# Derivation of snub disphenoid coordinates (no ontology)

The following shows how the coordinates of the snub disphenoid are derived.

Note that in this derivation we will not attempt to express the polynomial roots in terms of radicals; such expressions can obtained by using a CAS (computer algebra system) or by the well-known formula for solving cubic equations. Fast numerical algorithms exist for computing the numerical values of these roots to any desired precision via interpolation (such as Newton's method, or modern cubic-interpolation algorithms), and these numerical values are much easier to work with than the very unwieldy expressions in radicals produced by the cubic formula.

## Initial setup and constraints

First, we write the general form of the coordinates based on the hemitetragonal symmetry of the snub disphenoid, centered on the origin, as follows. We will assume an edge length of 2.

<0, A, ±1>
<±C, B, 0>
<0, -B, ±C>
<±1, -A, 0>

where A, B, C > 0, and A > B.

Based on the fact that edge lengths must be uniformly 2, we have the following constraints:

1. ||<C, B, 0> - <0, A, 1>|| = ||<C, B-A, -1>|| = 2
2. ||<0, -B, C> - <0, A, 1>|| = ||<0, -B-A, C-1>|| = 2
3. ||<0, -B, C> - <C, B, 0>|| = ||-C, -2B, C>|| = 2

where ||x|| denotes the vector length of x, that is, √(x12 + x22 + x32), where x = <x1, x2, x3>.

For convenience, we also note that the angle between the edge from <0, A, 1> to <C, B, 0> and the edge from <0, A, 1> to <0, -B, -C> must be 60°, since they form part of one of the equilateral triangles of the snub disphenoid. Thus, their dot product should equal 2*2*cos(60°) (since all edges have length 2):

(<C, B, 0> - <0, A, 1>)∙(<0, -B, -C> - <0, A, 1>) = 4*cos(60°) = 4*(1/2) = 2

Expanding the left-hand side and applying a little algebra, we arrive at:

A2 - B2 = C + 1

Strictly speaking, we don't need this equation, because it can be derived from the preceding three length constraints described above. However, it is in a convenient form for the algebra that will follow. It is also possible to derive more constraints by requiring the angle between the other edges to be 60°, but again, these are already subsumed by the three length constraints above, and they are not needed for the following derivation.

In summary, then, after applying some algebraic simplifications of the above constraints, our initial constraints are:

[Equation 1] A2 + B2 + C2 - 2AB = 3 A2 + B2 + C2 + 2AB - 2C = 3 C2 + 2B2 = 2 A2 - B2 = C + 1

subject to the inequalities:

A > B > 0
C > 0

While these inequalities are pretty obvious, it's useful to state them explicitly because they will come in handy later when we need to decide which of multiple possible polynomial roots are admissible.

## Solving for B

The first thing we notice about the initial system of equations is that equation 1 and equation 2 are very similar, and we can eliminate the squared terms (A2 + B2 + C2) by subtracting these two equations. Then applying a little algebraic rearrangement, we obtain the following relation between A, B, and C:

C = 2AB [Equation 7]

### Eliminating C

Now we can use equation 7 to eliminate C from equations 3 and 4. First, we substitute equation 7 into equation 4:

A2 - B2 = 2AB + 1
A2 - 2AB - B2 = 1

In order to separate the variables A and B, we complete the square for (A2 - 2AB) by adding the term +B2:

A2 - 2AB + (B2 - B2) - B2 = 1
(A2 - 2AB + B2) - B2 - B2 = 1
(A - B)2 - 2B2 = 1
(A - B)2 = 1 + 2B2
A - B = ±√(1 + 2B2)

But since A > B according to our initial constraints, (A - B) must be positive, so we select the positive root:

A - B = √(1 + 2B2)
A = √(1 + 2B2) + B [Equation 8]

### Eliminating A

Now, we try to eliminate A by deriving a different expression for A in terms of B by substituting equation 7 into equation 3:

(2AB)2 + 2B2 = 2

Solving for A and simplifying, we eventually obtain:

A2 = (1 - B2) / (2B2) [Equation 9b]

Note that this equation implies that (1 - B2) > 0, which in turn implies:

B2 < 1 [Inequality 9a]

This will be important later on in selecting the correct root of the solution polynomial.

