(See also considerations‘, a follow-up blog from December 2019)

Introduction

Description:

AN ortholateral is a(n orthogonal) quadrilateral formed of two pairs of perpendicular lines (crosses) in either the ‘blue’, ‘red’ or ‘green’ sense of chromogeometry(video clip). If two such ‘blue’ crosses are thus inclined in either the ‘green’ or the ‘red’ sense their lines commute into two distinct ‘green’ (or ‘red’) crosses that meet in a ‘blue’ sense.

The Ortholateral

An ‘orthogonal-quadrilateral’ (green lines) together with its triangle of diagonals (shaded) plus medial line forming the related ortholateral

AN equivalent description that produces this construction is to take an initial line, reflect it in a red (or green) ‘null-line’, reflect this combination in a green (or red) alternate null line and translate at least one line parallel to itself by an arbitrary amount, which could include no translation. (The apparent dependence of the ortholateral on Euclidean (blue) geometry is somewhat an illusion therefore.)

[ANOTHER equivalent description comes from looking at a suitably oriented triangle and its altitudes (six lines, perpendicular in pairs.) The ortholateral is then one altitude plus the associated side from the triangle plus the opposite side of its orthic triangle and the perpendicular bisector of that side:

side, altitude, [orthic] side, perpendicular bisector]

Complete Quadrilateral, Radical axis, mutual perpendicularity

Collinear orthocentres:

BY the Gauss-Bodenmiller Theorem[i], the midpoints of the three ‘diagonal’ segments completing a general quadrilateral are collinear, and (blue) circles whose diameters lie on these segments meet in precisely two points forming a radical-axis (blue) perpendicular to the medial line joining their centres. For an ortholateral (which is not the case with a general quadrilateral) however, the radical-axis aligns with the segment itself which joins the blue perpendiculars (‘crosses’). And in an ortholateral, with a triple form of perpendicularity present, this implies the remaining joins of crosses are respectively red and green perpendicular to the medial line and, transitively, that opposite vertices in the ortholateral form a mutually orthogonal set (a triply-right triangle in the language of chromogeometry.) Together with the medial line, therefore, all three joins of vertices forms a NEW ortholateral constructed on an existing one.

HOWEVER, also by the Gauss-Bodenmiller[ii] Theorem also, the four (blue) orthocentres of the four triangles associated with a general quadrilateral  are collinear and lie on the same (blue) radical-axis. In chromogeometry, with its three distinct perpendiculars, there will be three sets of collinear orthocentres forming three radical-axes mutally ‘perpendicular’ to the medial line. The Theorem thus establishes that every quadrilateral has an associated ortholateral and, since every ortholateral forms a new ortholateral, we naturally consider the ‘anti-ortholateral’ relation (c/f  the anti-orthic triangle) to a given ortholateral.

In search of an ‘anti-ortholateral’:

Projective geometry, Harmonic ratio, Dual constructions:

IN obtaining the ortholateral-of-ortholateral we drew lines completing the quadrilateral and formed a triply right triangle of diagonals*, plus the medial line. That being a general method in projective geometry, the vertices of an ortholateral must be in an harmonic ratio with the ‘diameters’ of its associated anti-ortholateral (which indicates a method for obtaining the latter’s construction).

TWO of the three sides of a triangle of diagonals in a quadrilateral cut off an interval along the third side which, for an ortholateral due to the presence of perpenducality of sides, can be divided harmonically (per the construction for the circles of Apollonius) by a blue circle centred at the midpoint of a ‘red’ vertex and a ‘green’ vertex. (In other words, the interval to be divided will be the side of the triangle of diagonals, with the associated circle diameter being the side of the ortholateral on which it lies.)

Dual anti-ortholaterals

From the same ortholateral as before (green lines) the circle-construction is harmonic to sides of two triangles of diagonal lines and gives an anti-ortholateral (broken red, blue lines) for each

THE circle along one side is met by the line blue-perpendicular to it in two symmetric points. In addition, a ‘dual’ interval forms where the same lines, with their perpendicularity swapped, cross and meet the alternate side of the ortholateral and a circle with diameter on this side meets the line perpendicular to it in two new points. A third circle centred on the meet of both sides – at the acute ‘corner’ of the ortholateral – will pass through all four points and will meet the lines forming the sides of the ortholateral  at another four points along its own diameters.

THE ortholateral has generated a total of four chords on this circle – two as diameters and two as perpendicular chords (which commute to oblique chords) giving not one but two cyclic quadrangles (four points in general position) which each allow for four separate lines, forming an anti-orthoilateral. We thus have a new result: every ortholateral is associated with a pair of ‘dual’ figures (c/f four anti-orthic triangles of  a triangle being considered as an orthocentric system ) and to each ortholateral there is a ‘dual’.

Discussion points:

BY analogy with triangle geometry, the ortholateral may be considered a system of four ‘ortho-lines’, each side respectively perpendicular to its complement of three lines (which form ‘trilaterals‘ rather than ‘triangles‘), the medial line playing the role of the orthocentre in an orthic triangle (i.e. not the orthocentre of the triangle) that completes the ‘ortholineal’ system.

THE perspectivity of a triangle with its orthic triangle now becomes dual in the sense of Desargue’s Theorem: any three lines of the ortholateral are side-perspective to the lines of the triangle of  diagonals in its fourth line (representing an orthocentre). The orthic-of-orthic perspector is represented differently, by a side-perspectivity of the lines of the ortholateral representing the triangle of  diagonals to the triangle of diagonals forming part of the ortholateral-of-ortholateral.

