No Three in a Line Again
AO elaboration and other teaching resources
This activity has a logic and reasoning focus.
use symmetry and rotation to create geometric shapes that satisfy the no thee in aline condition
This problem is one of a series of 8 that builds up to some quite complicated maths based around the theme of no-three-in-a-line. These problems are Strawberry Milk, Strawberry and Chocolate Milk, Level 1; Three-In-A-Line, Level 2; No Three-In-A-Line, Level 3; More No-Three-In-A-Line, Level 4; No-More-In-A-Line Level 6; and No-Three-In-A-Line Game, Level 6
The no-three-in-a-line theme was not obvious in the two Level 1 problems, Strawberry Milk and Strawberry and Chocolate Milk, but it has started to come to the fore now. It might help if you have done the Level 3 and 4 problems with your class before you start on this problem. The theme continues in later lessons at Level 6.
In this problem we revisit the three steps that were introduced in Strawberry Milk and Strawberry and Chocolate Milk. You may recall that these are
- find some answers to a problem;
- think about whether there are any more answers or not;
- try to explain why there are no more answers.
The complete answer to this question is best done by making a systematic check of all possibilities. This approach has been used in No Three-In-A-Line, Level 3, More No-Three-In-A-Line, Level 4. In getting all the possibilities, we hope that the students will realise that there is a lot of symmetry in Mary’s square window. So their answers should take all the symmetries of a square into account.
In the Extension to this problem your whole class together might be able to come up with all possible arrangements of the red panes that fit Mary’s two conditions. At the same time, they should try to provide some systematic reason for why there are no more answers.
On the Statistics side, we are trying to count all possibilities. This is a precursor to determining probabilities, which is an important part of Statistics. On the Geometry side, we are concerned with the symmetry of a square.
There is a web site on the no-three-in-line problem. Its url is www.uni-bielefeld.de/~achim/no3in/readme.html.. This is still an open problem in mathematics and has an interesting number of sub problems relating to symmetry.
Mary is again making a square window with smaller red and white square panes of glass. This time her window will have 9 panes altogether. What is the smallest number of red panes she 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?
- Talk about square windows and their symmetry. Talk about similar problems and how they were tackled.
- Tell the class Mary’s problem and discuss any difficult ideas.
- Let the class work on the problem in their groups.
- Help the students that need it.
- Call them all together from time to time to see to check on their progress..
- Try to get them to see how to use a systematic approach to get all possible answers. Let the more able students work on the Extension problem.
- Let a few groups report back to the whole class. Try to choose groups that have used different approaches to the problem.
- Discuss their conclusions.
- Perhaps the class might like to make coloured windows.
- Get the students to write up what they did in their notebooks.
Mary had 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?
First of all Mary has to have at least four red panes. This is because, using symmetry, if she had 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.)
We now work systematically through all of the cases. First we note that with 4 red panes, there has to be a row with two red panes. We consider two cases. The 2 red panes are in the first row and the two red panes are in the second row. We don’t have to worry about two red panes being in the third row as we could use a rotation to bring the third row up to the first row.
All other cases come from the above by symmetry.
So we have two solutions for Mary. 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.
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: