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# Volume Solutions to Snub Dodecahedron

Mark S. Adams and Menno T. Kosters
5/9/2021

Abstract
Volume solutions to the Snub Dodecahedron, Thirteenth Archimedean Solid, are made by HSM Coxeter, EW Weisstein, MT Kosters, HC Rajpoot, and MS Adams. There are more than one closed form solution, each yielding different expressions. The Adams solution may possess discovery of an origin to one third powers of the golden ratio.
Keywords: Snub dodecahedron · Archimedean solids · Golden ratio
Mathematics Subject Classification (2000): 51M15, (51M20)

Contents
Closed form volume of the Snub Dodecahedron
Harold S. M. Coxeter
Eric W. Weisstein
Menno T. Kosters
Harish C. Rajpoot
Numerical
Origin to one third powers of the golden ratio
References

## Closed form volume of the Snub Dodecahedron

#### There are at least three closed form expressions for the volume of a snub dodecahedron of unit edge length:

$\text{Coxeter expression =}\; \frac{12\color{Crimson}{\eta^2}\color{Black}{(3}\color{DarkGoldenrod}{\varphi} \color{Black}{+1)-}\color{Crimson}{\eta} \color{Black}{(36}\color{DarkGoldenrod}{\varphi} \color{Black}{+7)-(53}\color{DarkGoldenrod}{\varphi} \color{Black}{+6)}}{6\sqrt{(3-\color{Crimson}{\eta^2}\color{black}{)^3}}} \\$

$\text{Kosters expression =}\; \frac{(3\color{DarkGoldenrod}{\varphi}+1)\color{Teal}{\xi^2}+(3\color{DarkGoldenrod}{\varphi}+1)\color{Teal}{\xi}- \color{DarkGoldenrod}{\varphi}/6-2}{\sqrt{3\color{Teal}{\xi^2}-\color{DarkGoldenrod}{\varphi^2}}}$

$$\text{Adams expression =}\; \frac{10\color{DarkGoldenrod}{\varphi}}{3}\sqrt{ \color{DarkGoldenrod}{\varphi^2} \color{Black}{+3}\color{Crimson}{\eta} \color{Black}{(}\color{DarkGoldenrod}{\varphi} \color{Black}{+}\color{Crimson}{\eta} \color{Black}{)}} \,+\,\frac{\color{DarkGoldenrod}{\varphi^2}}{2}\sqrt{ 5 + 5\sqrt{5}\color{DarkGoldenrod}{\varphi}\color{Crimson}{\eta} \color{Black}{(}\color{DarkGoldenrod}{\varphi} \color{Black}{+}\color{Crimson}{\eta} \color{Black}{)}} \\$$

$\text{where}\; \color{Crimson}{eta}\; \text{ is defined as} \quad \color{Crimson}{\eta\;\equiv\;\sqrt[3]{ \frac{ \color{DarkGoldenrod}{\varphi} \color{Crimson}{}}{2} + \frac{1}{2} \sqrt{ \color{DarkGoldenrod}{\varphi} \color{Crimson}{-}\frac{5}{27}}}\;+\;} \color{Crimson}{ \sqrt[3]{ \frac{ \color{DarkGoldenrod}{\varphi} \color{Crimson}{}}{2} - \frac{1}{2} \sqrt{ \color{DarkGoldenrod}{\varphi} \color{Crimson}{-}\frac{5}{27}}}} \quad \text{ and } \color{Teal}{xi} \quad \color{Teal}{\xi\,}=\, \frac{ \color{DarkGoldenrod}{\varphi}}{ \color{Crimson}{\eta} }$

## Harold S. M. Coxeter

One of the world's finest and most eloquent geometers Harold Scott MacDonald Coxeter in his "Uniform Polyhedra."1. Uniform Polyhedra, Section 10, The Snub Polyhedra: We construct $\vert$ p q r by regarding the spherical triangles (p q r) as being alternately white and black (see § 3, especially figure 6). The three white triangles that surround a black one contain corresponding points forming an equilateral triangle which we may called a ‘snub face' of $\vert$ pqr. One of these three white triangles is derived from another, sharing with it the vertex P (say), by a rotation through $\frac{2\pi}{p}$ about P. If this rotation takes the chosen point $C\prime\prime\prime$ in the first triangle to $C\prime\prime$ in the second, we have an isosceles triangle $C\prime\prime\prime PC\prime\prime$ whose base $C\prime\prime\prime C\prime\prime$ (opposite to the angle $\frac{2\pi}{p}$ at P) is one side of the snub face. Solving this isosceles triangle, we find. $\sin PC\prime\prime\sin\frac{\pi}{p}\;=\;\sin\frac{1}{2}C\prime\prime C\prime\prime\prime$ Besides the usual twelve pentagrams and sixty 'snub’ triangles, they have each forty more triangles, lying by pairs in twenty planes (the face-planes of an icosahedron).

$Let\;\color{Teal}{xi\;\xi}\;be\;the\;real\;root\;of\;the\;polynomial:\; X^3 + 2X^2 - \color{DarkGoldenrod}{\varphi^2} = 0$

## Eric W. Weisstein

MathWorld--A Wolfram Web Resource by Eric W. Weisstein shows in "Snub Dodecahedron."2 Coxeter's polynomial expression, volume is the real root of x:

$$187445810737515625 - 182124351550575000\,x^2 + 6152923794150000\,x^4 + 1030526618040000\,x^6 + \\ 162223191936000\,x^8 - 3195335070720\,x^{10} + 2176782336\,x^{12} = 0 \\$$

## Menno T. Kosters

The snub dodecahedron

In this SageMath notebook we will derive coordinates for the snub dodecahedron. We will start with a regular icosahedron, for which standard coordinates are well known. Then we will inscribe a snub dodecahedron in it. We will use the coordinates to obtain symbolic expressions and numeric values of various metric properties of the snub dodecahedron.

