# No three in a line again

The Ministry is migrating nzmaths content to Tāhurangi.
Relevant and up-to-date teaching resources are being moved to Tāhūrangi (tahurangi.education.govt.nz).
When all identified resources have been successfully moved, this website will close. We expect this to be in June 2024.
e-ako maths, e-ako Pāngarau, and e-ako PLD 360 will continue to be available.

For more information visit https://tahurangi.education.govt.nz/updates-to-nzmaths

Purpose

This problem solving activity has a logic and reasoning focus.

Achievement Objectives
GM5-9: Define and use transformations and describe the invariant properties of figures and objects under these transformations.
Student Activity

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 that Miriama can put into the window, so that no three of them are in a line, and there is no way to put in another red pane without there three being in a line?

Specific Learning Outcomes
• Use symmetry and rotation to create geometric shapes that satisfy the no three in a line condition.
Description of Mathematics

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

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

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 Logic and Reasoning problems: Strawberry MilkStrawberry 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.

Required Resource Materials
Activity

### The Problem

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

What is the smallest number of red panes that Miriama can put into the window, so that no three of them are in a line, and there is no way to put in another red pane without there being three in a line?

### Teaching Sequence

1. Pose Miriama’s problem to the students and check it is understood. There are three parts to understanding the problem:
• There are nine square panes in total, some of which are red and some of which are white?
• There can not be more than two red panes in a line - either horizontally, vertically, or diagonally.
• The problem asks students to find the smallest number of red panes that could be placed, so that you can not swap any of the white panes for a red pane without there being three red panes in a row.
2. After some discussion, have the students work on the problem in their groups. Encourage them to be systematic in their approach.
3. 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.
4. Ask
Are all of these arrangements different?
Do you see any symmetry in the arrangement? Does this matter?
How might we think of some of them as being the same?
5. Pose the extension problem as appropriate.
6. Have groups report back to the class. Choose groups that have used different approaches to the problem.
7. Students could construct their own windows using transparent coloured paper.

#### Extension

Miriama has a square window made up of nine 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). Students may think they have found a solution with only three red panes, but there will always be a place that another red pane can be added.

Work systematically through all of the cases. First note that with four red panes, there has to be a row with two red panes. Consider two cases. The two 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.

The images below show all the possible ways to arrange four red panes (labelled r) without three of them being in a line. There is an x in each place that adding a red pane would make three in a row. These places would need to have a white pane. Places that are left blank could be either red or white, so examples with blank spaces are not a solution to the problem using only four red panes.

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.

#### Solution to the 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 four, five and six red panes here. The four case is covered by the first part of this problem. The six case is covered by No Three-In-A-Line and the five case comes from the Extension of No Three-In-A-Line.

So we have the following different answers:

Attachments
Add to plan

Log in or register to create plans from your planning space that include this resource.

Level Five