 

 

THIS brief description of a new geometrical construction opens up an exciting set of connections between classical (Euclidean) triangle geometry with projective geometry via the (new) development of quadratic-metrical geometry known as ‘chromogeometry’ 

*Note the ‘triangle of diagonals’ of a quadrilateral here corresponds to the ‘diagonal triangle’ of the dual quadrangle in projective geometry

Further note There is a projective dual quadrangle which can be constructed from an orthlolateral (per quadrilateral). Although the figure (when constructed) is not in general orthogonal, we speculate that it could be rectifiable by a suitable change of ‘viewing angle’ (amounting to a linear transformation) so as to become an orthogonal system in four points in either the blue red or green sense, depending on whether the set is ‘covex’ (red or green) or ‘concave (blue).  (More to follow)

(See post ‘Blue, Red and Green quadrances on a circle‘)

IN Rational Trigonometry, the Triple Spread Formula amounts to a ‘function’ which plots all three spreads of a triangle as a co-ordinate:

'(s_1,s_2,s_3)'

IN Chromogeometry meanwhile, a triangle possesses three kinds of quadrance – associated to the three bilinear forms:

q_b = x^2 + y^2,

q_r =x^2-y^2,

q_g=2xy

This means we can represent all three bilinear forms of a particular triangle as the ‘vectors’:

(s_1,s_2,s_3)_b

(s_1,s_2,s_3)_r,

(s_1,s_2,s_3)_g

And in turn, as a consequence of the Spread Law we may use quadrances instead to obtain the ratio of two spreads and, by dividing the first two quadrances of a triangle by the third, a pair of such ‘bi-spreads’ yields a point (effectively projecting the Triple Spread function through the origin onto the ‘z = 1’ standard plane):

For example:

[(\frac{s_2}{s_1})_b,\,(\frac{s_3}{s_1})_b] = [(\frac{q_2}{q_1})_b,(\frac{q_3}{q_1})_b]= [\bar s_2,\bar s_3]_b,

similarly,

[\bar s_2,\bar s_3]_r = [(\frac{q_2}{q_1})_r,(\frac{q_3}{q_1})_r],

[\bar s_2,\bar s_3]_g = [(\frac{q_2}{q_1})_g,(\frac{q_3}{q_1})_g]

WITH these bispread values, we would have two ‘end points’ which will allow us to calculate the equation of the line – the polar line of the blue bispread point. Since the equation of a line is:

< y_1 - y_2 : x_2 - x_1 : x_1 y_2 - x_2 y_1>

we would have:

< (q_3/q_1)_r - (q_3/q_1)_g : (q_2/q_1)_g - (q_2/q_1)_r : (q_2/q_1)_r (q_3/q_1)_g - (q_2/q_1)_g (q_3/q_1)_r>

and our desired identity becomes:

((\frac{q_3}{q_1})_r - (\frac{q_3}{q_1})_g) x +((\frac{q_2}{q_1})_g - (\frac{q_2}{q_1})_r) y + (\frac{q_2}{q_1})_r (\frac{q_3}{q_1})_g - (\frac{q_2}{q_1})_g (\frac{q_3}{q_1})_r=0 

where

\frac{q_2}{q_1}|^d_r = x, \frac{q_3}{q_1}|^d_r = y

Clearing denominators, then gives:

((q_1)_g(q_3)_r - (q_1)_r(q_3)_g) (q_2)|^d_r +

((q_1)_r(q_2)_g - (q_1)_g(q_2)_r)(q_3)|^d_r +

((q_2)_r (q_3)_g - (q_2)_g (q_3)_r) (q_1)|^d_r =0

Using the “barred” form allows in-common blue quadrances to be factored also:****

\implies((\bar q_1)_g(\bar q_3)_r - (\bar q_1)_r(\bar q_3)_g) (\bar q_2)|^d_r +

((\bar q_1)_r(\bar q_2)_g - (\bar q_1)_g(\bar q_2)_r) (\bar q_3)|^d_r +

((\bar q_2)_r (\bar q_3)_g - (\bar q_2)_g (\bar q_3)_r) (\bar q_1)|^d_r =0

Now:

(\bar q_1)_g =\frac{2(t_2+ t_3)(t_2 t_3 - 1)}{(t_2^2 + 1 ) (t_3^2 + 1 )}, (\bar q_1)_r =\frac{(t_2+ t_3)^2-(t_2 t_3 - 1)^2}{(t_2^2 + 1 ) (t_3^2 + 1 )}

(etc) So, letting:

(\bar q_1)'_g =2(t_2+ t_3)(t_2 t_3 - 1), (\bar q_1)'_r =(t_2+ t_3)^2-(t_2 t_3 - 1)^2

\implies ((\bar q_1)'_g(\bar q_3)'_r - (\bar q_1)'_r(\bar q_3)'_g) =

2(t_2+ t_3)(t_2 t_3 - 1)((t_1+ t_2)^2-(t_1 t_2 - 1)^2)

-2(t_1+ t_2)(t_1 t_2 - 1)((t_2+ t_3)^2-(t_2 t_3 - 1)^2)

= 2(t_2^2 + 1)^2 (t_3 - t_1) (t_1 t_3 + 1)

and thus:

((\bar q_1)_g(\bar q_3)_r - (\bar q_1)_r(\bar q_3)_g) = \frac{2(t_3 - t_1) (t_1 t_3 + 1)}{(t_1^2+1)(t_3^2+1)}

Now:

\bar q_2|^d_r=\frac{((d^2-1) (t_1 + t_3) + 2 d ( t_1 t_3 - 1 ))^2-((d^2-1) (t_1 t_3 - 1) - 2 d ( t_1 + t_3 ))^2}{ (d^2 + 1 )^2(t_1^2 + 1 ) (t_3^2 + 1 )}

and using the same ‘prime’ notation as before, we define the numerator equivalent:

\bar q_2|'^d_r=((d^2-1) (t_1 + t_3) + 2 d ( t_1 t_3 - 1 ))^2-((d^2-1) (t_1 t_3 - 1) - 2 d ( t_1 + t_3 ))^2

and common denominator:

(\frac{1}{(d^2+1)(t_1^2+1)(t_3^2+1)})^2

Then, multiplying through by a factor of:

\frac{(d^2+1)(t_1^2+1)(t_2^2+1)(t_3^2+1))^2}{2}

leaves the denominator-free form of our identity:

(t_2^2 + 1)^2 (t_3 - t_1) (t_1 t_3 + 1)\cdot \bar q_2|'^d_r+

(t_3^2 + 1)^2 (t_1 - t_2)*(t_1 t_2 + 1)\cdot \bar q_3|'^d_r+

(t_1^2 + 1)^2 (t_2 - t_3) (t_2 t_3 + 1)\cdot \bar q_1|'^d_r=0

ALLOWING for sign change inserted at line ‘*’, this equation has been verfied in GGB CAS. The first plus the second equals the algebraic negation of the third line for both ‘red’ (difference of squares) and ‘green’ (product of terms) forms of the boosted bi-quadrance formulas (but only at ‘d’ =1 and ‘d’ = 0) This would be expected since these values correspond to an unboosted situation – or ‘degenerate’ case.

GGB boosts 3

OBSERVE the figure above.

It shows a fixed point ‘P‘ and points ‘L‘ and ‘M‘ moving along a fixed line. The triangle ‘HIJ‘ lies on the unit circle and is the result of applying a varying amount of turn or ‘boost’to each point of a circumscribed (hatched) triangle and rationally (re)parameterizing to produce a (‘blue’) rotation. Points ‘P‘, ‘L‘ and ‘M‘, meanwhile, are determined from the quadrances of ‘HIJ’ by dividing the quadrances of second (‘x‘) side and third side (‘y‘) respectively of ‘HIJ‘ by the quadrance of its first side – for each of the Blue (‘P‘), Red (‘L‘) and Green (‘M‘) bilinear forms associated to the triangle in each such position.

The seeming fact that the Red (‘L’) and Green (‘M‘) ‘poles’ observed remain on a fixed line under a rotation that fixes the Blue (‘P‘) pole results from the equivalence of ratios of spreads to ratios of quadrances, since:

\frac{s_2}{s_1} = \frac{q_2}{q_1}

Thus ratios of Red and Green spreads (qua their associated quadrances) from a triangle on a unit circle when projected through the origin will lie on a fixed polar line with respect to the corresponding ratio of Blue spreads when subject to Blue rotation (and similarly for Red and Blue ‘bi-spreads’ under Green rotation or Green and Blue ‘bi-spreads’ under Red rotation).

WE NOW seek to establish this important result algebraicially and begin by obtaining expressions for Blue, Red and Green quadrances on a unit circle

Blue quadrance (on circle +   =  1):

(q_3)_b =\frac{4(t_1 - t_2)^2}{(t_2^2 + 1)(t_1^2 + 1)}

Red:

(q_3)_r =\frac{4(t_1 - t_2)^2}{(t_2^2 + 1 ) (t_1^2 + 1 )} \cdot \frac{(t_1 + t_2)^2 -(t_1t_2 -1)^2}{(t_2^2 + 1 ) (t_1^2 + 1 )}

=(q_3)_b\cdot \frac{(t_1 + t_2)^2 -(t_1t_2 -1)^2}{(t_2^2 + 1 ) (t_1^2 + 1 )}

=(q_3)_b\cdot (\bar q_3)_r

Green:

(q_3)_g =\frac{4(t_1 - t_2)^2}{(t_2^2 + 1 )(t_1^2 + 1)}\cdot \frac{2(t_1+ t_2)(t_1 t_2 - 1)}{(t_2^2 + 1 ) (t_1^2 + 1 )}

=(q_3)_b\cdot \frac{2(t_1+ t_2)(t_1 t_2 - 1)}{(t_2^2 + 1 ) (t_1^2 + 1 )}

=(q_3)_b\cdot (\bar q_3)_g

Apply ‘boost’ d  by the substitution

t'_1 := \frac{t_1 + d}{1 - t_1 d},  t'_2 := \frac{t_2 + d}{1 - t_2 d}, x'_n := x(t_n'), y'_n:=y(t_n')

to give x and y differences (used to calculate the three quadrance types):

x'_2-x'_1=\frac{2(t_1 - t_2 )}{(t_2^2 + 1 )(t_1^2 + 1)}\cdot\frac{(d^2-1)(t_1 + t_2) + 2d (t_1 t_2 - 1) }{(d^2 + 1 )}

y'_2-y'_1=\frac{2(t_2 - t_1 )}{(t_2^2 + 1)(t_1^2 + 1)}\cdot \frac{(d^2 -1)(t_1 t_2 - 1) - 2d(t_1 + t_2) }{(d^2 + 1)}

From which we verify the expected ‘null’ result for blue quadrance(s):

q_3|^d_b = (x'_2-x'_1)^2+(y'_2-y'_1)^2

=\frac{4 (t_1 - t_2 )^2}{(t_2^2 + 1 ) (t_1^2 + 1 )}

=(q_3)_b

and get expressions for ‘boosted’ red quadrance(s):

q_3|^d_r = (x'_2-x'_1)^2-(y'_2-y'_1)^2

=\frac{4 (t_1 - t_2 )^2}{(t_2^2 + 1 ) (t_1^2 + 1 )}\cdot \frac{((d^2-1) (t_1 + t_2) + 2 d ( t_1 t_2 - 1 ))^2-((d^2-1) (t_1 t_2 - 1) - 2 d ( t_1 + t_2 ))^2}{ (d^2 + 1 )^2(t_2^2 + 1 ) (t_1^2 + 1 )}

=(q_3)_b\cdot \frac{((d^2-1) (t_1 + t_2) + 2 d ( t_1 t_2 - 1 ))^2-((d^2-1) (t_1 t_2 - 1) - 2 d ( t_1 + t_2 ))^2}{ (d^2 + 1 )^2(t_2^2 + 1 ) (t_1^2 + 1 )}

=(q_3)_b\cdot \bar q_3|^d_r

and boosted green quadrance(s):

q_3|^d_g = 2(x'_2-x'_1)\cdot(y'_2-y'_1)

=\frac {4(t_1 - t_2 )^2}{(t_1^2 + 1 )(t_2^2 + 1 )}\cdot \frac{ 2((d^2-1)( t_1 +  t_2) + 2 d( t_1 t_2 - 1)) ((d^2-1)(t_1 t_2 - 1) - 2d( t_1 + t_2)))}{(d^2 + 1 )(t_1^2+1)(t_2^2+1)}

=(q_3)_b\cdot \frac{ 2((d^2-1)( t_1 +  t_2) + 2 d( t_1 t_2 - 1)) ((d^2-1)(t_1 t_2 - 1) - 2d( t_1 + t_2)))}{(d^2 + 1 )(t_1^2+1)(t_2^2+1)}

=(q_3)_b\cdot \bar q_3|^d_g

When multiplied out again we observe that:   q|ᵈb ² = qb ² = q|ᵈr² + q|ᵈg ², which is the basic identity of Chromogeometry!