The icosahedron

The number $\varphi = (1 + \sqrt{5})/2$ is called the golden ratio. It satisfies $\varphi^2 = \varphi + 1$. It plays an important role in the decription of the regular icosahedron. Indeed, the 12 points $(0, \pm 1/\varphi, \pm 1)$, $(\pm 1, 0, \pm 1/\varphi)$ and $(\pm 1/\varphi, \pm 1, 0)$ are the vertices of a regular icosahedron with edge length $2/\varphi$, centered at the origin. We want to do some symbolic computations with the number $\varphi$:

In [1]:
%display latex

In [2]:
var('phi')
k.<phi> = NumberField(phi^2 - phi - 1)


Let $P$, $Q$, and $R$ be three vertices, forming one of the triangular faces of the icosahedron:

In [3]:
P = vector([0, -1/phi, 1])
Q = vector([1, 0, 1/phi])
R = vector([0, 1/phi, 1])


Let $R_3$ be the $120^\circ$ rotation sending $P$ to $Q$, $Q$ to $R$, and $R$ back to $P$:

In [4]:
R3 = matrix([Q, R, P]).transpose()*matrix([P, Q, R]).inverse().transpose(); R3

Out[4]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} -\frac{1}{2} \phi + \frac{1}{2} & -\frac{1}{2} \phi & \frac{1}{2} \\ \frac{1}{2} \phi & -\frac{1}{2} & -\frac{1}{2} \phi + \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \phi - \frac{1}{2} & \frac{1}{2} \phi \end{array}\right)$

Now let $A = (x, y, z)$ be a point inside the triangle $PQR$. Note that $x = \varphi^2(1-z)$, because the $x$- and $z$- coordinates of $P$, $Q$ and $R$ satisfy this relation and $A$ lies in the plane through these three points.

In [5]:
r.<y, z> = PolynomialRing(k)
x = phi**2*(1 - z)
A = vector([x, y, z])


Let $B$ be obtained from $A$ by applying $R_3$, and let $C$ be the point $(-x, -y, z)$.

In [6]:
B = R3*A
C = vector([-x, -y, z])


We want to choose $y$ and $z$ so that $ABC$ is an equilateral triangle. Then $A$, $B$ and $C$ will be vertices of a snub dodecahedron centered at the origin. The other vertices can then be found by applying the 60 rotational symmetries of the icosahedron. For this, the dot products $(A-B,A-B)$, $(A-C,A-C)$ and $(B-C, B-C)$ need to be equal. Because $(A,A) = (B,B) = (C,C)$ this can be simplified to $(A,B) = (B,C) = (A,C)$, or $(A-C,B)=0$ and $(A,B-C)=0$:

In [7]:
eq1 = (A - C).dot_product(B); eq1

Out[7]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}-y^{2} + \left(-\phi + 1\right) y z + \left(-3 \phi - 2\right) z^{2} + \left(5 \phi + 3\right) z - 2 \phi - 1$

(The right-hand side of our equations will always be $0$, so we omit them.)

In [8]:
eq2 = 2*A.dot_product(B - C); eq2

Out[8]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}y^{2} + \left(3 \phi - 1\right) z^{2} + \left(-6 \phi - 4\right) z + 4 \phi + 3$

These two equations are simple to solve. Adding them removes the $y^2$ part:

In [9]:
eq = eq1 + eq2; eq

Out[9]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(-\phi + 1\right) y z - 3 z^{2} + \left(-\phi - 1\right) z + 2 \phi + 2$

This is linear in $y$, so we now we can express $y$ in $z$:

In [10]:
y = eq(0, z)/(eq(0, z) - eq(1, z)); y

Out[10]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{-3 z^{2} + \left(-\phi - 1\right) z + 2 \phi + 2}{\left(\phi - 1\right) z}$

Substituting in eq2, and multiplying by $z^2$ to get rid of the denominator:

In [11]:
eq2 = z**2*eq2(y, z); factor(eq2)

Out[11]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(12 \phi + 8\right) \cdot (z - 1) \cdot (z^{3} + 2 z^{2} - \phi - 1)$

We now have an equation for $z$ alone. $z = 1$ is clearly not the zero we want. It corresponds to the degenerate solution where $A=P$, $B=Q$ and $C=R$. So, finally, we find that $z$ is a zero of the following polynomial:

In [12]:
factor(eq2)[1][0]

Out[12]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}z^{3} + 2 z^{2} - \phi - 1$

Let $\xi$ be a zero of this polynomial. For the time being we do not specify which of the three zeroes.

Metric properties

We now want to use the coordinates we found to get symbolic expressions for lengths, volumes, etcetera. To this end we need to be able to do symbolic computations with $\xi$:

In [13]:
l.<xi> = k[]
k.<xi> = k.extension(xi^3 + 2*xi^2 - phi^2 )


We express the coordinates of $A$ in $\varphi$ and $\xi$:

In [14]:
x, y, z = phi**2*(1 - xi), y(0, xi), xi
A = vector([x, y, z]); A

Out[14]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(\left(-\phi - 1\right) \xi + \phi + 1,\,2 \phi \xi^{2} + \phi \xi - 2 \phi - 1,\,\xi\right)$

Now we can compute the squared circumradius of our snub dodecahedron:

In [15]:
A.dot_product(A)

Out[15]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}4 \xi^{2} - \phi - 1$

The edge length, squared:

In [16]:
B = R3*A
(A - B).dot_product(A - B)

Out[16]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}12 \xi^{2} - 4 \phi - 4$

As a volume, the snub dodecahedron can be seen as consisting of $80$ triangular pyramids and $12$ pentagonal pyramids, with apex at the origin $O=(0, 0, 0)$. The volume $V_3$ of one triangular pyramid:

In [17]:
D = R3*B
V3 = 1/6*det(matrix([A, B, D])); V3

Out[17]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\phi \xi^{2} - \frac{2}{3} \phi - \frac{1}{3}$

To handle the pentagonal pyramids we need an additional vertex of the icosahedron we started with: $S=(1/\varphi,1,0)$. $R_5$ will be the $72^\circ$ rotation around the axis $OR$ taking $P$ to $Q$ and $Q$ to $S$ (and fixing $R$):

In [18]:
S = vector([1/phi, 1, 0])
R5 = matrix([Q, S, R]).transpose()*matrix([P, Q, R]).inverse().transpose(); R5

Out[18]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(\begin{array}{rrr} \frac{1}{2} \phi - \frac{1}{2} & -\frac{1}{2} \phi & \frac{1}{2} \\ \frac{1}{2} \phi & \frac{1}{2} & \frac{1}{2} \phi - \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \phi - \frac{1}{2} & \frac{1}{2} \phi \end{array}\right)$

The volume $V_5$ of one pentagonal pyramid:

In [19]:
E = R5*A
G = R5*E
V5 = 1/6*(2 + phi)*det(matrix([A, E, G])); V5

Out[19]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(\frac{1}{3} \phi - \frac{8}{3}\right) \xi^{2} + \left(\frac{28}{3} \phi + \frac{16}{3}\right) \xi - 8 \phi - \frac{13}{3}$

The total volume is equal to $80 V_3 + 12 V_5$

In [20]:
V = 80*V3 + 12*V5; V

Out[20]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(84 \phi - 32\right) \xi^{2} + \left(112 \phi + 64\right) \xi - \frac{448}{3} \phi - \frac{236}{3}$

If we want the volume of a snub dodecahedron with unit edge length we have to divide by the cube of the edge length. We first divide by the edge length squared:

In [21]:
V/(A - B).dot_product(A - B)

Out[21]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(6 \phi + 2\right) \xi^{2} + \left(6 \phi + 2\right) \xi - \frac{1}{3} \phi - 4$

This, together with the expression for the edge length, gives the following fairly simple formula for the volume of a snub dodecahedron of unit edge length:

$$$\frac{(3\varphi+1)\xi^2+(3\varphi+1)\xi-\varphi/6-2}{\sqrt{3\xi^2-\varphi^2}}.$$$

The snub dodecahedron has two inscribed spheres: one touching the triangular faces, and one, slightly smaller, touching the pentagonal faces. We can compute the squares of their radii, by first computing the centers $C_3$ and $C_5$ of a triangular and pentagonal face:

In [22]:
C3 = 1/3*sum([R3**i*A for i in range(3)])
C3.dot_product(C3)

Out[22]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\frac{1}{3} \phi + \frac{1}{3}$

This is not surprising: it is the squared inradius of the icosahedron we started with.

In [23]:
C5 = 1/5*sum([R5**i*A for i in range(5)])
C5.dot_product(C5)

Out[23]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(-\frac{12}{5} \phi - \frac{4}{5}\right) \xi^{2} + \frac{11}{5} \phi + \frac{7}{5}$

We can also compute the (cosine of) the angle $\phi_{33}$ between adjacent triangular faces. Let $C_3^\prime$ be the center of the triangle $ABC$. Then the cosine of the angle between the vectors $C_3$ and $C_3^\prime$ equals $(C_3,C_3^\prime)/(|C_3||C_3^\prime|) = (C_3,C_3^\prime)/(C_3,C_3)$, because $|C_3|=|C_3^\prime| = \sqrt{(C_3, C_3)}$. So for the angle $\phi_{33}$ between the faces $ABD$ and $ABC$ (with normals $C_3$ and $C_3^\prime$) we have:

In [24]:
C = vector((-x, -y, z))
C31 = 1/3*(A + B + C)
cosphi33 = -C31.dot_product(C3)/C3.dot_product(C3); cosphi33


Out[24]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{2}{3} \xi - \frac{1}{3}$

Similarly, (the square of the cosine of) the angle $\phi_{35 }$ between a triangular and an adjacent pentagonal face:

In [25]:
C32 = 1/3*(A + D + E)
v1 = C32.dot_product(C5)
v2 = C32.dot_product(C32)
v3 = C5.dot_product(C5)
cosphi35square = v1**2/(v2*v3); cosphi35square

Out[25]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(-\frac{4}{15} \phi - \frac{8}{15}\right) \xi^{2} + \left(-\frac{4}{15} \phi - \frac{8}{15}\right) \xi + \frac{4}{5} \phi + \frac{19}{15}$

Numerical values

We now do all the preceding computations once again, this time not symbolically but numerically. From now on, $\xi$ will be the real number satisfying $\xi^3+2\xi^2=\varphi^2$. We start by getting numerical values for $\varphi$ and $\xi$:

