Attractive tablecloths

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. 

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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: 

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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).

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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 $\frac{n+1}{2} \times (\frac{n+1}{2}-1)=\frac{n+1}{2} \times \frac{n-1}{2}$. 

Adding in the central square, the maximum number of colours is  $\frac{n+1}{2} \times \frac{n-1}{2} + 1$

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 $\frac{n^2-1}{4}$

Adding in the colour C7, the maximum number of colours is thus $\frac{n^2-1}{4}+1$

Wednesday tablecloth

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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 $9$. 

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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 $1+3+5+...+(2k-1)=k^2$. Noting that $5=2\times3-1$, 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 $6$.

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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.