Elizabeth factorised the expression as follows:

r2 - r + 1/4 sin2 4q = 0 r2 - r + 1/4(2(sin2qcos2q))2 = 0 r2 - r + sin2 2qcos2 2q = 0 (r - sin2 2q)(r- cos2 2q) =0.

and plotted the graphs of r = sin² 2q and r=cos² 2q.

Andrei factorised the expression differently

(2r+cos4q -1)(2r cos4q - 1)

and plotted the graphs of r = ¹/₂(1 + cos 4q) and r = ¹/₂(1 - cos4q) but you should be able to see that the results are equivalent.

Our friend from Winchester College used the difference of two squares to arrive at the same form of this result as Andrei and another method would be substitution in the formula for the solution of a quadratic equation.

Consider first the graph of r- sin² 2q = 0 in polar coordinates where r is the length of the line segment from the point from the origin and q is the angle measured counter clockwise between this line segment and the positive x axis. For points on this graph, as q increases from 0 to

p 4

we have sin² 2q increases from 0 to 1. Between

q =

p 4

and

p 2

the value of r decreases from 1 to 0 so that the graph in the first quadrant is a 'petal' symmetrical about the line

q =

p 4

.

Similarly the graph in the second quadrant is a 'petal' for q between

p 2

and p where r takes positive values corresponding to sin² q. The graph in the third quadrant is a 'petal' for q between p and

3p 2

and the graph in the fourth quadrant is a 'petal' for q between

3p 2

and 2p.

Next consider the graph of r- cos² 2q = 0. This will be of the same form but rotated by

p 4

corresponding to the phase shift between the graphs of sin2q and cos2q.

For points on this graph, as q increases from 0 to

p 4

we have cos² 2q decreases from 1 to 0. Between

q =

p 4

and

p 2

the value of r increases from 0 to 1 and between

p 2

and

3p 4

, as r decreases from 1 to 0, the 'petal' symmetrical about the y-axis is completed.

The next petal, symmetrical about the negative x-axis, is drawn for q between

3p 4

and

5p 4

. The next petal, symmetrical about the negative y-axis, is drawn for q between

5p 4

and

7p 4

and the remaining petal is completed for q between

7p 4

and 2p.