In [26]:
phi = .5 + .5*5.**.5; phi

Out[26]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}1.61803398874989$
In [27]:
from numpy import array, roots

def is_real(x):
eps = .0001
return (abs(x.imag) < eps)

def first_real_root(p):
for x in roots(p):
if is_real(x):
return x.real
raise Exception('No real element')

p = array([1., 2., 0., -phi**2])
xi = first_real_root(p); xi

Out[27]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}0.9431512592438817$
In [28]:
P = vector([0., -1./phi, 1.])
Q = vector([1., 0., 1./phi])
R = vector([0, 1./phi, 1.])
R3 = matrix([Q, R, P]).transpose()*matrix([P, Q, R]).inverse().transpose()
x, y, z = (-phi - 1)*xi + phi + 1, 2*phi*xi^2 + phi*xi - 2*phi - 1, xi
A = vector([x, y, z])
B = R3*A


Edge length:

In [29]:
l = (A - B).dot_product(A - B)**.5; l

Out[29]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}0.4497506184100908$

For the numerical results it seems more natural to have an edge length equal to $1$. For the remainder of this section we will therefore divide lengths by $l$, areas by $l^2$ and volumes by $l^3$.

In [30]:
A.dot_product(A)**.5/l

Out[30]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}2.155837375115645$

The volume of one triangular pyramid:

In [31]:
D = R3*B
V3 = 1./6.*det(matrix([A, B, D]))/l^3; V3

Out[31]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}0.299802068545933$

The volume of one pentagonal pyramid:

In [32]:
S = vector([1./phi, 1., 0.])
R5 = matrix([Q, S, R]).transpose()*matrix([P, Q, R]).inverse().transpose()
E = R5*A
G = R5*E
V5 = 1./6.*(2. + phi)*det(matrix([A, E, G]))/l^3; V5

Out[32]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}1.13604037325491$

The total volume:

In [33]:
80.*V3 + 12.*V5

Out[33]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}37.6166499627336$

In [34]:
C3 = 1./3.*sum([R3**i*A for i in range(3)])
C3.dot_product(C3)**.5/l

Out[34]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}2.07708965974321$

In [35]:
C5 = 1./5.*sum([R5**i*A for i in range(5)])
C5.dot_product(C5)**.5/l

Out[35]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}1.98091594728185$

The triangle-triangle dihedral angle (degrees):

In [36]:
C = vector((-x, -y, z))
C31 = 1/3*(A + B + C)
cosphi33 = -C31.dot_product(C3)/C3.dot_product(C3)

from numpy import arccos, pi
arccos(cosphi33)*180./pi

Out[36]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}164.1753660560339$

The triangle-pentagon dihedral angle (degrees:)

In [37]:
C32 = 1/3*(A + D + E)
v1 = C32.dot_product(C5)
v2 = C32.dot_product(C32)
v3 = C5.dot_product(C5)
cosphi35 = -v1/(v2*v3)**.5; arccos(cosphi35)*180./pi

Out[37]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}152.929920275835$

We can show that the Adams and Kosters expressions are equal.

In [38]:
r.<phi> = QQ[]

In [39]:
k.<phi> = NumberField(phi^2 - phi - 1)

In [40]:
r.<xi> = QQ[]

In [41]:
k.<xi> = NumberField(xi^3 + 2*xi^2 - phi^2)


From Adams solution, $\eta$:

In [42]:
eta = phi/xi


It satisfies $\eta^3 -2\eta - \varphi = 0$:

In [43]:
eta^3 - 2*eta - phi

Out[43]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}0$

Now write Adams expression as $$$\frac{10\varphi}{3}\sqrt{A} + \frac{\varphi^2}{2}\sqrt{B}$$$ (write $\sqrt{5}= 2\varphi-1$):

In [44]:
A = (phi^2 + 3*eta*(phi + eta)); A

Out[44]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}3 \xi^{2} + 9 \xi + \phi + 7$
In [45]:
B = (5 + 5*(2*phi -1)*phi*eta*(phi + eta)); B

Out[45]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(5 \phi + 10\right) \xi^{2} + \left(15 \phi + 30\right) \xi + 10 \phi + 25$

Now put $A^\prime = (3\xi^2 - \phi^2)A$ and $B^\prime = (3\xi^2 - \phi^2)B$:

In [46]:
A1 = (3*xi^2 - phi^2)*A; A1

Out[46]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}1$
In [47]:
B1 = (3*xi^2 - phi^2)*B; B1

Out[47]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}-20 \phi \xi^{2} + 15 \phi + 10$

Now $\sqrt{B^\prime}=(4\varphi - 2)\xi^2 + (4\varphi - 2)\xi - 3\varphi - 1$:

In [48]:
sqrtB1 = (4*phi - 2)*xi^2 + (4*phi - 2)*xi - 3*phi - 1; sqrtB1^2 - B1

Out[48]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}0$

Now Adams expression can be written as

$$$(\frac{10\varphi}{3}\sqrt{A^\prime} + \frac{\varphi^2}{2}\sqrt{B^\prime})/\sqrt{3\xi^2-\varphi^2}$$$

Computing the numerator of this expression shows that Adams expression is indeed equal to Kosters expression:

