Harry explained why we get a Pentagon:

I firstly calculated the length of YS and the equations of the circle using Pythagoras' theorem on the triangle YOS (and by symmetry YOR) to get:

YS= Ö5
2
Equation of C1:
x2+y2=1
Equation ofC4:
x2+(y+1)2= (Ö5-1)2
4
Equation of C5:
x2+(y+1)2= Ö5+1)2
4

At the points of intersection of two circles, the points satisfy the equations both both circles and so we have a set of two simultaneous equations to solve. Solving these gives

at the intersection of C1 and C4 (so for the points C and D),
y= -(Ö5+1)
4

and at the intersection of C1and C5,
y= Ö5-1
4

.

But we can now look at the hint, and find we have the same y coordinates for all our points as those of a regular hexagon. But since we are on the circle, we can work out the x coordinates from the y coordinates. So we have the same points as the regular pentagon in the notes section, and so this is a regular pentagon.

Tom from Bristol Grammar School then suggested a method to construct a regular decagon using the pentagon that we've already constructed.

  1. Construct the regular pentagon using the prescribed technique.
  2. Bisect the angle ÐA C E by drawing a circle centre A and a circle of the same radius (perhaps E C) centre E and drawing a straight line between one of the points at which the circles intersect and point C. (This works because A C=E C, as the pentagon is regular - it is a fact that is obvious and easily proven using SAS congruence, and therefore it is equivalent to the classroom-taught angle bisection technique.)
  3. Let us call the point (other than C) at which this line crosses the pentagon's circumcircle P. Join A to P, and join E to P. Essential to our method is that now A P=P E, which is clearly true by SAS congruence of the triangles A C P and E C P (A C=E C was used above, ÐA C P = ÐP C E holds because P C is an angle bisector, C P is common).
  4. Construct the circle centre A through P and label the point (other than P) at which it crosses the circumcircle of the pentagon Q. Draw in the line segments A Q and Q B. Similarly construct the circle centre B through Q, join up the line segments, and repeat this process for C and D.
  5. The 10-sided shape we now have inscribed in the circle is a regular decagon.