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Attractive Rotations

Stage: 3 Challenge Level: Challenge Level:1

Thank you to Sally and Emily from Parrs Wood for sending us pictures of their rotations. Here are their findings.

Sally noticed a pattern with the coordinates of our shape. 

coord
In the table below, the coordinates (x',y') describe where the points (x,y) end up after they have rotated by $90^\circ$ around (0,0):  


(x,y) (x',y')
(-4,-5) (-5,4)
(-6,2) (2,6)
(4,2) (2,-4)
(-5,4) (4,5)
(2,6) (6,-2)
(2,-4) (-4,-2)
(4,5) (5,-4)
(6,-2) (-2,6)
(-4,-2) (-2,4)
(5,-4) (-4,5)
(-2,6) (2,-6)
(-2,4) (4,2)
  

So the coordinates (x',y'), which are a rotation of (x,y) by $90^{\circ}$ around (0,0), are (y, -x).

$180^{\circ}$ rotation is just two $90^{\circ}$ rotations so

(x,y) $\longrightarrow$ (x',y') = (y,-x) $\longrightarrow$ (y',-x') = (-x, -y).
         $90^{\circ}$                            $90^{\circ}$


Here are Emily's rotated shapes and mathematical discoveries.
 
I first rotated shapes by $60^{\circ}$ and found the rotational symmentry was of order 6 because $360\div60 = 6$. But when I coloured them in sometimes the rotational symmentry was of order less than 6. I noticed they were factors of 6.
 
 smaller

I predicted rotations of $30^{\circ}$ would have rotational symmetry of order 12 and $72^{\circ}$ would have rotational symmetry of order 5 because $360\div30 = 12$ and $360\div72 = 5$. These were true for the pictures I drew. I also knew I couldn't colour in my $72^{\circ}$ rotation shape because 5 is prime so its only factors are 5 and 1.

 smallernow

$360\div80 = 4.5$ and I had to go round twice to complete my shape. This then gave me rotational symmetry of order 9 because $360\times2=720$ and $720\div80 = 9$.

As $360\div135 = 2{2\over3}$, I worked out $360\times3=1080$ and $1080\div135 = 8$ so I knew I had to go round 3 times to get a shape with rotational symmetry.

Do continue sending us pictures of your rotation patterns along with any interesting mathematical discoveries you make.