In [49]:
10/3*phi + phi^2/2*sqrtB1

Out[49]:
$\renewcommand{\Bold}[1]{\mathbf{#1}}\left(3 \phi + 1\right) \xi^{2} + \left(3 \phi + 1\right) \xi - \frac{1}{6} \phi - 2$

\begin{align*} laevo\;and\;dextro \end{align*}

## Harish C. Rajpoot

Harish Chandra Rajpoot published “Optimum Solution of Snub Dodecahedron"3 HCR's Theory of Polygon & Newton-Raphson Method is used to calculate the volume of the Snub Dodecahedron. After only 7 iterations, the calculated volume matches the closed-form solutions to 50 digits of accuracy. \begin{align*} \\& Iterate\,to\,find\,Circumradius\quad C_{0} = 2.3\quad C_{n+1}=\frac{f(C_{n})}{f\prime (C_{n})} \\& f(x)= 256(3-\sqrt{5})x^8 - 128(13-2\sqrt{5})x^6 + 32(35-3\sqrt{5})x^4 - 16(19-\sqrt{5})x^2 +(29-\sqrt{5}) \\& f\prime(x)= 2048(3-\sqrt{5})x^7 - 768(13-2\sqrt{5})x^5 + 128(35-3\sqrt{5})x^3 - 32(19-\sqrt{5})x \\& Volume = \left( \frac{20\sqrt{3C^2-1}}{3} + \sqrt{ \frac{10(5+2\sqrt{5})C^2 - 5((7+3\sqrt{5})}{2}} \right) \end{align*}

## Numerical

Jupyter notebook Snub Dodecahedron Volume Calculated Six Ways4 is a supplement to this essay. Six Python classes, each calculating the volume of the Snub Dodecahedron in a different way, the five methods above plus a 3D Numerical method. 3D distances drive a numerical root finder. Two triangle objects are defined as adjacent triangles on a regular icosahedron. The algorithm is applied to one point on the plane of each triangle so that the distance of a side of an inscribed snub triangle is equal to the side of a non inscribed snub triangle.

Mark Shelby Adams published "Archimedean Platonic Solids"7. Whereas Kosters' solution inscribes a Snub Dodecahedron on to a regular icosahedron with an edge length of $\frac{2}{\color{DarkGoldenrod}{\varphi}}$, Adams' solution inscribes a Snub Dodecahedron on to a regular icosahedron with an edge length of $1$.

$\text{Equations 1 and 2 are second order equations of} \;\color{green}{D}.$

$\quad \color{red}{ \overline{\color{red}{g}\color{green}{d}}^2} \,-\, \color{green}{\frac{D^2}{4}} \,=\, \color{red}{ \overline{\color{red}{g}\color{green}{f}}^2} \,+\, \color{blue}{ ( \overline{\color{blue}{a}\color{red}{b}}} \,-\, \color{blue}{ \overline{\color{blue}{a}\color{green}{f}}} \,-\, \color{green}{ \overline{\color{green}{e}\color{red}{b}} )^2 } \,+\, \color{green}{ \overline{\color{green}{e}\color{green}{d}}^2} \,-\, \color{green}{\frac{D^2}{4}} \,=\,0$

ParseError: KaTeX parse error: Invalid color: 'blue ' at position 376: …overline{\color{̲b̲l̲u̲e̲ ̲}̲{a}\color{green…

$\quad \color{red}{ \frac{D^2}{3} \,+\, \frac{1}{12} \,-\, \frac{D}{3} \, \cos\alpha \,+\, } \color{red}{ \sin^2\beta \, \frac{1\,-\,2\,D\cos\alpha}{2\sqrt{3}}} \color{red}{ \left[ \frac{2\,D\cos\alpha}{\sqrt{3}} \,-\, \frac{1}{\sqrt{3}} \,+\, \frac{1\,-\,2\,D\cos\alpha}{2\sqrt{3}} \right] } \,-\, \color{green}{\frac{D^2}{4}} \,=\,0$

$\color{red}{ D^2 \,-\, 4D\,\cos\alpha \,+\, 1 \,-\, \sin^2\beta \,(1 \,-\, 2\,D\,\cos\alpha)^2 \,=\, 0 \quad(Eq.1)}$

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$\quad \color{blue}{ \frac{D^2}{12} \,+\, \frac{1}{12} \,-\, \frac{D}{12} \, (\cos\alpha \,+\, \sqrt{3}\sin\alpha) \,+\, } \color{blue}{ \sin^2\beta \, \frac{1\,-\,2\,D\cos\alpha}{2\sqrt{3}}} \color{blue}{ \left[ \frac{D(\cos\alpha \,+\, \sqrt{3}\sin\alpha)}{2\sqrt{3}} \,-\, \frac{1}{\sqrt{3}} \,+\,\frac{1\,-\,2\,D\cos\alpha}{2\sqrt{3}}\right]}\,-\, \color{green}{\frac{D^2}{4}} \,=\,0$

$\color{blue}{ -2D^2 \,-\, D\,(\cos\alpha \,+\, \sqrt{3}\sin\alpha) \,+\, 1 \,-\, \sin^2\beta \,(1 \,-\, 2\,D\,\cos\alpha) \,} \color{blue}{ \left[ 1 \,+\, D\,(\cos\alpha \,+\, \sqrt{3}\sin\alpha) \right] \,=\, 0 \quad(Eq.2)}$

$\text{Equations 3 and 4 are trigonometric operators defining gamma letters:} \;\color{red}{\gamma}\;\text{and}\;\color{blue}{\Gamma}.$

