Harmonious Pythagorean Tetrahedra – The Apex Singularity Path | Steve Wait – April 4, 2022

Finding the apex singularity D₁ for the Harmonious Pythagorean Tetrahedron ABCD₁ .

Note that the z negative solution D₂ is not explicitly shown here but may be inferred.

For this example of the base right triangle ABC in the xy plane:

$(\frac{cathetus\ {b}}{4})<(cathetus\ a)>(4\left(cathetus\ b\right))$

$(hypotenuse\ c)=\sqrt{{a}^2+{b}^2}$

${tetrahedron\ ABC{D}_1}_{vertice\ A}=\left(0,b,0\right)$

${tetrahedron\ ABC{D}_1}_{vertice\ B}=\left(a,0,0\right)$

${tetrahedron\ ABC{D}_1}_{vertice\ C}=\left(0,0,0\right)$

Graphing line g as y=x and the following function**:

$f\left(x\right)=\left(\sqrt{\left(4a^2+\left(\sqrt{\left(4c^2+\left(c-\sqrt{\left(4b^2+\left(b\pm\sqrt{\left(x^2-4b^2\right)}\right)^2-4c^2\right)}\right)^2-4a^2\right)-a}\right)^2\right)}\right)$

Yields the x (or if you prefer, y) component at their intersection as h_x . This being the length of CD₁ .

$Point\ {x}_{{D}_1}=if\ a\ <\sqrt{65}\ then\ -\sqrt{\left({{h}_x}^2-\left(2a\right)^2\right)}\ else\ \sqrt{\left({{h}_x}^2-\left(2a\right)^2\right)}$

$Point\ {y}_{{D}_1}=if\ a\ <1.98455575342734…\ \ then\ \sqrt{\left(-\left(4{b}^2-4{a}^2\right)+{x}^2\right)}\ else\ -\sqrt{\left(-\left(4{b}^2-4{a}^2\right)+{x}^2\right)}$

$Point\ {z}_{{D}_1}=\sqrt{\left(4{b}^2-{x}^2\right)}$

Point D₁ then traverses the x-axis at $a=\sqrt{65}$ and the y-axis at $a=1.98455575342734…$

Thus, the volume of the tetrahedron can be realized by:

${V}_{ABC{D}_1}=\frac{\left(\left(.5ab\right){z}_{{D}_1}\right)}{3}$

I’ve created a Geogebra animation (Harmonious Pythagorean Tetrahedra – GeoGebra) to illustrate the path of D₁ between the limits of the a:b ratio 1:4 and 4:1 where the two dimensional degenerates occur. *Note that the animation is computationally demanding. Thus, it is rather slow and coarse in nature.