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

*via the proportions*

**s**[* r*² –

*² : 2*

**s***:*

**rs***² +*

**r***²] since (*

**s***² –*

**r***²)² + (2*

**s***)² = (*

**rs***² +*

**r***²)².*

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

##### ( *d₁ + d₂ + d₃*) = **r**² – **s**² + 2**rs** + **r**² + **s**² = 2**r**² + 2**rs = 2r ****(****r + s****)**

**r**

**s**

**rs**

**r**

**s**

**r**

**rs = 2r**

**r + s**

##### (-*d₁ + d₂ + d₃*) = **s**² – **r**² + 2**rs** + **r**² + **s**² = 2**s**² + 2**rs = 2s ****(****r + s****)**

**s**

**r**

**rs**

**r**

**s**

**s**

**rs = 2s**

**r + s**

##### ( *d₁ *–* d₂ + d₃*) = **r**² – **s**² – 2**rs** + **r**² + **s**² = 2**r**² – 2**rs = 2r ****(****r – s****)**

**r**

**s**

**rs**

**r**

**s**

**r**

**rs = 2r**

**r – s**

##### ( *d₁+ d₂ – d₃*) = **r**² – **s**² + 2**rs** – **r**² – **s**² = 2**rs **– 2**s**² **= 2s ****(****r – s****)**

**r**

**s**

**rs**

**r**

**s**

**rs**

**s**

**= 2s**

**r – s**

Thus

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

**2r**

**r + s**

**2s**

**r + s**

**2r**

**r – s**

**2s**

**r – s**

**16r**

**s**

**r – s**

**r + s**

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 ‘**2 rs**‘ 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 2*rs* = 2 *r’s’*. This gives a triangle with a perpendicular height of 2**abcd** 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 *= (**a**² + **d**²)(**b**² – **c**²)

(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)**.