These are the raw ingredients which we must use to establish our observation of  polarity in Chromogeometry: that ‘boosted’ spreads, when viewed projectively, lie on a line joining unboosted spreads, similarly projected.

[insert text here]

 

THIS is a bit of a departure. Responding to Paul Vanderklay’s dicussion with guest Carl in this video today, I’m going to develop my ‘simplest case’ of mathematical chaos in the spirit of rational trigonometry.

Angles are the wrong way to measure relative directions – contrary to what every child in school learns. But! … isn’t the “angle-sum” just the greatest discovery of every young learner? Doesn’t this open up the world of polygonal figures .. (like pentagons and hexagons?) .. of polyhedral figures composed of polygonal figures (like tetrahedra and octahedra?) No.

The truth is, angles don’t really work and are bascially a cheat. Take the simplest case of the angle sum of a triangle. Every triangle, we learn, sums to 180 degrees internally. So if you are given two of the internal angles (say ’36’ degrees and ’48’ degrees) it’s a cinch to find the third (’96’ degrees) Yet, later in our education radian-measure comes along and we are told the angle sum is ‘pi‘. Is that a problem? No, we are assured it just leads to stories of angles like ‘pi/24′ and ’11/57ths pi‘ (say) and so the final angle must be pi (1 – 1/24 – 11/57) and with almost no new technology we have our answer: (‘1047/1368ths’ pi)

But what if, instead you measure a couple of angles with a radian based protracter? Then the answers you give me will be ‘real numbers’ like ‘0.27562 rad‘ or ‘0.60754 rad‘. (It has to be that way because you start with a physical object specified by nature rather than by Plato.) And the third angle? Hmmm isn’t it going to be pi – (0.27562 + 0.60754) or ‘pi minus 0.88316′? This is supposedly exact as was our previous calculation, but for the fact we haven’t reduced our result to a number of any kind yet – it’s still in the form of a sum or addition – meaning there is still work to be done.

So, the unsuspecting student takes his scientific calculator and either converts ‘pi‘ to a decimal and performs the subtraction – or converts ‘0.88316’ to a fraction of ‘pi‘ and does the same thing. But in either case the result is no longer exact. This is because ‘pi‘ is an irrational number and all measurements from the real world exist as finite decimals (sometimes expressed as fractions) Houston, we have a problem!

When something basic breaks down like this the only alternative is start over and try to spot where the problem begins.

What we can’t do is ‘turn’ by a precise amount that brings us back to where we start after a predetermined number of such equal ‘turns’. Think how ‘turning’ is achieved. You need something like a constantly turning object (and how is that going to be established?) and then a metronome-like clock to put down a syllus to record our cumulative change in direction. There is actually no (controlled) relationship between ‘time’ and ‘turn’. It is ficitve – only appearing to be true within this limited example. It can’t be taken out ‘on the road’.

What we can do is apply a initial change of direction (such as over ‘2 over 1 up’) over a co-ordinatised plane (of fixed axes) and use ‘composition’ (which is a truly algebraic operation) to calculate multiples of this change. That at last is reproducible – and that is how we make progress. And that’s the subject of a more techcial blog post to follow.

In 2017 we introduced the ‘orthogonal quadrilateral’ ( or ortholateral ) which arises from the Gauss-Bodenmiller theorem for a complete quadrilateral:

“The Circles on the Diagonals of a Complete Quadrilateral as Diameters are Coaxal. [T]he Orthocenters of the four Triangles of a Complete Quadrilateral are Collinear on the Radical Axis of the Coaxal Circles.”  Source

together with the consequences of three perpendicular forms in Chromogeometry, which give not one but three radical axes mutually perpendicular to the ‘medial’ line (known as the  ‘Newton-Gauss line‘)

Therefore:

For every general quadrilateral an ortholateral exists. And for every ortholateral (being a quadrilateral) a successor exists directly by completing it and including the Newton-Gauss line

‘Ante’ ortholaterals were next considered as logical precursor objects and these would be one of two kinds:

i)  an ortholateral,  or

ii) a general quadrilateral.

A construction for first type of object was given in the previous post from November 2017. Here, we present a construction of the general ‘antiorthic’ quadrilateral derived from the chromogemetry present. We conclude by asking whether such constructions are ‘proper’ (in the sense of being rational or transcendental)

In chromogeometry, an omega triangle is the set of three orthocentres of a given triangle with the property that both ‘Red’ and ‘Green’ orthocentres lie on the ‘Blue’ (i.e usual) circumcircle. We begin therefore with three coloured orthocentres, P, Q and S of a unknown triangle placed on each line lying ‘perpendicular’ with a fourth side of a given ortholateral. Next, we define that triangle’s Red, Green and Blue circumcircles (shaded regions, below) selecting pairs P S , Q S and P Q to form their diameters. The ‘anteomegal’ triangle, if it exists, will be circumscribed where these circles meet triply (in three of six places). Using Geogebra ‘intersection’ commands and set operations we can define this logical set and join their threefold intersections to give three lines (enclosing the hatched region in the figure below) of the four sides of the ‘antiorthic’ quadrilateral  with the fourth side being found by a further construction:

AntiOL via blue circumcircle plus AOLs beginning with AOL
Constructing an ‘antiorthic’ quadrilateral via the anteomegal triangle

(click for animated view)

The unused side of our ortholateral becomes a medial (Newton-Gauss) line with respect to the new object. To find the fourth side in the antiorthic quadrilateral, we reflect each vertex in our designated NG line, drawing parallels to it through reflected positions and note where these meet the extension of the side opposite that vertex. By a ‘similar triangles’ argument, these intersections will lie at equal distance from the NG line opposite to their related vertex and, by perspectivity/Desargue’s Theorem, will be collinear, as required.

Finally the Red and Green orthocentres of the three addtional triangles could be found by direct construction to lie on the revelant lines. But using the fact that, for any quadrilateral, all four Blue circumcircles meet in a single point (Miquel’s Thm) – and our initial centres lie on lines passing through this point having the required pendicular directions – the additional orthocentres must lie on the same Red or Green ‘radical axes’ at those circles’ additional points of intersection. Incidence of the Blue orthocentres with the axis through the original Blue centre is similarly assured.

Discussion
While satisfying as discoveries in ‘constructed’ geometry, the claimed existence of this new ‘antiorthic’ quadrilateral as an algebriac object depends on number theoretic considerations, as does the constructed ‘anti-ortholateral’ of the previous post. Both questions require further examination and investigation.

The least problematic is the claim that every quadrilateral in rational co-ordinates must give rise to a single ortholateral in rational co-ordindates – which is effectively assured by the Gauss-Bodenmiller Theorem. We now see that the anti-ortholateral can only be a valid construction if the circle centred at one vertex used in its construction meets the two circles with diameters raised on each side-segment  (as well as its own diameters)  in rational points. – see the figure below

Anteortholateral side view

The first consideration is satisfied when the perpendicular extensions raised on those sides (‘BE’ shown above) meet their auxiliary circle rationally – which requires the associated spreads at a cicumference to be a number of the form    s = x(1 - x) = y^2     (a natural square, in other words.)

The second consideration: for the major circle centred at A  with a radius given by the geometric mean of the leftmost segment b and the diameter a of the auxiliary circle,  is that the product ab be a square. Finally for the perpendicular meeting the auxiliary circle to also lie on the major circle, its  left segment times that circle’s right segment must be a square. (To ensure that both b(a – b) and ab are squares is equivalent to a, b and (a – b) forming a Pythagorean triple)

Now, for any pair of relatively prime nautral numbers m, n

(m^2 + n^2)^2 = (2 mn)^2 + (m^2 - n^2)^2

thus we let

a = (m^2 + n^2)^2, (a - b) = (2 mn)^2, b = (m^2 - n^2)^2

\sqrt{ab} = (m^2 + n^2)(m^2 - n^2)

\sqrt{ab} + b = (m^2 + n^2)(m^2 - n^2) + (m^2 - n^2)^2 = 2m^2(m^2 - n^2)

\sqrt{ab} - b = (m^2 + n^2)(m^2 - n^2) - (m^2 - n^2)^2 = 2n^2(m^2 - n^2)

Hence

\sqrt{(\sqrt{ab} + b) (\sqrt{ab} - b)} = 2mn(m^2 - n^2)= \sqrt{(a-b)b}

(as required)

We conclude we have at least six rational points lying on the major circle to use in the construction of our two putitative anti-ortholaterals.  What about the last two?

Now as we can seen from the next figure, the major and minor segments a and b in our triple have projections to corresponding segments, b’ and a’ on the ortholateral’s adjacent side via the ‘cosine’, cv at vertex BAB’:

Anteortholateral rational points2

a = b'/c_v : a' = b/c_v ,

b = a' * c_v : b' = a * c_v =>a b = a' b'

and form an equal product to give the squared radius of the major circle as required. But to satisfy the Pythagorean triple condition, a‘, b‘ and (b‘ – a‘) being squares, we wish to show (for some other pair of relatively prime natural numbers, m‘ and n‘):

a' = (m'^2 + n'^2)^2, (a' - b') = (2m'n')^2,b'=(m'^2 - n^2)^2

=>(m'^2 + n'^2)^2 = (2m'n')^2 + (m'^2 - n'^2)^2

<=> (m'^2 + n'^2)^2 = (m^2 - n^2)^2/c_v

<=> (m'^2 - n'^2)^2 = (m^2 + n^2)^2*c_v

<=>(2m'n')^2 = (m^2 - n^2)^2/c_v - (m^2 + n^2)^2*c_v

The only non-zero solution to which is

c_v = (m^2 - n^2)^2/(m^2 + n^2)^2 <=> m=n', n=m'

Which can be considered the degenerate (i.e. ‘parallel line’) case in ortholateral terms since it cannot be continued by completing the quadrilateral.

Concluding Remarks
With the impossibility of forming a Pythagorean triple on an adjacent side, we conclude that the antiortholateral (just as the ‘antiorthic’ quadrilateral) construction is, in general, transcendental. While we could construct, in rational coordinates, a single anti-ortholateral from an ortholateral specified with a Pythagorean triple (such as a = 25, b = 16 giving (ab) = 9) no ‘random’ given ortholateral will have a true Pythagorean triple division of either side.

Analysing the ‘x’ and ‘y’ components, we can expect to find a similarly symmetric algebraic form:

x:              ½( (ya –  yb)(xc² – xb² + yc² – yb²)  –  (y – yc)(xb² – xa² + yb² – ya²)) / norm

y:              ½( (xb –  xc)(xb² – xa² + yb² – ya²)  –  (xa – xb)(xc² – xb² + yc² – yb²)) / norm

norm:        [(x –  xb)(y –  yc)  –  (xb  –  xc)(ya  –  yb)]

The normalising term, as we have seen, becomes

:                  (xa yb – xb y +  xb yc –  xc yb  +  xc ya –  xa yc)

While, distributing component ‘x‘ we obtain:

x   =  ½ [  (ya xc² – ya xb² + ya yc² – ya yb²)  –  (yb xc² – yb xb² + yb yc² – yb yb²)

:       – (yb xb² – yb xa² + yb yb² – yb ya²)  +  (yc xb² – yc xa² + yc yb² – yc ya²) ]

:    =    ½ [   ( ya xc²  –  ya xb²  –  yb xc²  +  yb xa² +  yc xb² –  yc xa² )

:          +      ( ya yc²  –   ya yb²  –  yb yc²  +  yb ya² +  yc yb²  –  yc ya² ) ]

:     =    ½ [   ya (xc²  –  xb²)  +  yb (xa²  –  xc²) +  y(xb²  –  xa²)