Continuing, we take square roots on both sides to express A in terms of B:

A = ±√((1 - B2) / (2B2))

But since we require A > 0, we again select the positive root:

A = √((1 - B2) / (2B2)) [Equation 9]

Equations 8 and 9 let us eliminate A completely, by equating their respective right-hand sides, thus obtaining an equation in a single variable B:

√((1 - B2) / (2B2)) = √(1 + 2B2) + B [Equation 10]

Now we try to solve for B. First, we square both sides in order to remove the radical on the left-hand side of equation 10:

(1 - B2) / (2B2) = (√(1 + 2B2) + B)2

Expanding the square on the right-hand side and rearranging the equation so that the radical √(1 + 2B2) appears on the left-hand side, we eventually obtain:

4B3√(1 + 2B2) = 1 - 3B2 - 6B4

Now we square both sides again to eliminate the last radical:

16B6(1 + 2B2) = (1 - 3B2 - 6B4)2

### Selecting the correct root

Since the exponents on B are all even, we make the substitution v=B2 in order to keep the equations manageable. Note that inequality 9a implies that v < 1. Thus we have:

16v3(1 + 2v) = (1 - 3v - 6v2)2

Expanding the square on the right-hand side and rearranging the equation, we eventually arrive at the following quartic in v:

4v4 + 20v3 - 2v2 - 6v + 1 = 0 [Equation 13]

It turns out that this quartic is divisible by (2v - 1), yielding (via polynomial long division, left as an exercise for the reader) the factorization:

(2v - 1)(2v3 + 11v2 + 4v - 1) = 0

The factor (2v - 1) yields the solution v = 1/2, implying B2 = 1/2. However, if we substitute this into equation 9, we obtain:

A = √((1 - 1/2) / (2*1/2)) = ... = √(1/2)
A2 = 1/2 = B2

which contradicts our construction that A > B.

Therefore, v = 1/2 is an inadmissible solution. This leaves us with the cubic:

2v3 + 11v2 + 4v - 1 = 0 [Equation 14]

Since v = B2 is a square, it must be positive; furthermore, by inequality 9a, B2 < 1. Therefore:

0 < v < 1

As it turns out, equation 14 has exactly one positive root between 0 and 1, at approximately v=0.169022229424176. Extracting the square root, we obtain the value of B as approximately 0.4111231317.

## Solving for C

Going back to equation 3, we can rearrange it as:

B2 = 1 - (1/2)C2

Rewriting in terms of v=B2, we have:

v = 1 - (1/2)C2

For ease of subsequent algebraic manipulation, we perform the substitution w=C2, obtaining:

v = 1 - w/2

Substituting this into equation 14 and expanding and rearranging the equation, we eventually obtain:

w3 - 17w2 + 64w - 64 = 0

Now, since v>0, we have:

1 - w/2 > 0
1 > w/2
w < 2

It turns out that the above cubic has exactly one positive root less than 2, at around w=1.661955541. Extracting the square root, we obtain the value of C as approximately 1.289168546448310.

## Solving for A

We proceed in a similar vein to find a polynomial expression for A. Starting with equation 9b and making the substitution u=A2, we obtain:

u = (1 - v) / (2v)

Solving for v:

v = 1/(2u + 1)

Substituting this into equation 14 and expanding and rearranging the equation, we eventually obtain:

2u3 - u2 - 8u - 4 = 0

It turns out that this cubic has only one positive root at approximately u=2.458190775872486. This yields the value of A as approximately 1.567861848465127.

## Summary

In summary, the values of A, B, and C can be obtained by computing the indicated roots of the following cubic polynomials:

2u3 - u2 - 8u - 4 = 0; (0 < u < 3)
2v3 + 11v2 + 4v - 1 = 0; (0 < v < 1)
w3 - 17w2 + 64w - 64 = 0; (0 < w < 2)

where u=A2, v=B2, w=C2.

Numerically, the values of A, B, and C are approximately:

A = 1.567861848465127...
B = 0.411123131706519...
C = 1.289168546448310...