This problem solving activity has a logic and reasoning focus.
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 any row or column?
How many ways can this be done?
This problem requires students to use a systematic approach to justify that they have considered all possibilities. 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 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.
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 any row or column?
How many ways can this be done?
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 any row or column? 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 row or column. So we can 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.
With 25 possible squares we can put at most two red panes in a row so that no three are in a row or column. 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.
Printed from https://nzmaths.co.nz/resource/more-no-three-line at 7:05am on the 26th April 2024