:          +        ya (yc²  –  yb²)  +  yb (ya²  –  yc²) +  y(yb²  –  ya²) ]

:     =  ½ [ ya ((xc²+yc²)(xb²+yb²)) + yb ((xa²+ya²)(xc²+yc²)) + yc ((xb²+yb²)(xa²+ya²))]

Now ‘xc² + yc²‘ is equal to the squared-distance or quadrance from the orgin, ‘O’ often writen as ‘Q0,c‘ or, more conveniently as’Qc‘. Thus

x     =       ½ [ ya(Qc  –  Qb) + yb(Q –  Qc) + yc(Q –  Qa) ] / norm

Meanwhile, ‘y’ (above) is the negated form with xi s replacing the yi s

y      =       ½ [ xa(Qb  –  Qc) + xb(Q –  Qa) + xc(Qa –  Qb) ] / norm

 

Now the Equation of the Circle having a normalised centre with symmetric form at:

[x / norm , y / norm]

will have a quadrance (squared radius) given by sum of squares of the differences of the centre from any of the three points, from which we expect to find a symmetric (point-independent) form also.

For example, using: (xa , ya) as our exemplar point we have:

(Point co-ordinate minus circumcentre co-ordinate) squared:

xa²

–  2 xa (½ (ya(Qc  –  Qb) + yb (Q –  Qc)+ yc (Q –  Qa))) / norm

+ ¼ ( ya (Qc  –  Qb) + yb (Q –  Qc) + yc (Q –  Qa))² / norm²

+ ya² 

–  2 ya (½ (xa(Qb  –  Qc) + xb (Q –  Qa)+ xc (Q –  Qb))) / norm

+ ¼ ( xa (Qb  –  Qc) + xb (Q –  Qa) + xc (Q –  Qb))² / norm²

 

= [  (xa² + ya²) norm²

–  (xa yb (Q –  Qc) + xayc (Q –  Qa) + (xb ya(Q –  Qa) + xc ya(Q –  Qb)) norm

+ ¼ (xa (Qb  –  Qc) + xb (Q –  Qa) + xc (Q –  Qb))² 

+ ¼ ( ya (Qc  –  Qb) + yb (Q –  Qc) + yc (Q –  Qa))²] // norm²

 

=  [  (Qa norm –  ((xa yb xb ya)(Q –  Qc) + (xayc xc ya)(Q –  Qa)) norm

+ ¼ (xa (Qb  –  Qc) + xb (Q –  Qa) + xc (Q –  Qb))² 

+ ¼ ( ya (Qc  –  Qb) + yb (Q –  Qc) + yc (Q –  Qa))²] // norm²

 

= [  (Qa (xa yb – xb ya   +  xb yc  –  xc yb   +  xc y –  xa yc)

 –  ((xa yb xb ya)(Qa  –  Qc) + (xayc xc ya)(Q –  Qa) ) norm

+ ¼ (xa (Qb  –  Qc) + xb (Q –  Qa) + xc (Q –  Qb))² 

+ ¼ ( ya (Qc  –  Qb) + yb (Q –  Qc) + yc (Q –  Qa))²] // norm²

 

WE are nearly there! We observe cancellations involving the duplicated appearance of ‘Qa(from our choice of point ‘a’) yielding a natural, symmetric triple form for:

NOW    (xa(Qb – Qc) + xb(Qc – Qa) + xc(Qa – Qb))   =   Qa(xcxb)  + Qb(xaxc) + Qc(xbxa)

and       (ya(Qc – Qb) + yb(Qa – Qc) + yc(Qb – Qa))   =   Qa(ybyc)  + Qb(ycya) + Qc(ya – yb)

hence

The quadrance of a circle through three points (‘a’,’b’,’c’):

=>  [  ¼ (Qa(xcxb)  +  Qb(xaxc) + Qc(xbxa))²                    +

(Qa(xbyc – xcyb) + Qb(xcya – xayc) + Qc(xayb – xbya)) *

     ((xayb – xbya+   (xbyc – xcyb)   +   (xcya – xayc))    +

          ¼  (Qa(ybyc)  + Qb(ycya) + Qc(ya – yb))²  /

      ((xayb – xbya)   (xbyc – xcyb)   (xcya – xayc))²

 

AS would expect!

FINALLY, although now in symmetric terms, this formula is not one of a perfect ‘algebraic square’ i.e. a single, even-factorable expression for the circumquadrance. We could look for a suitable rearrangment of this type but, instead, we consider the implication of assuming no such general condition applies.

‘CONSIDER’ that we have obtained a circle centre in a rational point, starting with three other rational points. Whilst the ‘x’ and ‘y’ displacements to these points are rational, the sums of squares of the distances from the centre (necessarily equal) are not necessarily square. The condition for the circumquadrance be square is for the slopes of the bisectors to be ‘spread numbers’ (i.e. a rational number ‘s’ such that ‘s(1-s)’ is a square) which is the equivalent of making all three points rationally parameterized on a circle. We can accomplish this outcome by making the slopes between the points on the putative circle be spread values (the recipocral of a spread number being a spread number)

 

[Post Script: Our conclusion above has a corollary/predction: that where the three points are related by circular parametrization (and we are not told so) we should obtain an algebriac square expression from them. Assuming we complete the square on the already squared ‘end’ terms by adding a suitable product term, this term will be quadratic in quadrances (like: Qa*Qa, Qa*Qb etc) and will have to be subtracted from the existing ‘middle’ term which is linear in quadrances.  If these two terms can be reconciled into a square, this would be a negative quantity. (The condition for the resulting difference of squares to be a square is the formation of a Pythagorean triple, of course!)]

 

 

 

THIS is a small example in mathematics of how understanding context can generate predictions: the equation of the centre of a circle through three points in terms of the points (I had aimed higher: for the equation of the circle itself but that has defeated my abilities so far!)

How to proceed:

We have a ‘method’ in mind

a) find the (equations of the) lines through two pairs (of three) points

b) find the (equation of the) perpendicular bisector of each line

c) calculate the intersection of these two bisectors

While this is a perfectly reasonable way to proceed, note some (possibly?) slight incongruence: we have a choice of which pair of sides of the triangle formed we use. Why should that be so if the result is the same calculated position?

This generates an expectation that the final result ought to have a symmetric form even if it is generated from an arbitrary selection.

Here we go:

a) a line between two points has proportion : ( y_b - y_a  :  x_a - x_b  :  x_b y_a - x_a y_b)

That is the line has the more usual equation:

(y_b - y_a) x + (x_a - x_b) y + (x_b y_a - x_a y_b) = 0

b) the perpendicular bisector becomes: ( x_a - x_b : y_a - y_b : \frac {1}{2}(x^2_b - x^2_a + y^2_b - y^2_a))

with standard equation:

(x_a - x_b)x + (y_a - y_b)y + \frac {1}{2}(x^2_b - x^2_a+y^2_b-y^2_a))=0

c) a second line has the proportion:            ( y_c - y_b : x_b - x_c : x_c y_b - x_b y_c)

d) has perpendicular bisector:                     (x_b - x_c : y_b - y_c : \frac {1}{2}(x_c^2 - x_b^2 + y_c^2 - y_b^2))

To find the intersection of the lines we take the cross product of the first proportion and the second and ‘normalise’ (divide) the first two entries in the product by the third entry. The use of proportions here (making everything homogeneous) makes points and lines become interchangeable – an instance of symmetry in mathematics

Intersection of bisectors from cross product (using points ‘a’, b’ and ‘b’,’c’):

1st component (x):            \frac {1}{2}(y_a - y_b)(x_c^2 - x_b^2 + y_c^2 - y_b^2) - \frac {1}{2}(y_b - y_c)(x_b^2 - x_a^2 + y_b^2 - y_a^2),

2nd component (y):          \frac {1}{2}(x_b - x_c)(x_b^2 - x_a^2 + y_b^2 - y_a^2) - \frac {1}{2}(x_a - x_b)(x_c^2 - x_b^2 + y_c^2 - y_b^2)

3rd component (‘norm’):  ((x_a - x_b)(y_b - y_c) - (x_b - x_c)(y_a - y_b))

We’ll start with the divisor ( or ‘norm’) – when expanded we would expect four products from the first half and four from the second. Meanwhile what we get is:

:            (x_a y_b - x_a y_c - x_b y_b + x_b y_c) - (x_b y_a - x_b y_b - x_c y_a + x_c y_b)

:      =   (x_a y_b - x_a y_c - x_b y_b + x_b y_c - x_b y_a + x_b y_b + x_c y_a - x_c y_b)

:      =   (x_a y_b - x_a y_c + x_b y_c - x_b y_a + x_c y_a - x_c y_b)

Here is another instance of symmetry: the repeated subscripts disappear and the result no longer depends on which choice of two lines were made. This, of course, is something we would expect from understanding the context which gave rise to the algebra.

The norm which appears is then capable of being put into a geometric context: it corresponds in form to a determinant giving the signed area of our triangle. We worked ‘projectively’, and therefore the first two components giving the location of the circle’s centre form a (projective) line which much be normalised by the measure of separation of the vertices ‘from’ that line (think ‘dispersion’ of the points about their centre)

The ‘x’ and ‘y’ components hold simliar delights of the appearance of symmetry – but we propose to pick that up in a subsequent posting.

PLEASE IGNORE (TEST PURPOSES ONLY):

test

(a^2+b_2-c^2)(a^2-b^2+c^2) : (a^2+b^2-c^2)(-a^2+b^2+c^2) : (a^2-b^2+c^2)(-a^2+b^2+c^2)

CEVA’S THEOREM elegantly describes why a point of perspectivity, joined to the vertices of a triangle, divides the opposite sides in such proportions that the product of corresponding left hand segments matches the product of corresponding right hand segments and, conversely, that if we have such a division of sides by Cevian lines through the vertices the lines must meet in a (Cevian) point.

[WHEREAS we will generally have concerns about the exact description of ‘lengths’ of segments in planar situations, provided all points remain rational (number pairs) we will have rational proportions. Regarded as displacements only therefore, our segments may still be compared as ratios – which is all we rely on when applying the Theorem.]

WE NOW take advantage of this fact to establish the remarkable result – sometimes referred to as a ‘Cevian Nest‘ – and here termed the ‘Cevian of Cevian Rule‘ that when a second Cevian triangle is ‘nested’ inside a first that its vertices (in contact with the sides of the first Cevian triangle) may be joined to the corresponding vertices of the original given triangle to give three lines which meet in a new Cevian point!

Cevian of Cevian Rule 6

(Click here for a larger image)

A TRIANGLE ABC (above) contains a Cevian point X with associated Cevian triangle A’B’C’. In turn, A’B’C’ contains a Cevian point Y with Cevian triangle (sides not shown) A”B”C”. If sides AB and CA are divided in the affine proportions β : (1-β) and α : (1-α) at C’ and B’ respectively then, since

           \frac{(\alpha)( \beta)}{(1 - \alpha - \beta + 2 \alpha \beta)} + \frac{(1 - \alpha )(1 - \beta )}{(1 - \alpha - \beta+ 2 \alpha \beta)} = 1

it follows from Ceva’s Theorem that side BC is divided in proportion

           CA’ : A’B = (αβ)/(1-α-β+2αβ) : (1-α)(1-β)/(1-α-β+2αβ) at A’.

