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I've looked for ways to construct a tetartoid pentagon. The coordinates for the five points are given with three numbers each. I realize I'm probably looking at the wrong nomenclature but it seems that five points described with two numbers each would give points for the pentagon on a plane.

At first I thought that a, b, c were going to be sides of the pentagon aabbc but that does not appear to be the case.

Any assistance that you can give me as to how I'm looking at this wrong would be greatly appreciated.

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It happens that the $3$-dimensional coordinates are far nicer than their $2$-dimensional counterparts. For another example of this kind of thing, consider the equilateral triangle: in the plane, sample coordinates are: $(1,0)$, $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$, $\left(\frac{1}{2},-\frac{\sqrt{3}}{2}\right)$; not too bad, but space offers us the ultra-simple $(1,0,0)$, $(0,1,0)$, $(0,0,1)$.

As for a tetartoid pentagon $ABCDE$, we have (slightly modified) Ed Pegg's parameterization of these $3$d coordinates: $$A = (a, b, c) \qquad B = (−a, −b, c) \qquad C = \frac{cm}{p}(-1,-1,1) \\ D = (−c, −a, b) \qquad E = \frac{cm}{q}(-1,1,1) \tag{1}$$ where $0 \leq a \leq b \leq c$ and $$m := b c - a^2 \geq 0 \qquad p := m + ( b - a ) ( c - b ) \geq 0 \qquad q := m + ( a + b ) ( c-b ) \geq 0$$ (As you note, the $a$, $b$, $c$ here are not used in the same way as in the "rhyme scheme" $abbcc$ for the congruence of edge-lengths in the pentagon. They're simply independent parameters.) The coordinates above are reasonably uncomplicated; converting them to $2$d isn't too difficult a process, but yields messier expressions: $$ A = (-r,0) \qquad B = (r, 0) \tag{2}\\ C = \frac{c}{pr} \left(\; m ( a + b ), s ( b - a )\;\right) \quad D = \frac{1}{r}\left(\; a ( b + c ), s \;\right) \quad E = \frac{c}{qr}\left(\; -m ( b - a ), s ( a + b ) \;\right) $$ where $r := \sqrt{a^2+b^2}$ and $s := \sqrt{( a^2 - b c )^2 + ( a^2 + b^2 ) ( b - c )^2}$.

We can simplify the coordinates in $(2)$ slightly by magnifying the pentagon by a factor of $\sqrt{a^2+b^2}$ (that is, by multiplying everything by $r$, which eliminate our need for "$r$" altogether): $$ A = (-(a^2+b^2),0) \qquad B = (a^2+b^2, 0) \tag{3}\\[8pt] C = \frac{c}{p} \left(\; m ( a + b ), s ( b - a )\;\right) \quad D = \left(\; a ( b + c ), s \;\right) \quad E = \frac{c}{q}\left(\; -m ( b - a ), s ( a + b ) \;\right) $$ It's possible that further simplification is possible ---perhaps in terms of parameters other than $a$, $b$, $c$--- but $(3)$ is the best I have at the moment.

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