This activity has a logic and reasoning focus.
This problem requires students to use a systematic approach in order to be able to justify that all possibilities have been considered. 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.
Miriama is making a square window using 16 smaller red or white square panes.
What is the biggest number of red panes that Miriama can put in the window so that no three of them are in a line?
How many ways can this be done?
- 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.
- 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.
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.
Miriama is now working on a square window that has 25 smaller red and white square panes. What is the biggest number of red panes that she can put in the window so that no three of them are in a line? In how many ways can this be done?
The method of solution here is the same as for No Three-In-A-Line, Level 3
The first thing to note is that we cannot have more than two red panes in each row of the window. This means that we can’t fit in more than 8 red panes without forcing three to be in a line. So we systematically try to insert two red panes in each row to see if that is the largest number of red panes that Miriama can use. Here is the full set of answers.
Note that 1 and 2 are the right- and left-hand forms of the same situation. We can flip 1 over about an axis of symmetry in the plane of the window, so that it becomes 2. Hence for windows these two arrangements are the same. So there are 4 different solutions to Miriama’s problem: 1, 3, 4, and 5.
Extension: With 25 possible squares we can put at most two red panes in a row so that no three are in a line. So let’s try to put 10 red panes in Miriama’s window. This can be done in 5 different ways. Any other way that you can find can be rotated or reflected into one of these 5.