This activity has a logic and reasoning focus.

Miriama is making a square window using smaller red or white square panes of glass.

Her window has 9 panes altogether.

What is the smallest number of red panes Miriama can put into the window so that no three of them are in a line and so that she can’t put in another red pane without there three being in a line?

use symmetry and rotation to create geometric shapes that satisfy the no three in a line condition

This problem requires students to use a systematic approach in order to be able to justify that all possibilities have been considered.

- find
answers to a problem;*some* - think about whether there are any
answers or not;*more* - try to explain why there are
answers.*no more*

The problem also challenges students to recognise the symmetry in a figure, and to see that by rotating a figure through a quarter turn either clockwise or anticlockwise, two 'answers' are essentially the same. Symmetry through a line in the plane of the square is therefore important.

See also these problems. Strawberry Milk, Strawberry and Chocolate Milk, Level 1; Three-In-A-Line, Level 2; No Three-In-A-Line, Level 3; No-Three-In-A-Line Again, Level 5; No-More-In-A-Line Level 6; and No-Three-In-A-Line Game, Level 6.

This no-three-in-line problem is still an open problem in mathematics and has an interesting number of sub-problems relating to symmetry. See www.uni-bielefeld.de/~achim/no3in/readme.html.

Copymaster of the problem (English)

Coloured pens and paper.

Bottles tops.

Copymaster of 3 by 3 windows.

## Problem

Miriama is making a square window using smaller red or white square panes of glass. Her window has 9 panes altogether.

What is the smallest number of red panes Miriama can put into the window so that no three of them are in a line and so that she can’t put in another red pane without there three being in a line?

### Teaching sequence

- Pose Miriama’s problem to the students and check it is understood.
- After some discussion, have the students work on the problem in their groups. Encourage them to be systematic in their approach.
- As solutions emerge, have students share ideas, showing and explaining their arrangements. Give a student's name to each arrangement for convenience in referring to it.
- Ask
*Are all of these arrangements different?**How might we think of some of them as being the same?* - Pose the Extension problem as and when appropriate.
- Have groups report back to the class. Choose groups that have used different approaches to the problem.
- Students could construct their own windows using transparent coloured paper.

#### Extension

Miriama has a square window made up of 9 red and white smaller square panes. How many ways can she put red panes in her window so that no three are in a line and if she puts in one more red pane in the place of a white one, she is forced to put three in a row?

#### Solution

First of all Miriama has to have at least four red panes. This is because, using symmetry, if she has only three red panes there would be one row that had no red pane at all. (It’s possible that there would be a column, rather than a row, which had no red pane. However, by using a rotation that column becomes a row.) It’s always possible to put a red pane in this row. (You need to check out the cases.)

Work systematically through all of the cases. First note that with 4 red panes, there has to be a row with two red panes. Consider two cases. The 2 red panes are in the first row and the two red panes are in the second row. Don’t worry about two red panes being in the third row as rotation can be used to bring the third row up to the first row.

All other cases come from the above by symmetry.

There are two solutions for Miriama. These are 1 and 16. It is clear that no symmetry of the square will take 1 into 16 and so these answers are really different.

#### Extension:

From work already done in the first part of this problem (see No Three-In-A-Line Level 3, More No-Three-In-A-Line, Level 4) we know that we can only use 4, 5 and 6 red panes here. The 4 case is covered by the first part of this problem. The 6 case is covered by No Three-In-A-Line. And the 5 case comes from the Extension of No Three-In-A-Line.

So we have the following different answers: