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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).

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