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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.
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}
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.
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.
Mr Bakker sent us a photo of the work some of his students did on this task:
Do continue sending us pictures of your rotation patterns along with any interesting mathematical discoveries you make.
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