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]

BY the Gauss-Bodenmiller Theorem[i], the midpoints of the three ‘diagonal’ segments completing a 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), this radical-axis aligns with the segment joining the blue perpendiculars (or crosses).  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 form a NEW ortholateral.

HOWEVER, also by Gauss-Bodenmiller[ii], the four (blue) orthocentres of the triangles associated to a general quadrilateral  are collinear and lie on the (blue) radical-axis. In chromogeometry, with its three distinct perpendiculars, there are therefore three sets of collinear orthocentres forming three radical-axes ‘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 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 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’.

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 may be constructed from the starting orthlolateral (per quadrilateral) but has no affine properties inherited from it. We speculate that such a figure could be rectified by a suitable change of ‘viewing angle’ so as to become an orthogonal system of four points in either the blue red or green sense. (More to follow)

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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

Problem:

IN Exer 129.3 of Norman Wildwerger’s Math Foundations video: MF129: The projective line, circles, and a proof of the CQQ theorem we are asked to show that the quadrea of a triangle (necessarily ‘cyclic’) can be obtained from the signed area formula:

½ (x₁y₂ – x₂y₁ + x₂y₃ – x₃y₂ + x₃y₁ – x₁y₃)

Using the substitutions for a rational point lying on the unit circle centred at the origin :

[x(u,t)i , y(u,t)i] = [ui2ti2/ ui2+ ti2 , 2uiti / ui2+ ti2]

we must show the same formula becomes:

2 (ut₂-ut₁)(ut₃-ut₂)(ut₁-ut₃)/(u₁²+ t₁²)(u₂²+ t₂²)(u₃²+ t₃²)

 

Solution:

BEGIN by factoring a single ‘y‘ (say y₁ ) in the area formula, make substitutions and place over a common denominator:

½ ( x₃ – x₂ )y₁

   =  (½) 2ut₁((u₃²- t₃²)(u₂²+ t₂²)  –  (u₂²- t₂²)(u₃²+ t₃²))/(u₁²+t₁²)(u₂²+t₂²)(u₃²+t₃²)

   =          ut₁((u₃²- t₃²)(u₂²+ t₂²)  –  (u₂²- t₂²)(u₃²+ t₃²))/(u₁²+t₁²)(u₂²+t₂²)(u₃²+t₃²)

EXPAND the numerator and collect like terms:

               ut₁((u₃² –  t₃²)(u₂² +  t₂²)  –  (u₂² –  t₂²)(u₃² +  t₃²))/(u₁² + …

   =         ut₁(u₃²u₂²t₃²t₂² + u₃²t₂²- t₃²u₂²)-(u₂²u₃²t₂²t₃² + u₂²t₃²- t₂²u₃²)/(u₁² + …

   =        2ut₁(u₃²t₂² – t₃²u₂²)/(u₁² + t₁²)(u₂² + t₂²)(u₃² + t₃²)

WE notice a difference of squares in the numerator, but rather than factoring, instead distribute allowing the collection and permuting of terms:

      2ut₁(u₃²t₂² – t₃²u₂²)

   =        2ut₁(u₃²t₂²) –  2ut₁(u₂²t₃²)

   =        2 (ut₁)(ut₂)(ut₂) –  2(ut₁)(ut₃)(ut₃)

By commutativity (of multiplication) we amend the two distinct factors to beome three:

   =        2(ut₂)(ut₁)(ut₂) –  2(ut₃)(ut₁)(ut₃)

[in simpler notation, we write ‘(ut₂)’ as ‘(12)’ – signifying a ‘u before t‘ convention ]

CONTINUING, we have shown:

½ ( x₃ – x₂ )y₁

   =       2 ((ut₂)(uti)(ut₂) –  (ut₃)(uti)(ut₃))/(u₁² + …

   =       2 ((12)(31)(32) – (13)(21)(23))/…

AND thus:

½ (x₁y₂ – x₂y₁ + x₂y₃ – x₃y₂ + x₃y₁ – x₁y₃)

   =       2 ((12)(31)(32) – (13)(21)(23)) +

  ((23)(12)(13) – (21)(32)(31)) +

((31)(23)(21) – (32)(13)(12))/…

NOTE how ‘(32)’ and ‘(23)’ each appear beside two matching pairs in the second and third lines:

[ ‘(12)(13)’ = ‘(13)(12)’ & ‘(21)(31)’ = ‘(31)(21)’ ]

but beside two unmatched pairs in the first line:

[ (12)(31)(32)  –  (13)(21)(23) ]

However adding the following terms, as the second line:

+ (13)(21)(32*) –  (12)(31)(23*)

will give a ‘balanced’ expression (minus the adjustment term) of eight ‘triple’ terms which (by the binomial theorem) can then be factorised.

FROM  the ‘enlarged’ expression:

        2 ((12)(31)(32) – (13)(21)(23)) +

(13)(21)(32*) – (12)(31)(23*) +

  ((23)(12)(13) – (21)(32)(31)) +

((31)(23)(21) – (32)(13)(12))/…

–  [(13)(21)(32) – (12)(31)(23)]/…

We note that algebraically, since ‘like’  terms in corresponding positions still commute, the adjustment terms can be ignored. That is:

[(13)(21)(32*) – (12)(31)(23*)]/… =  0

THUS, for the area of a triangle, we have:

½ (x₁y₂ – x₂y₁ + x₂y₃ – x₃y₂ + x₃y₁ – x₁y₃)

=      2 (((23) – (32))((12)(13) – (12)(31)) –  ((23) – (32))((21)(13) – (21)(31)))/…
=      2 ( (23) – (32) ) ( (12)(13) – (12)(31) –  (21)(13) – (21)(31) )/…
=      2 ( (23) – (32) ) ( (12)( (13) – (31) ) –  (21)( (13) – (31)) )/…
=      2 ( (23) – (32) ) ( (12) –  (21) )( (13) – (31) )/…
=      2 ( (12) – (21) ) ( (23) –  (32) )( (13) – (31) )/…

=     – 2 (ut₂-ut₁)(ut₃-ut₂)(ut₁-ut₃)/(u₁²+ t₁²)(u₂²+ t₂²)(u₃²+ t₃²)


WHICH is (almost!) what we required to show since there is a ‘negative’ sign upon inspection of the terms in the original product.

 

 

NORMAN Wilberger’s ‘Math Foundations’ video MF124: Heron’s formula, Archimedes’ function, and the TQF [Triple Quad Formula] presents the rational trigonometry equivalent of the famous Heron’s formula for the area of a triangle, giving instead its quadrea Æ, (16 Area²) in terms of sides lengths d₁, d₂ and d₃

Æ(Q₁,Q₂,Q₃) = (d₁ + d₂ + d₃)(-d₁ + d₂ + d₃)(d₁ – d₂ + d₃)(d₁ + d₂ – d₃)

As we have seen in Irrational triangles? if the quadrea is a square, the vertical height built on rational base side will be rational also. Archimedes formula for the quadrea quickly shows when a triangle will posess a squared value of Æ:

For example, choosing d₁ = 3, d₂ = 4, d₃ = 5 (a known Pythagorean triple and therefore with commensurable sides) The formula becomes

Æ(9,16,25) = (12)(6)(4)(2)  = (2.2.3)(2.3)(2.2)(2)

which is square. Note the geometric aspect of the first pair of brackets being a multiple (here 3 times) the second pair of brackets.

Now Pythagorean triples can be found from parameters r and s via the proportions

[r² – s² : 2rs : r² + s²] since (r² – s²)² + (2rs)² = (r² + s²)².

Let’s examine how sides with these proportions apply to Archimedes formula:

( d₁ + d₂ + d₃) = r² – s² + 2rs + r² + s² = 2r² + 2rs = 2r (r + s)
(-d₁ + d₂ + d₃) = s² – r² + 2rs + r² + s² = 2s² + 2rs = 2s (r + s)
( d₁   d₂ + d₃) = r² – s² –  2rs + r² + s² = 2r² –  2rs = 2r (r –  s)
( d₁+ d₂ – d₃) = r² – s² +  2rs –  r² – s² = 2rs –  2s² = 2s (r –  s)

Thus

Æ(Q₁,Q₂,Q₃) = 2r (r + s).2s (r + s).2r (r –  s).2s (r –  s) = 16r²s²(r – s)²(r + s)²

Moreover, the triangle area is given by  a(n integral) rational expression: rs(r – s)(r + s) as desired and ,since this produces a right angled figure with integral sides, all three vertices can lie on rational points.

THE FINAL STEP in producing a general (scalene) triangle with integer sides whose vertices are rational points is to join two such triangles along their medial side length ‘2rs‘ whilst making these sides equal. This is accomplished by making the parameters themselves products of smaller values and interchanging in pairs:

Let r = ab, s = cd, r’ = ac and s’ = bd. Hence 2rs = 2 r’s’. This gives a triangle with a perpendicular height of 2abcd and sides consisting of the separate hypotenuses from the right angled triangles:

d₁   =   (ab)²  +  (cd)²
d₂   =   (ac)²  +  (bd)²

combined with either the sum: [(ab)² – (cd)²]  +  [(ac)² – (bd)²]

>d₃ =  (a² –  d²)(b² + c²)

or difference: [(ab)² – (cd)²]  –  [(ac)² – (bd)²]

<d₃ =  (a² + d²)(b² – c²)

(a>d b>c)

Rational triangle example:

Let a =4, b = 3, c = 2 and d = 1

i)  common height               = 2 (4 x 3 x 2 x 1)  =   48

ii)  d₁  = (4 x 3)² + (2 x 1)²    =     144  +  4         = 148

iii) d₂   = (4 x 2)² + (3 x 1)²   =       64  +  9         =   73

iv) >d₃ =  (4² – 1²)(3² + 2²)   =     (15)(13)          = 195

 v) <d₃ =  (4² + 1²)(3² – 2²)   =     (17)( 5 )          =   85

Find quadreas and verfiy they are squared values, giving rational value areas:

vi) Æ(d₁² ,d₂² ,>d₃² )            =  (148+73+195) (-148+73+195) (148-73+195) (148+73-195)

     =   (416)                 (120)                   (270)                 (26)

     =   (2.2.2.2.2.13)  (2.2.2.3.5)          (2.3.3.3.5 )        (2.13)

     = 2⁴ 2⁶ 3⁴ 5² 13²

     = 16 (2³ 3² 5 13)²

Likewise, for the obtuse triangle formed from a difference of lengths/areas:

vii) Æ(d₁² ,d₂² ,<d₃² )          =  (148+73 + 85) (-148+73 + 85) (148-73 + 85) (148+73 – 85)

     =   (306)                 (10)                    (160)                 (136)

     =   (2.3.3.17)         (2.5)                   (2.2.2.2.2.5 )    (2.2.2.17)

    = 2⁴ 2⁶ 3² 5² 17²

    = 16 (2³  3  5 17)².

Each triangle has a vertex at (0,48) and shares the x axis as base. The difference (110) between the acute and obtuse cases of the triangle divides equally about the orgin as alternate positions: either (-55,0) or (55,0) to give a second vertex, whilst the final vertex consists of the shorter side length (85) added to (or subtracted from) that postion giving: either (-140,0) or (140,0).

While specifying triangles using rational-valued distances between vertices ensures the quadrances are at least squares, valid solutions to the spreads of a triangle do not depend on this property. Thus a triangle with quadrances 18, 32 and 50 still has spreads of 3/5, 4/5 and 1 even though it would formally have side lengths of 3√2, 4√2 and 5√2, and therefore would be non-constructible.

EDIT 19/10/2015: Actually, such a triangle is constructible!

IT0

A slightly weaker irrationality occurs when one quadrance is a square, but not the other two. For instance a triangle with unit quadrance AC, having an obtuse spread of 5/7  (and therefore lying 1 – 5/7 or 2/7 outside the referent semi-circle, as shown in the figure below) has non constructible vertex B at (-2/7,√(10/49)) and quadrance AB of 14/49.

IT 1

The third quadrance BC (from the larger right angled triangle) is then (9/7)² + 10/49 = 91/49.

Multiplying by 49 to clear denominators on these three values 1 : 14/49 : 91/49 gives quadrances of 49, 14 and 91. Then using the formula for the quadrea (16 times square of signed area) of this similar triangle will give the same rational spreads:

49² Æ (A,B,C)   =  Æ (a,b,c)

        Æ (a,b,c)   =   49² [ (QAB + QAC + QBC)²  – 2  (QAB² + QAC² + QBC²) ]

                            =          (Qab + Qac + Qbc)²     – 2  (Qab² + Qac² + Qbc²)

      ≡           (14  +  49  +  91)²        – 2  (14²  + 49²  +  91²)

     =           1960

SA  = Æ (a,b,c)/(4 QabQac)   ≡   1960/2744      =   (5 x 392)/(7 x 392)        =  5/7

SB  = Æ (a,b,c)/(4 QabQbc)   ≡   1960/5096      =   (35 x 56)/(91 x 56)        =  35/91

SC  = Æ (a,b,c)/(4 QacQbc)   ≡   1960/17836    =   (10 x 196)/(91 x 196)    =  10/91

Triangle abc (not shown) would therefore have rational quadrances (only one a square) and spreads, with vertices dilated by √49 (x7) at (say) (0,0), (7,0) and   (-2,√10) and a non-square qudarea of 1960. It is apparent therefore that no rescaling of the basic triangle ABC can make all of its points rational.

What will make a triangle completely ‘rational’, geometrically as well as metrically?

Provided all quadrances of a triangle are square AND the quadrance of one vertex to an opposite side is also a square, the resulting triangle will contain three rational points and thus be constructible.

This is shown considering two right triangles, with common height, meeting to form a third triangle. Without loss of generality, the base of this triangle can be a unit, as shown in the figure below (the semi-circle, given for reference, shows the triangle is itself non-right):

IT 3

For vertex D to be rational only requires that y² exists (given the prior assumption that x and 1 – x are rational lengths)

QAD = x² + y²,   QCD = (dist(A,C) – x)² + y² = QAC – 2dist(A,C)x + x² + y²

Thus

QCD – QAD = QAC – 2dist(A,C)x

⇒ (QCD – QAD – QAC)² = 4QACx²

x² = (QCD – QAD – QAC)²/4QAC

y² =  QAD – x² = [4QACQAD – (QCD – QAD – QAC)²]/4QAC

The denominator is a square, so we require the numerator [4QACQAD – (QCD – QAD – QAC)²] be square also.

We note this numerator is the non-symmetric expression for the quadrea of the triangle Æ (A,B,C), 16 times the square of its area, which we have already seen above. That is:

4QABQAC – (QAB – QBC – QAC)² ≡ (QAB + QAB + QAC)²  –  2 (QAB² + QAB² + QAC²)

Thus Q(y) is square ⇒ dist(D,E) is rational ⇒ Æ (A,B,C) is square.

The square property of the quadrea assures only that  y is a square therefore, but constraints still exist on choosing (square) quadrances QAD and QCD to give all three vertices rational coordinates.

Rational Trigonometry:

deriving the TSF from a standard trigonometric identity

The Triple Spread Formula (TSF)

(SA + SB + SC)² = 2 (SA² + SB² + SC²) + 4 SASBSC

features prominently in rational trigonometry, but its derivation (via the Cross Law) may not satisfy everyone coming across it from a standard trigonometrical background, steeped in angles. In particular, how is the TSF really ‘equivalent’ to the familiar sum of angles condition?

Well, the TSF is derived from the sine-ratios of vertices, not the angular measures of those vertices, so the derivation must use a known relationship on the sines in a triangle to stand in for the simpler (but less tractable) angles themselves:

Since the angles in a triangle sum to 180º and the sine of an angle equals the sine of its supplement, it follows that:

Sin(A) = Sin(B+C) = Sin(B)Cos(C) + Sin(C)Cos(B).

Square both sides, noting Cos²(B) = 1 – Sin²(B) = 1 – SB etc

⇒ S = SB(1 – SC) + 2 Sin(B)Sin(C)Cos(B)Cos(C) + SC(1 – SB)

Distribute and collect like terms (i.e. spreads) on one side:

⇒ S – (SB + SC) + 2 SBSC  = 2 Sin(B)Sin(C)Cos(B)Cos(C)

Square both sides a second time to ‘clear’ the ordinary trigonometric ratios:

⇒ (S – (SB + SC))² + 4 (S – (SB + SC))SBSC + 4 (SBSC)² = 4 SBSC(1 – SB)(1 – SC)

 But we want a symmetric form to appear in the final formula, hence we substitute:

     (S + (SB + SC))² – 4 SA(SB + SC)  ≡  (S – (SB + SC))²

⇒ (S +  SB + SC)²  –  4 SA(SB + SC) + 4(S – (SB + SC))SBSC + 4 (SBSC

= 4 SBSC(1 – SB)(1 – SC)

= 4 SBSC(1 – (S+ SC) + SBSC)

Distribute both sides and cancel any like terms we find:

⇒ (S +  SB + SC)² – 4 SASB – 4 SASC + 4 SASBSC – 4 SBSC(SB + SC) + 4 (SBSC

= 4 SBSC                                                                            – 4 SBSC(S+ SC) + 4 (SBSC

⇒ (S +  SB + SC)² – 4 SASB –  4 SASC + 4 SASBSC

= 4 SBSC

We’re almost there! Take everything except the perfect square to the right hand side:

⇒ (S +  SB + SC)²  = 4 SASB + 4 SASC + 4 SBSC – 4 SASBSC

Note the three cross terms appearing together and recall the identity

(S +  SB + SC)²  ≡ (SA² +  SB² + SC²) + 2  (SASB + SASC+ SBSC)

⇒ 4 (SASB + SASC + SBSC)  ≡  2 (S +  SB + SC)²  –  2 (SA² +  SB² + SC²)

Therefore our equation becomes:

(S +  SB + SC)²  = 2 (S +  SB + SC)²  –  2 (SA² +  SB² + SC²) – 4 SASBSC

and simple rearrangement (subtract twice the perfect square and switch sign) gives

(S +  SB + SC)²  =   2 (SA² +  SB² + SC²) + 4 SASBSC

Which is what we required to show.

Hopefully this derivation is of some use to the reader since (after obtaining it one time to satisfy myself) I couldn’t easily recreate it, so it can sit here as a reference to what we ‘know’ to be the case but may not have at our ‘fingertips’