LIKEWISE if sides B’C’, C’A’ of triangle A’B’C’ are divided in affine proportions

             B’A” : A”C’ = γ : (1-γ) and

            C’B” : B”A’ = δ : (1-δ) it follows that

            A’C” : C”B’ = (1-γ)(1-δ)/(1 – γ -δ + 2γδ) : (cδ)/(1 – γ -δ + 2γδ)

NOW the line AA” being a barycentric combination of A and A” necessarily contains B’ and C’ in fixed proportion. Whilst every point on AA” will contain varying proportions of both A and C and A and B respectively, only the proportion of C (in B’) and B (in C’) determine how AA” divides BC (where the weight of A becomes ‘0’.) Line AA”therefore contains B and C in fixed proportion given by

           [B : C]AA”   = (β)(γ) : (1-α)(1-γ)

LIKEWISE BB” and CC” contain the fixed proportions

           [C : A]BB”   = (1-α)(1-β)(δ)/(1-α-β+2αβ) : (1-β)(1-δ)

           =  (1-α)(δ)/(1-α-β+2αβ) : (1-δ)

           [A : B]CC”   = (α)*(1-γ)(1-δ)/(1-γ-δ+2γδ) : (α)(β)/(1-α-β+2αβ)*(γ)(δ)/(1-γ-δ+2γδ)

           =  (1-γ)(1-δ) : (β)(γ)(δ)/(1-α-β+2αβ)

NOW lines AA”, BB” and CC” are concurrent precisely when we have

           [C : A]BB” x [A : B]CC”  =

           (1-α)(δ)/(1-α-β+2αβ)*(1-γ)(1-δ) : (1-δ)* (β)(γ)(δ)/(1-α-β+2αβ) =

           (1-α)(1-γ) : (β)(γ) = 1/[B : C]AA”

In other words, along the new Cevian lines constructed through points A”, B” and C” the proportions of  B : CC : A and A : B  commute just as they would along AA’, BB’ and CC’ and the reference triangle ABC is perspective with any Cevian triangle A”B”C” of its own Cevian triangle A’B’C’.

(This completes the proof )

 

PLEASE IGNORE (TEST ONLY):

(a^2+b_2-c^2)(a^2-b^2+c^2) : (a^2+b^2-c^2)(-a^2+b^2+c^2) : (a^2-b^2+c^2)(-a^2+b^2+c^2)

THE core circle, x2 + y2x = 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.

PENTAGON

PENTAGON (approximately regular)

First five spread polynomials (in the interval [0,1]

First five spread polynomials (in the interval [0,1])

SPREAD polynomials arise as solutions to the triple spread formula (or TSF)

(sa + sb + sc)2 – 2 (sa2 + sb2 + sc2) – 4 sa sb sc = 0.

By fixing two of the spreads, parametrized in the variable s, the formula reduces to a quadratic equation in the third spread. Being quadratic there are two such solutions in the spread variable corresponding, geometrically, to cases where the angle measured by the first spread is oriented positively or negatively with respect to the second. For the simplest case, using sa = sb = s, the TSF becomes:

(2s + Sc(s))2 – 2 (2s2 + Sc(s)2) – 4s2 Sc(s) = 0

   =>  Sc(s) (Sc(s) – 4s(1 – s)) = 0,

giving Sc(s) = 0 or 4s(1 – s).

S(s)  = 0 is indeed the spread when a positive angle corresponding to s combines with a negative angle of equal measure, while 4s(1 – s) (the ‘double angle formula’) is the spread produced when equal, like oriented angles are combined.

Using sa = s,  sb = 4s(1 – s), the TSF then becomes:

(s +  4s(1 – s) + Sc(s))2 – 2 (s2 + 16s2(1 – s)2 + Sc(s)2) – 64s2(1 – s) Sc(s) = 0

  =>  (Sc(s) – s)(Sc(s) – s(3 – 4s)2) = 0,

giving Sc(s) = s or s(3 – 4s)2.

The TSF also gives the ‘solution-set’ of every spread triple where three relations, rather than a single spread relation, are considered at once.

If we take the spread s plotted as  (s, 0, 0)  and its double angled spread relation  4s(1 – s)  plotted as  (s, 4s(1 – s), 0)  and allow s to run through all values in the interval  [0,1]  the familiar Logistic Curve is drawn in the unit square in the first quadrant of the x-y plane.

The pair of spread polynomials (first and second) then combine in the solution of the TSF to form an appropriate triple:

1st, 2nd and 3rd { {s, 4s(1 – s)} , s(3 – 4s)2}  or  1st, 2nd and 1st { {s, 4s(1 – s)} , s}

each entry a polynomial in s which, plotting all three entries at once, describe co-ordinates:

(s, 4s(1 – s), s(3 – 4s)2)  or  (s, 4s(1 – s), s)

lying on separate curves in three dimensions. The ambient space is referred to as a phase-space (sometimes, in 2-d, as phase portrait) when the variables plotted represent the effects of changes occurring in some actual space (rotations of angles in 2-d planar space, for example). In the phase-space either solution forms the locus of a curve whose projection in x-y and x-z planes are respectively the spread polynomials

(s, Sn(s), 0) and (s, 0, Sn-1(s))

and whose projection in the y-z plane is then an implicit curve of the form

(0, Sn(s), Sn-1(s)).

First and second spread triples - in x0y, x0z and y0z projection

First and second spread triple functions – in x0y, x0z and y0z projection

A recursive formula for spread polynomials (obtainable from the TSF) gives, for each new polynomial:

Sn+1(s) = 2(1 – 2s)Sn(s) – Sn-1(s) + 2s

This provides a means of successively generating the spread triples that make up these separate spread triple ‘functions’ S(s,n):

(s, s, S2(s)).. (s, S2(s), S3(s)).. (s, S3(s), S4(s)).. (s, S4(s), S5(s)).. (s, S5(s), S6(s))

These functions can either be varied in the spread variable or the indexing parameter ‘n’. When the latter is carried out the successive values

(s, S2(s)).. (S2(s), S3(s)).. (S3(s), S4(s))…

will all lie on an ellipse in the plane x = s. Developing these ellipses along the x direction traces out the surface resembling an inflated regular tetrahedron and called the Ellipson.

Developing the Ellipson from recurrence relation:

Developing the Ellipson from recurrence relation for spread polynomials

Alternatively, by varying the spread parameter, fixing on particular spread triples and indexing on these, the same surface is developed .

Four Spread Tiples in phase space indicating the form of the Ellipson

First four Spread Tiples in phase-space indicating the form of the Ellipson

Four spread triples with inscribed regular tetrahedron and xoz projection for comparison

Same four Spread Triples with inscribed regular tetrahedron and x0z projection

An Animated Ellipson:

Ellipson 3-D100ms

The Ellipson, f(a,b,c) = (a + b +c)² - 2(a² + b² + c²) - 4abc = 0

Ellipson 2.gif