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 = \frac{\sqrt{5}}{2}

Equation of

C_{1}

x^{2} + y^{2} = 1

Equation of

C_{4}

x^{2} + (y + 1)^{2} = \frac{(\sqrt{5} - 1)^{2}}{4}

Equation of

C_{5}

x^{2} + (y + 1)^{2} = \frac{\sqrt{5} + 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

C_{1}

and

C_{4}

(so for the points C and D),

y = \frac{- (\sqrt{5} + 1)}{4}

and at the intersection of

C_{1}

and

C_{5}

y = \frac{\sqrt{5} - 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.

Construct the regular pentagon using the prescribed technique.
Bisect the angle $∠ A C E$ 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.)
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, $∠ A C P = ∠ P C E$ holds because $P C$ is an angle bisector, $C P$ is common).
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 $D$ .
The 10-sided shape we now have inscribed in the circle is a regular decagon.