This problem solving 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 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?
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 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 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?
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?
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.
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:
Printed from https://meaningfulmaths.nt.edu.au/mmws/nz/resource/no-three-line-again at 8:59pm on the 26th February 2024