THE core circle, x2 + y2 – x = 0, introduced in Norman Wildberger’s video MF143, has a [u : t] ‘projective’ parametrization of:
[u2/(u2 + t2), u*t/(u2 + t2)]
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 + t2), t/(1 + t2)]
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+ t2 then we find:
s(1 – s) = (1/(1 + t2))((1 + t2) – 1)/(1 + t2) = t2/(1 + t2)2
The y co-ordinate, y(t) given by this parametrization is t/(1 + t2) whose square is x(t)(1 – x(t)) and thus just the x values given by this paramterization on the core circle will be spread numbers. Recall from the Spread in a circle (part 1) discussion that the square of the height of this particular circle above or below the x axis was ‘x(1 – x)’. A secant line through the origin and meeting the circle at [x(t), y(t)] thus forms a spread:
s(x(t)) = (1 – x(t))
with the x-axis and has quadrance:
Q([0, 0], [x(t), y(t)]) = x(t).
Compared with spread ‘s‘, the (double-angle) formula – the second spread polynomial – gives
S2(s) = 4s(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(S2(s)) for example:
4 s(t) ( 1 – s(t))
= 4 (1/(1 + t2)) *( 1 – 1/(1 + t2))
= 4 ((1 + t2) – 1)/(1 + t2)2
x(S2(s)) = 4 t2/(1 + t2)2
and for y(S2(s)):
4 t2/(1 + t2)2 (1 – 4 t2/(1 + t2)2)
= 4 t2((1 + t2)2 – 4t2)/(1 + t2)4
= 4 t2((1 + 2t2 + t2 – 4t2)/(1 + t2)4
= 4 t2((1 – t2)2/(1 + t2)4
=> y(S2(s)) = 2 t(1 – t2)/(1 + t2)2
A line through points P1 = [0, 0] and P2 = [x(S2(s)), y(S2(s))] thus makes a spread of 1 – S2(s) to the x-axis (a spread of S2(s) to the y-axis) and has quadrance S2(s). Furthermore Q(S(s)) ≡ x(S(s)) so is always measured by the x values of points on the circle.
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 P1, and P2 thus forms the third side in an isosceles triangle [P0, P1, P2] 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[P4, P0] equalling the others depends on S4(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.