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https://archive.org/download/musing_math/Volume_Solutions_To_Snub_Dodecahedron.ipynb
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Kernel: SageMath 9.2

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
  Mark S. Adams
  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:



Coxeter expression =  12η2(3φ+1)η(36φ+7)(53φ+6)6(3η2)3 \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}}} \\


Kosters expression =  (3φ+1)ξ2+(3φ+1)ξφ/623ξ2φ2\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}}}


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


where  eta   is defined asη    φ2+12φ5273  +  φ212φ5273 and xiξ=φη\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 2πp\frac{2\pi}{p} about P. If this rotation takes the chosen point CC\prime\prime\prime in the first triangle to CC\prime\prime in the second, we have an isosceles triangle CPCC\prime\prime\prime PC\prime\prime whose base CCC\prime\prime\prime C\prime\prime (opposite to the angle 2πp\frac{2\pi}{p} at P) is one side of the snub face. Solving this isosceles triangle, we find. sinPCsinπp  =  sin12CC \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  xi  ξ  be  the  real  root  of  the  polynomial:  X3+2X2φ2=0Let\;\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 φ=(1+5)/2\varphi = (1 + \sqrt{5})/2 is called the golden ratio. It satisfies φ2=φ+1\varphi^2 = \varphi + 1. It plays an important role in the decription of the regular icosahedron. Indeed, the 12 points (0,±1/φ,±1)(0, \pm 1/\varphi, \pm 1), (±1,0,±1/φ)(\pm 1, 0, \pm 1/\varphi) and (±1/φ,±1,0)(\pm 1/\varphi, \pm 1, 0) are the vertices of a regular icosahedron with edge length 2/φ2/\varphi, centered at the origin. We want to do some symbolic computations with the number φ\varphi:

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

Let PP, QQ, and RR 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 R3R_3 be the 120120^\circ rotation sending PP to QQ, QQ to RR, and RR back to PP:

In [4]:
R3 = matrix([Q, R, P]).transpose()*matrix([P, Q, R]).inverse().transpose(); R3
Out[4]:
(12ϕ+1212ϕ1212ϕ1212ϕ+121212ϕ1212ϕ)\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)A = (x, y, z) be a point inside the triangle PQRPQR. Note that x=φ2(1z)x = \varphi^2(1-z), because the xx- and zz- coordinates of PP, QQ and RR satisfy this relation and AA 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 BB be obtained from AA by applying R3R_3, and let CC be the point (x,y,z)(-x, -y, z).

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

We want to choose yy and zz so that ABCABC is an equilateral triangle. Then AA, BB and CC 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 (AB,AB)(A-B,A-B), (AC,AC)(A-C,A-C) and (BC,BC)(B-C, B-C) need to be equal. Because (A,A)=(B,B)=(C,C)(A,A) = (B,B) = (C,C) this can be simplified to (A,B)=(B,C)=(A,C)(A,B) = (B,C) = (A,C), or (AC,B)=0(A-C,B)=0 and (A,BC)=0(A,B-C)=0:

In [7]:
eq1 = (A - C).dot_product(B); eq1
Out[7]:
y2+(ϕ+1)yz+(3ϕ2)z2+(5ϕ+3)z2ϕ1\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 00, so we omit them.)

In [8]:
eq2 = 2*A.dot_product(B - C); eq2
Out[8]:
y2+(3ϕ1)z2+(6ϕ4)z+4ϕ+3\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 y2y^2 part:

In [9]:
eq = eq1 + eq2; eq
Out[9]:
(ϕ+1)yz3z2+(ϕ1)z+2ϕ+2\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 yy, so we now we can express yy in zz:

In [10]:
y = eq(0, z)/(eq(0, z) - eq(1, z)); y
Out[10]:
3z2+(ϕ1)z+2ϕ+2(ϕ1)z\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 z2z^2 to get rid of the denominator:

