Rhombi, Trapezia, and a Case of Pythagorean (et al.) Expansion | Steve Wait – July 4, 2022

For any triangle A₁ B₁ C₁ where c** is the long leg, an exponent n exists that will satisfy a ⁿ + b ⁿ = c ⁿ of non-Diophantine concern. Vertex trisection of A₁ B₁ C₁ through the proportional center yields areas where B₁ C₁ D + A₁ C₁ D = A₁ B₁ D. Subsequent dilation about the proportional center produces dissimilar quadrilaterals. Ad infinitum, these trapezia can exhibit corresponding conservation of the Pythagorean (et al.) Theorem if for a right triangle and the analogous for those absent perpendicularity.

**Alternatively, a ⁿ + b ⁿ = c ⁿ can be re-written b ⁿ – a ⁿ = c ⁿ at the isosceles inversion point to maintain constancy of convention. This is necessitated as ∠C₁ changes between the degenerate triangle limits of zero and π radians, the transition from obtuse scalene to acute scalene. Constrained by this equation, as triangles of a:b < 1 approach isosceles, n becomes extreme. In the case of a:b = 1, the equation is invalid at the equilateral where c fails to be the long leg.

 

To better illustrate this, I have created a Geogebra animation found here Pythagorean Expansion.

 

Maintaining the Pythagorean area analogy as the interdependent n and ∠C₁ deviate from the orthogonal, necessitates a departure from the conventional expression of area. Used here, rhombic or non-orthogonal area as defined by .5ab and given in units n, maintains harmony with the area of a right triangle while offering distinction from ½ base × height as well Heron.

Thus, the following assertion can be made regarding the requisite trapezia area of “Pythagorean” conformance:

$\left[a^n{=\ Rhombic\ Area\ of\ Trapezium}_{B_1C_1C_2B_2}^\ast\right]+\left[{b}^n{=\ Rhombic\ Area\ of\ {Trapezium}}_{{A}_1{C}_1{C}_2{A}_2}^*\right]=\left[{c}^n{=\ Rhombic\ Area\ of\ {Trapezium}}_{{A}_1{B}_1{B}_2{A}_2}^*\right]$

$*Shape\ is\ Square\ if\ n=2$

 

Validity for this statement is found where the a:b ratio is expressed as the compliance exponent n and ∠C₁ from:

$\sqrt{\left({a}^2+{b}^2-\left(2ab\times\cos{C}\right)\right)}=\sqrt[{n}]{{a}^n+{b}^n}$

Via a curve of function:

$f\left(x\right)=\cos^{-1}{\left(\frac{-\left(\sqrt[{n}]{{a}^n+{b}^n}\right)^2+{a}^2+{b}^2}{2ab}\right)}$

 

Further, while knowing the proportional center lies along a segment between the mid-points of a and b, I have yet to define it for other than right triangles and therefore use lateral trapezia height to establish the dilated triangle size and proportional center.

${h}_a=\frac{2{a}^n\left(\frac{\sqrt{\left(\left(\frac{a+b+{c}}{2}\right)\left(\left(\frac{a+b+{c}}{2}\right)-a\right)\left(\left(\frac{a+b+{c}}{2}\right)-b\right)\left(\left(\frac{a+b+{c}}{2}\right)-c\right)\right)}}{.5ab}\right)}{a+\left(\frac{{a}}{b}\right)\sqrt{\left(\frac{b\left(.5ab\right)+{a}^n+{b}^n+{c}^n}{.5a}\right)}}$

$h_b=\frac{2{b}^n\left(\frac{\sqrt{\left(\left(\frac{a+b+c}{2}\right)\left(\left(\frac{a+b+c}{2}\right)-a\right)\left(\left(\frac{a+b+c}{2}\right)-b\right)\left(\left(\frac{a+b+c}{2}\right)-c\right)\right)}}{.5ab}\right)}{b+\sqrt{\left(\frac{b\left(.5ab\right)+a^n+b^n+c^n}{.5a}\right)}}$