Question

Can we have rotational symmetry of order more than 1 whose angle of rotation is \[{{45}^{o}}\].

Answer 1

Hint: To check the conditions given in the question, we are going to assume a polygon which has an angle of rotational symmetry equal to \[{{45}^{o}}\]. After determining the polygon which satisfies this condition, we will be checking whether there exists an angle greater than \[{{45}^{o}}\]at which rotational symmetry exists. So, if an angle greater than \[{{45}^{o}}\] exists at which rotational symmetry exists then order of rotational symmetry would be greater than one. So, let us solve the question.

Complete step-by-step answer:
The polygon which exhibits rotational symmetry at an angle of \[{{45}^{o}}\] is octagon. Octagon is a type of polygon which has 8 sides. We are taking a regular octagon in this case. So, we can clearly see that the next angle of rotation at which it will exhibit rotational symmetry is \[{{90}^{o}}\]. The next angles where the same would happen are: \[{{135}^{o}},{{180}^{o}},{{225}^{o}},{{270}^{o}},{{315}^{o}}\] and \[{{360}^{o}}\]. These are the angles at which rotational symmetry would exist. Thus, we can conclude from here that the order of the octagon would be 8 (which is greater than 1). Thus we have proved that if a figure has an angle of rotation equal to \[{{45}^{o}}\], then the order of rotational symmetry can be greater than 1.

Note: We could have also proved it by taking a hexadecagon (a polygon with 16 sides) because the hexadecagon has one of the angle of rotational symmetry equal to \[{{45}^{o}}\]. We could have also taken a polygon which has sides equal to the integral multiple of 8. Note that these polygons should be regular. Otherwise the rotational symmetry would not be possible.