Attractive Tablecloths
Problem
Attractive Tablecloths printable worksheet  tablecloths
Attractive Tablecloths printable worksheet  templates
Charlie has been designing square tablecloths for each weekday. He likes to use as many colours as he possibly can but insists that his tablecloths have some symmetry.
The $5$ by $5$ tablecloths below each satisfy a different symmetry rule.
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Monday's $5$ by $5$ tablecloth has just $1$ line of symmetry. Design some square tablecloths of other odd by odd sizes with just $1$ line of symmetry.
Check you agree that a $7$ by $7$ tablecloth can have at most 28 colours. Can you find a way of working out the maximum number of different colours that can be used on an n by n tablecloth (where n is odd), following Monday's rule?

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Tuesday's $5$ by $5$ tablecloth has rotational symmetry of order $4$, and no lines of symmetry. Design some square tablecloths of other odd by odd sizes with rotational symmetry of order $4$, and no lines of symmetry.
Check you agree that a $7$ by $7$ tablecloth can have at most 13 colours.
Can you find a way of working out the maximum number of different colours that can be used on an n by n tablecloth (where n is odd), following Tuesday's rule?

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Wednesday's $5$ by $5$ tablecloth has $2$ lines of symmetry (horizontal and vertical), and rotational symmetry of order $2$. Design some square tablecloths of other odd by odd sizes with $2$ lines of symmetry, and rotational symmetry of order $2$.
Check you agree that a $7$ by $7$ tablecloth can have at most 16 colours.
Can you find a way of working out the maximum number of different colours that can be used on an n by n tablecloth (where n is odd), following Wednesday's rule?

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Thursday's $5$ by $5$ tablecloth has $2$ (diagonal) lines of symmetry and rotational symmetry of order $2$.
Design some square tablecloths of other odd by odd sizes with $2$ (diagonal) lines of symmetry and rotational symmetry of order $2$.
Check you agree that a $7$ by $7$ tablecloth can have at most 16 colours.
Can you find a way of working out the maximum number of different colours that can be used on an n by n tablecloth (where n is odd), following Thursday's rule?

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Friday's $5$ by $5$ tablecloth has $4$ lines of symmetry and rotational symmetry of order $4$.
Design some square tablecloths of other odd by odd sizes with $4$ lines of symmetry and rotational symmetry of order $4$.
Check you agree that a $7$ by $7$ tablecloth can have at most 10 colours.
Can you find a way of working out the maximum number of different colours that can be used on an n by n tablecloth (where n is odd), following Friday's rule?