$\color{red}{\gamma \,\equiv\,\sqrt{3}\,\tan\alpha} \quad \color{blue}{\Gamma \,\equiv\,3\cos\alpha-\sqrt{3}\,\sin\alpha}$

$\color{blue}{\cos^2\alpha} \,+\, \color{red}{\sin^2\alpha} \,=\, 1 \,=\, \color{blue}{\cos^2\alpha \left(1 \,+\, \color{red}{\frac{\gamma^2}{3}}\right) } \quad or \quad \color{blue}{ \cos^2\alpha\,=\,\frac{1}{ 1 \,+\, \color{red}{ \frac{\gamma^2}{3}} }}$

$\color{blue}{ \Gamma\,\cos\alpha \,=\, (\cos\alpha -\color{red}{\sqrt{3}\sin\alpha}) \cos\alpha }\,=\, \color{blue}{ (3 \,-\, \color{red}{\gamma}) \, \cos^2\alpha \,=\, \frac{3\,-\,\color{red}{\gamma}}{1 \,+\, \color{red}{ \frac{\gamma^2}{3}} }}$

$\color{red}{ 3 \left( 1 \,+\, \frac{\gamma^2}{3} \right)} \color{red}{ \left[ \color{blue}{\Gamma\,\cos\alpha\,-\,} \color{blue}{ \frac{3\,-\,\color{red}{\gamma}}{1 \,+\, \color{red}{ \frac{\gamma^2}{3}} }} \right]} \,=\, \color{red}{ \overbrace{\Gamma\,\cos\alpha}^{a}\;\gamma^2 \,+\, \overbrace{3}^{b}\;\gamma + \overbrace{3( \Gamma\,\cos\alpha\,-\,3)}^{c} }\,=\,0$

$Positive\,Root:\;\color{red}{ \gamma \,=\, \frac{-b+\sqrt{b^2-4\,a\,c}}{2\,a} } = \color{red}{ \frac{ -3 + \sqrt{ 9-12\,\Gamma\cos\alpha(\Gamma\cos\alpha -3)}}{2\Gamma\,\cos\alpha} \quad(Eq.3) }$

$\color{blue}{ \Gamma^2 \,=\, \Gamma\,\cos\alpha \,(3 - \gamma) }\,=\, \color{blue}{ 3\Gamma\,\cos\alpha - \frac{1}{2}\left[ -3 + \sqrt{ 9-12\,\Gamma\cos\alpha(\Gamma\cos\alpha -3)} \right] \quad (Eq. 4)}$

$\text{Combine Equations 1 and 2 with variable y to solve between D and }\alpha:$

$\color{Goldenrod}{ Golden\,Ratio\,phi:\;\varphi\,\equiv\, \frac{1+\sqrt{5}}{2} } \quad\quad\color{red}{ Icosa\,symmetry:\;\sin^2\beta\,=\,\frac{1}{3\varphi^2}}$

$\color{blue}{Combine:}\quad\color{blue}{ 3\varphi^2(Eq.2)} \;+\; \color{red}{3\varphi^2(Eq.1) \color{green}{y}} \,=\,0$

$\color{green}{\overbrace{\left[3\varphi^2(y-2)+2\left((1-2y)\cos\alpha-\sqrt{3}\sin\alpha\right)\cos\alpha\right]}^{i}\,D^2 \,-\,} \color{green}{\overbrace{\left((4y+1)\cos\alpha+\sqrt{3}\sin\alpha\right)\varphi^4}^{j}\,D\;+} \color{green}{\overbrace{(y+1)\varphi^4}^{k}}\,=\,0$

$\color{green}{Define}\,\color{red}{mu}\,and\,\color{blue}{lambda:}\; \color{green}{j^2-4ik \,\equiv\,(} \color{red}{\mu}\color{green}{\cos\alpha\,+\,} \color{blue}{\lambda}\color{green}{\sqrt{3}\sin\alpha)^2\;=\;} \color{blue}{\mu^2}\color{green}{\cos^2\alpha\,+\,} \color{red}{(\mu\lambda)} \color{green}{2\sqrt{3}\cos\alpha\sin\alpha +} \color{blue}{\lambda^2}\color{green}{3\sin^2\alpha}$

$\color{blue}{ \underbrace{ \left[ \varphi^2(4y+1)^2-4\varphi^2(y+1)\left(3\varphi^2(y-2)+2(1-2y)\right)\right]}_{\mu^2} } \color{green}{\cos^2\alpha \,+\,} \color{red}{ \underbrace{ \left[ \varphi^8(4y+1)+4\varphi^4(y+1) \right]}_{(\mu\lambda)} } \color{green}{2\sqrt{3}\cos\alpha\sin\alpha \,+\,} \color{blue}{ \underbrace{ \left[ \varphi^8-4\varphi^6(y+1)(y-2) \right]}_{\lambda^2} } \color{green}{3\sin^2\alpha}$

$\color{Goldenrod}{Sum\,components} \quad\color{red}{(\mu\lambda)^2} \color{blue}{-\mu^2\lambda^2}\,=\,0 \quad\color{Goldenrod}{using\;identities\quad \varphi^{n+1}-\varphi^{n-1}=\varphi^{n}\quad and \quad \varphi^{n+2}+\varphi^{n-2}=3\varphi^{n}}$

