we can fit 21 triominoes onto the board, then there will be a one square gap.

This is possible.

Page 6: Activity  2-16:  Triominoes

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

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

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Proof 1: Colouring

So the gap must be white.

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So the only possible solution has the gap as it is below:

We should note that we know this solution IS possible!

Page 13: Activity  2-16:  Triominoes

Proof 2: PolynomialsLet us put an algebraic termlogically into each square.

Page 14: Activity  2-16:  Triominoes

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

What happens if we put down a triomino horizontally?

We get (1+x+x2)f(x, y)

What happens if we put down a triomino vertically?

We get (1+y+y2)g(x, y)

Page 16: Activity  2-16:  Triominoes

So adding all the triominosplus 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

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

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

So possible values for a + b are

4, 7, 10, 13.

Page 20: Activity  2-16:  Triominoes

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

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

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

With thanks to:

Bernard Murphy, and Nick MacKinnon.

Carom is written by Jonny Griffiths, [email protected]