Rational Wedges for Selected Polycons

Abstract

Motivation for generalization of the quadrilateral construction is stimulated by examination of the wedge associated with vertex 1 of this element. The simplest rational function that vanishes on (2;3)₂ is

$$\displaystyle{ \mathrm{f}_{1}(\mathrm{x,y})\; =\;\mathrm{ k}_{1}(1 -\mathrm{ x}^{2} -\mathrm{ y}^{2})/\mathrm{Q}_{ 1}\,(\mathrm{x,y})\;, }$$

(3.1)

and we seek Q₁ to make this function linear on sides (1;2) and (3;1) of the 3-con. Let Q₁ = a + bx + cy. On (3;1), y = 0 and

$$\displaystyle{ \mathrm{f}_{1}(\mathrm{x,y})\; =\; [(1 -\mathrm{ x}^{\mathrm{2}})/(\mathrm{a}\; +\;\mathrm{ bx})]\;\;\bmod \;\;(3;1)\;. }$$

(3.2)

For f₁ to be linear on (3;1) with f₁(x₃, y₃) = 0, we must have a = b. On side (1;2), x = 0 and

$$\displaystyle{ \mathrm{f}_{1}(\mathrm{x,y})\; \equiv \; [(1 -\mathrm{ y}^{\mathrm{2}})/(\mathrm{a}\; +\;\mathrm{ cy})]\;\;\bmod \;\;(1;2)\;. }$$

(3.3)

For f₁ to be linear on (1;2) and f₁(x₂,y₂) = 0, we must set a = c. When a = b = c:

$$\displaystyle{ \begin{array}{rl} \mathrm{f}_{1}(\mathrm{x,y})\;& \equiv \; (1 -\mathrm{ x})\;\;\bmod \;\;(3;1) \\ \;& \equiv \; (1 -\mathrm{ y})\;\;\bmod \;\;(1;2)\;.\end{array} }$$

(3.4)

Thus Q₁ is the line on which 1 + x + y = 0, and a candidate for W₁ is the function

$$\displaystyle{ \mathrm{f}_{1}(\mathrm{x,y})\; =\; (1 -\mathrm{ y}^{2} -\mathrm{ x}^{2})/(1\, +\;\mathrm{ x}\; +\;\mathrm{ y}). }$$

(3.5)