$\begin{array}{l} \\ \color{Goldenrod}{\varphi^{16}} &\color{Goldenrod}{|}&&\color{Goldenrod}{|}&&\color{Goldenrod}{|}\; \color{red}{16}-\color{blue}{16} &&\color{Goldenrod}{|}\; \color{red}{8}-\color{blue}{8} &&\color{Goldenrod}{|}\; \color{red}{1}-\color{blue}{1} \\ \color{Goldenrod}{\varphi^{14}} &\color{Goldenrod}{|}\; \color{blue}{64} &&\color{Goldenrod}{|}\; \color{blue}{-32}\quad &&\color{Goldenrod}{|}\; \color{blue}{-144} &&\color{Goldenrod}{|}\; \color{blue}{-80} & \color{green}{144} &\color{Goldenrod}{|}\; \color{blue}{-32} & \color{green}{-144} \\ \color{Goldenrod}{\varphi^{12}} &\color{Goldenrod}{|}\; \color{blue}{-48}\quad & \color{green}{144}\; &\color{Goldenrod}{|}\; \color{blue}{96} & \color{green}{0}\; &\color{Goldenrod}{|}\; \color{red}{32} \color{blue}{+128}\quad & \color{green}{-288} &\color{Goldenrod}{|}\; \color{red}{40} \color{blue}{-200} &&\color{Goldenrod}{|}\; \color{red}{8} \color{blue}{-184}\quad & \color{green}{\;144} \\ \color{Goldenrod}{\varphi^{10}} &\color{Goldenrod}{|}\; \color{blue}{64} &&\color{Goldenrod}{|}\; \color{blue}{-32} &&\color{Goldenrod}{|}\; \color{blue}{-192} &&\color{Goldenrod}{|}\; \color{blue}{-32} &&\color{Goldenrod}{|}\; \color{blue}{64} \\ \color{Goldenrod}{\varphi^{8}} &\color{Goldenrod}{|}&&\color{Goldenrod}{|}&&\color{Goldenrod}{|}\; \color{red}{16} &&\color{Goldenrod}{|}\; \color{red}{32} &&\color{Goldenrod}{|}\; \color{red}{16} \\ \color{Goldenrod}{\div144\varphi^{12}} &\color{Goldenrod}{|}& \color{green}{y^4} &\color{Goldenrod}{|}& &\color{Goldenrod}{|}& \color{green}{-2y^2\;} &\color{Goldenrod}{|}& \color{green}{-\varphi^2 y\;} &\color{Goldenrod}{|}& \color{green}{\;-\varphi} \end{array}$

$\begin{array}{lc} \color{red}{First\,root\,of\,y:\;-1} &&&\color{blue}{y^3} &\color{blue}{\overbrace{-y^2}^{p=-1}} &\color{blue}{\overbrace{-y}^{q=-1}} &\color{blue}{\overbrace{-\varphi}^{r=-\varphi}} \\ \color{blue}{a=\frac{-p^2}{3} \color{Goldenrod}{\large{\subset}}+q\;=\; \frac{-1}{3}- \color{Goldenrod}{\large{\supset}} 1\;=\;\frac{-4}{3} } & \color{red}{(y+1)} & \color{green}{\overline{|\;y^4}} &&\quad\color{green}{-2y^2} &\quad\color{green}{-\varphi^2y} &\color{green}{-\varphi} \\ \color{blue}{b=\frac{2p^2}{27}-\frac{pq}{3}+r} && \color{green}{y^4} & \color{green}{+y^3} \\ \color{blue}{b=-\frac{2}{27}-\frac{1}{3}-\varphi \;=\; \frac{-49-27\sqrt{5}}{54}} && \color{green}{0} & \color{green}{-y^3} & \color{green}{-y^2} \\ \color{blue}{Second\,root\,of\,y:} &&& \color{green}{0} & \color{green}{-y^2} & \color{green}{-y} \\ \color{blue}{\frac{-p}{3} -\sqrt[3]{ \frac{b}{2}+\sqrt{ \frac{b^2}{4}+\frac{a^3}{27}}} -\sqrt[3]{ \frac{b}{2}-\sqrt{ \frac{b^2}{4}+\frac{a^3}{27}}} \;=\; \color{Crimson}{\eta^2}-\frac{1}{3}} &&&& \color{green}{0} & \color{green}{-\varphi y} & \color{green}{-\varphi} \\ \color{Crimson}{ eta:\,\eta\,\equiv\, \sqrt[3]{ \frac{ \color{Goldenrod}{\varphi} \color{Crimson}{}}{2} + \frac{1}{2} \sqrt{ \color{Goldenrod}{\varphi} \color{Crimson}{-}\frac{5}{27}}}\;+\;} \color{Crimson}{ \sqrt[3]{ \frac{ \color{Goldenrod}{\varphi} \color{Crimson}{}}{2} - \frac{1}{2} \sqrt{ \color{Goldenrod}{\varphi} \color{Crimson}{-}\frac{5}{27}}}} &&&&& \color{green}{0} & \color{green}{0} \end{array}$

$\color{Goldenrod}{\textbf{First Golden Circle}:\;If\;} \color{blue}{a}\; \color{Goldenrod}{is\,taken\,as\;} \color{blue}{\frac{2}{3}} \color{Goldenrod}{,\;then\;} \color{blue}{y\;} \color{Goldenrod}{becomes\,an\;\textbf{Origin}\,to\,one\,third\,powers\,of\,the\,\textbf{Golden Ratio}.}$

$\color{red}{From\,the\,first\,root\,of\,y:} \frac{ \color{blue}{(Eq.2)} -\color{red}{(Eq.1)}}{\color{green}{D}}\;=\;0$

