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: (cathetus b4)<(cathetus a)>(4(cathetus b))(\frac{cathetus\ {b}}{4})<(cathetus\ a)>(4\left(cathetus\ b\right)) (hypotenuse c)=a2+b2(hypotenuse\ c)=\sqrt{{a}^2+{b}^2} tetrahedron ABCD1vertice A=(0,b,0){tetrahedron\ ABC{D}_1}_{vertice\ A}=\left(0,b,0\right) tetrahedron ABCD1vertice B=(a,0,0){tetrahedron\ ABC{D}_1}_{vertice\ B}=\left(a,0,0\right) tetrahedron ABCD1vertice C=(0,0,0){tetrahedron\ ABC{D}_1}_{vertice\ C}=\left(0,0,0\right) Graphing line g as y=x and the following function**: f(x)=((4a2+((4c2+(c−(4b2+(b±(x2−4b2))2−4c2))2−4a2)−a)2))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 hx . This being the length of CD1 . Point xD1=if a <65 then −(hx2−(2a)2) else (hx2−(2a)2)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 yD1=if a <1.98455575342734… then (−(4b2−4a2)+x2) else −(−(4b2−4a2)+x2)Point\ {y}_{{D}_1}=if\ a\...
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