THE core circle, *x*^{2} + *y*^{2} – ** x** = 0, introduced in Norman Wildberger’s video MF143, has a [u : t] ‘projective’ parametrization of:

[u^{2}/(u^{2} + t^{2}), u*t/(u^{2} + t^{2})]

This is a little general for our purposes, which is to show that (on the core circle at least) spread polynomials of succeeding degree meet the circle at *equi-quadrant* points. Instead of general proportion [u : t], therefore, we specialise on proportion [1 : t] and the parametrization becomes:

[1/(1 + t^{2}), t/(1 + t^{2})]

As previously noted in the post *Irrational Triangles?* we can encounter a situation when a geometric spread like ‘2/7’ will not meet the circle in a *pair* of rational values (a *rational point* in other words). The source of the difficulty was the choice of spread value, which needs to be a *spread number*. Spread numbers are values, **s** such that **s**(1 – **s**) is always square. For example ‘1/2’ is a spread number since

1/2( 1 – 1/2) = 1/4 = (1/2)^{2}

A value like 2/7 however is *not* a spread number since

2/7 (1 – 2/7) = 10/49

and ’10’ is not a square.

Here the parametrization of the core circle provides an *ansatz* (an educated guess) to the problem of finding spread numbers. If **s** = 1/1+ t^{2} then we find:

**s**(1 – **s**) = (1/(1 + t^{2}))((1 + t^{2}) – 1)/(1 + t^{2}) = t^{2}/(1 + t^{2})^{2}

The ** y** co-ordinate,

**(t) given by this parametrization is t/(1 + t**

*y*^{2}) whose square is

**(t)(1 –**

*x***(t)) and thus**

*x**just*the

**values given by this paramterization on the core circle will be spread numbers. Recall from the**

*x**Spread in a circle*(part 1) discussion that the square of the height of this particular circle above or below the

**axis was ‘**

*x***(1 –**

*x***)’. A**

*x**secant line*through the origin and meeting the circle at [

**(t),**

*x***(t)] thus forms a spread:**

*y***s**(*x*(t)) = (1 – *x*(t))

*x*

*x*

with the ** x**-axis and has quadrance:

**Q**([0, 0], [*x*(t), *y*(t)]) = x(t).

*x*

*y*

Compared with spread ‘**s**‘, the (double-angle) formula – the second spread polynomial – gives

*S*_{2}(**s**) = 4**s**(1 – **s**).

We note from this that a ‘valid’ spread always corresponds to the ‘double-angle’ of another spread. The converse however is not true: as we have seen that ‘2/7’ is not really a valid spread as no secant line through either end of the unit diameter can meet the core circle, as parametrized, at *x* = 2/7, and neither can it at *x* = 40/49. This implies that all valid spreads correspond to bisect-able angles and this, in turn, can be understood from the need for any *ray* through the centre of the circle to be the perpendicular bisector to some secant line (one meeting the circle in two rational points, and necessarily possessing *rational* inverse slope and midpoint.)

We can use this paramterization **s**(t), however, to parametrize each spread polynomial to which it gives rise. For ** x**(

*S*

_{2}(

**s**)) for example:

4 **s**(t) ( 1 – **s**(t))

= 4 (1/(1 + t^{2})) *( 1 – 1/(1 + t^{2}))

= 4 ((1 + t^{2}) – 1)/(1 + t^{2})^{2}

*x*(*S*_{2}(**s**)) = 4 t^{2}/(1 + t^{2})^{2}

*x*

and for ** y**(

*S*

_{2}(

**s**)):

4 t^{2}/(1 + t^{2})^{2} (1 – 4 t^{2}/(1 + t^{2})^{2})

= 4 t^{2}((1 + t^{2})^{2} – 4t^{2})/(1 + t^{2})^{4}

= 4 t^{2}((1 + 2t^{2} + t^{2} – 4t^{2})/(1 + t^{2})^{4}

= 4 t^{2}((1 – t^{2})^{2}/(1 + t^{2})^{4}

##### => *y*(*S*_{2}(**s**)) = 2 t(1 – t^{2})/(1 + t^{2})^{2}

*y*

A line through points *P*_{1} = [0, 0] and *P*_{2 }= [** x**(

*S*

_{2}(

**s**)),

**(**

*y**S*

_{2}(

**s**))] thus makes a spread of 1 –

*S*

_{2}(

**s**) to the

**-axis (a spread of**

*x**S*

_{2}(

**s**) to the

**-axis) and has quadrance**

*y**S*

_{2}(

**s**). Furthermore

**Q**(

*S*(

**s**)) ≡

**(**

*x**S*(

**s**)) so is always measured by the

**values of points on the circle.**

*x*As we have seen, quadrance and spreads are equivalent on the core circle, so the triple spread formula is satisfied for inscribed triangles using quadrances as spread. The secant line through points *P*_{1}, and *P*_{2 }thus forms the third side in an isosceles triangle [*P*_{0}, *P*_{1}, *P*_{2}] with equal quadrance of **s**.

Using parametrization from ‘t’ to **s **in this way we can continue to add further spread polynomials, each one degree higher, forming a ‘fan’ of secant lines through the origin, that meet the circle at equi-quadrant intervals forming part of a larger, inscribed polygon where all but the final quadrances equal **s**.

The first four spread polynomials suitably expressed in ‘t’ give five points by including the origin. The fifth quadrance **Q**[*P*_{4}, *P*_{0}] equalling the others depends on *S*_{4}(**s**) being symmetric (and equal) with **s** which would imply a construction of a regular pentagon.

Because the regular pentagon is famously based on the golden ratio **Φ** however it *cannot* be inscribed on a circle in rational points. The best we can do with rational numbers is create a semi-regular polygon of the four equal quadrances and a fifth side of slightly greater (or lesser) quadrance and spread.