$\color{green}{ \left[ 2(3\cos\alpha-\sqrt{3}\sin\alpha)\cos\alpha-9\varphi^2\right] D \;+\; \varphi^4(3\cos\alpha-\sqrt{3}\sin\alpha)}\;=\; \color{green}{\left[ 2\Gamma\cos\alpha-9\varphi^2\right] D \;+\; \varphi^4\Gamma}\;=\;0$

$\color{green}{D\;=\;\frac{ \varphi^4\Gamma}{9\varphi^2 - 2\Gamma\cos\alpha} \quad(Eq.5)}$

$\color{red}{\frac{ \Gamma(9\varphi^2 - 2\Gamma\cos\alpha)}{ 3\varphi^2D} (Eq.1)} \;=\;0 \quad\color{green}{(substitute\,D\,with\,Eq.5)}$

$\color{red}{ \frac{ \Gamma(9\varphi^2 - 2\Gamma\cos\alpha)}{ 3\varphi^2}} \color{red}{\left[ (3\varphi^2-4\cos^2\alpha) \color{green}{ \left( \frac{ \varphi^4\Gamma}{9\varphi^2-2\Gamma\cos\alpha} \right)} -4\varphi^4\cos\alpha+\varphi^4 \color{green}{ \left( \frac{9\varphi^2-2\Gamma\cos\alpha}{ \varphi^4\Gamma} \right)} \right]} \;=\;0$

$\color{red}{4(\Gamma\cos\alpha)^2 - 36\varphi^2\Gamma\cos\alpha + 27\varphi^2 + \varphi^4\Gamma^2} \;=\;0 \quad\color{blue}{(substitute\,\Gamma^2\,with\,Eq.4)}$

$\color{red}{4(\Gamma\cos\alpha)^2 + 3\varphi^2(\color{blue}{\varphi^4}-12)\Gamma\cos\alpha + \frac{3}{2}\varphi^2(\color{blue}{\varphi^2}+18)}\;=\; \color{blue}{ \frac{\varphi^4}{2}\sqrt{9-12\Gamma\cos\alpha(\Gamma\cos\alpha - 3)}}$

$\color{green}{\text{Square both sides and subtract}}$

$\color{green}{16(\Gamma\cos\alpha)^4 + 24\varphi^2(\varphi^2-12)(\Gamma\cos\alpha)^3 + 36\varphi^2(21\varphi^2 + 11)(\Gamma\cos\alpha)^2 + 54\varphi^4(\varphi^2-36)\Gamma\cos\alpha + 81\varphi^4(\varphi^2+9)}\;=\;0$

$\color{green}{ Define\;x:\quad x\equiv\frac{2}{3}\Gamma\cos\alpha \quad \text{ and divide by 81:}}$

$\color{green}{x^4 + \varphi^2(\varphi^2-12)x^3 + \varphi^2(21\varphi^2+11)x^2 + \varphi^4(\varphi^2-36)x + \varphi^4(\varphi^2+9)}\;=\;0$

$\begin{array}{lc} &&\color{blue}{x^3} &\color{blue}{\overbrace{-9\varphi^2}^{p=-9\varphi^2}x^2} &\color{blue}{\overbrace{+\varphi^2(21\varphi^2+2)}^{q=\varphi^2(21\varphi^2+2)}x} &\color{blue}{\overbrace{-\varphi^4(\varphi^2+9)}^{r=-\varphi^4(\varphi^2+9)}} \\ \color{red}{(x-1)} &\quad \color{green}{\overline{|\;x^4}} \quad&\quad\color{green}{+\varphi^2(\varphi^2-12)x^3} \quad&\quad\color{green}{+\varphi^2(21\varphi^2+11)x^2} \quad&\color{green}{+\varphi^4(\varphi^2-36)x} \quad&\color{green}{+\varphi^4(\varphi^2+9)} \\ & \color{green}{x^4} & \color{green}{-x^3} \\ & \color{green}{0} & \color{green}{-9\varphi^2x^3} & \color{green}{+9\varphi^2x^2} \\ && \color{green}{0} & \color{green}{\varphi^2(21\varphi^2+2)x^2} & \color{green}{-\varphi^2(21\varphi^2+2)x} \\ &&& \color{green}{0} & \color{green}{-\varphi^4(\varphi^2+9)x} & \color{green}{\varphi^4(\varphi^2+9)} \\ &&&& \color{green}{0} & \color{green}{0} \\ \end{array}$

$\color{red}{First\,root\,of\,x:\;1} \\ \color{blue}{a=\frac{-p^2}{3}+q\;=\; -27\varphi^4+\varphi^2(21\varphi^2+2) \;=\; -2\varphi^6 } \\ \color{blue}{b=\frac{2p^2}{27}-\frac{pq}{3}+r} \\ \color{blue}{b=-54\varphi^6 + 3\varphi^4(21\varphi^2+2) - \varphi^4(\varphi^2+9)\;=\;\large{\color{Goldenrod}{\varphi^{10}}}} \\ \\ \color{blue}{Second\,root\,of\,x:} \\ \color{blue}{\frac{-p}{3} -\sqrt[3]{ \frac{b}{2}+\sqrt{ \frac{b^2}{4}+\frac{a^3}{27}}} -\sqrt[3]{ \frac{b}{2}-\sqrt{ \frac{b^2}{4}+\frac{a^3}{27}}} \;=\; 3\varphi^2-\varphi^3\eta}$