EXTENSION
At weekends Charlie likes to use tablecloths with an even number of squares. Investigate the number of colours that are needed for different types of symmetric $n$ by $n$ tablecloths where $n$ is even.
Getting Started
How many extra colours do you need to add to the $7$ by $7$ pattern to make the $9$ by $9$ pattern for each type of symmetry?
Where do those extra colours go?
How many extra colours would you need for the $11$ by $11$ design?
Where do those extra colours go?
Student Solutions
Monday tablecloth
Well done to Joseph and Dylan from Wilson's Grammer School and Joe from Leventhorpe. Each found a pattern which they then investigated to find a general formula.
Start with the example below, where the solid grey line is the line of symmetry. The red square forces the square symmetrically opposite to also be red.
So the squares on one side of the line of symmetry determine the colours on the other side.
The squares with different colours to each other form a rectangle of size $n \times d$, where $n$ is the number of rows in the tablecloth and $d$ is the number of columns with different colours (including the squares on the line of symmetry).
Komal from India illustrated this pattern by writing each colour as a different pronumeral:
Observe that the orange section (squares with unique colours) has area $15=5\times 3 =5\times \frac{5+1}{2}$
So $d=\frac{n+1}{2}$, and the maximum number of colours for 1 line of symmetry is $n \times \frac{n+1}{2}$
Tuesday tablecloth
The squares with unique colours form a rectangle of area $x\times y$ in one quadrant of the tablecloth, plus the colour in the central square (shaded in orange).
In the $5\times5$ case, the area of this rectangle is $6=3\times 2 = \frac{5+1}{2} \times (\frac {5+1}{2} 1) $.
So the maximum number of colours in this rectangle is $\frac{n+1}{2} \times (\frac{n+1}{2}1)=\frac{n+1}{2} \times \frac{n1}{2}$.
Adding in the central square, the maximum number of colours is $\frac{n+1}{2} \times \frac{n1}{2} + 1$
Another explanation, using the rotational symmetry of order $4$, was given by Nathan, Jack, Joseph, Harry and Daren from Wilson's Grammar School.
Except for the central square, each colour appears $4$ times in the tablecloth due to the rotational symmetry of order $4$. So excluding the central square, the number of colours is $\frac{n^21}{4}$
Adding in the colour C7, the maximum number of colours is thus $\frac{n^21}{4}+1$
Wednesday tablecloth
The number of different colours in each tablecloth is a square number, say $d^2$. Since $d$ is the length of a quadrant (including the squares on the line of symmetry) we have $d=\frac{n+1}{2}$.
So the maximum number of colours is $(\frac{n+1}{2})^2$
Thursday tablecloth
In the $5\times 5$ case the maximum number of different colours is $9$.
From the diagram above, the orange section (squares with unique colours) has area $9=1+3+5$.
Recall that the sum of consecutive odd numbers is $1+3+5+...+(2k1)=k^2$. Noting that $5=2\times31$, we then get $9=3\times3=(\frac{5+1}{2})^2$
So the maximum number of colours is $(\frac{n+1}{2})^2$
Are you able to predict this result? Compare the types of symmetries of this tablecloth with the Wednesday tablecloth; why is the formula for the number of colours the same?
Friday tablecloth
In the $5\times 5$ case the maximum number of different colours is $6$.
From the diagram above, the orange section (squares with unique colours) has area $6=1+2+3$.
Recall that the sum of consecutive numbers is $1+2+3+...+k=\frac{k(k+1)}{2}$, and by writing the area in terms of $n=5$, we get $6=3\times \frac{3+1}{2}=\frac{5+1}{2} \times \frac{\frac{5+1}{2}+1}{2}$
So the maximum number of colours is $(\frac{n+1}{2})^2\times \frac{\frac{n+1}{2}+1}{2}=\frac{(n+1)(n+3)}{8}$
Well done to Joe and Komal who also thought about what would happen if $n$ were even, and to Karrar from Michaela Community School who considered how to write an nth term rule which covered both the odd and even cases in one formula.
Teachers' Resources
Why do this problem?
Rather than deducing an $n^{th}$ term rule by patternspotting, this problem encourages students to look at the structure of several symmetric patterns and to explain how to generate rules for finding the number of colours required.
Possible approach
These printable resources may be useful: Attractive Tablecloths,
Attractive Tablecloths,
Templates.
Display these $5$ by $5$ tablecloth designs showing each of the five symmetry rules:
Monday's rule
Tuesday's rule
Wednesday's rule
Thursday's rule
Friday's rule
"This tablecloth is symmetric. Can you describe the types of symmetry it has?"
Once students have established what the five symmetry rules are:
"For each symmetry rule, I want you to find out the maximum number of colours you can use to colour in a $5$ by $5$, a $7$ by $7$ and a $9$ by $9$ tablecloth."
"In a while, I'll be asking you to work out the number of colours needed for a much larger tablecloth, one that's too big to draw. So you will need to record your results and think about how your work on these smaller cases could be generalised."
Hand out these templates and invite students to work in pairs using colours or numbering to create each design. As they finish each one, the designs could be pinned up together according to each symmetry rule, and students could look at each other's work to check there is agreement on the maximum number of colours needed.
"Imagine I wanted to create a $99$ by $99$ tablecloth. How many colours would I need for each type of symmetry? You need to develop a convincing argument to explain how to work out the number of colours needed. You might also like to consider how many colours would be needed for an $n$ by $n$ tablecloth."
Allow students time to work in pairs on this challenge. Then combine pairs and ask each group of four to refine their convincing argument for one or two of the designs. Finally, invite groups of four to present their convincing argument to the rest of the class, who could act as critical friends and suggest improvements to their explanations.
Possible support
Seven Squares and Steel Cables both offer students the chance to deduce algebraic rules by looking at the structure of different patterns.
Possible extension
Invite students to consider $n$ by $n$ tablecloths where $n$ is even.
A very challenging extension is to come up with a combined rule that works for both odd and even sized tablecloths, for each type of symmetry.