In [11]:
eq2 = z**2*eq2(y, z); factor(eq2)
Out[11]:
(12ϕ+8)(z1)(z3+2z2ϕ1)\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 zz alone. z=1z = 1 is clearly not the zero we want. It corresponds to the degenerate solution where A=PA=P, B=QB=Q and C=RC=R. So, finally, we find that zz is a zero of the following polynomial:

In [12]:
factor(eq2)[1][0]
Out[12]:
z3+2z2ϕ1\renewcommand{\Bold}[1]{\mathbf{#1}}z^{3} + 2 z^{2} - \phi - 1

Let ξ\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 ξ\xi:

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

We express the coordinates of AA in φ\varphi and ξ\xi:

In [14]:
x, y, z = phi**2*(1 - xi), y(0, xi), xi A = vector([x, y, z]); A
Out[14]:
((ϕ1)ξ+ϕ+1,2ϕξ2+ϕξ2ϕ1,ξ)\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]:
4ξ2ϕ1\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]:
12ξ24ϕ4\renewcommand{\Bold}[1]{\mathbf{#1}}12 \xi^{2} - 4 \phi - 4

As a volume, the snub dodecahedron can be seen as consisting of 8080 triangular pyramids and 1212 pentagonal pyramids, with apex at the origin O=(0,0,0)O=(0, 0, 0). The volume V3V_3 of one triangular pyramid:

In [17]:
D = R3*B V3 = 1/6*det(matrix([A, B, D])); V3
Out[17]:
ϕξ223ϕ13\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/φ,1,0)S=(1/\varphi,1,0). R5R_5 will be the 7272^\circ rotation around the axis OROR taking PP to QQ and QQ to SS (and fixing RR):

In [18]:
S = vector([1/phi, 1, 0]) R5 = matrix([Q, S, R]).transpose()*matrix([P, Q, R]).inverse().transpose(); R5
Out[18]:
(12ϕ1212ϕ1212ϕ1212ϕ121212ϕ1212ϕ)\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 V5V_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]:
(13ϕ83)ξ2+(283ϕ+163)ξ8ϕ133\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 80V3+12V580 V_3 + 12 V_5

In [20]:
V = 80*V3 + 12*V5; V
Out[20]:
(84ϕ32)ξ2+(112ϕ+64)ξ4483ϕ2363\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]:
(6ϕ+2)ξ2+(6ϕ+2)ξ13ϕ4\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:

(3φ+1)ξ2+(3φ+1)ξφ/623ξ2φ2.\begin{equation} \frac{(3\varphi+1)\xi^2+(3\varphi+1)\xi-\varphi/6-2}{\sqrt{3\xi^2-\varphi^2}}. \end{equation}

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 C3C_3 and C5C_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]:
13ϕ+13\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]:
(125ϕ45)ξ2+115ϕ+75\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 ϕ33\phi_{33} between adjacent triangular faces. Let C3C_3^\prime be the center of the triangle ABCABC. Then the cosine of the angle between the vectors C3C_3 and C3C_3^\prime equals (C3,C3)/(C3C3)=(C3,C3)/(C3,C3)(C_3,C_3^\prime)/(|C_3||C_3^\prime|) = (C_3,C_3^\prime)/(C_3,C_3), because C3=C3=(C3,C3)|C_3|=|C_3^\prime| = \sqrt{(C_3, C_3)}. So for the angle ϕ33\phi_{33} between the faces ABDABD and ABCABC (with normals C3C_3 and C3C_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]:
23ξ13\renewcommand{\Bold}[1]{\mathbf{#1}}-\frac{2}{3} \xi - \frac{1}{3}

Similarly, (the square of the cosine of) the angle ϕ35\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]:
(415ϕ815)ξ2+(415ϕ815)ξ+45ϕ+1915\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, ξ\xi will be the real number satisfying ξ3+2ξ2=φ2\xi^3+2\xi^2=\varphi^2. We start by getting numerical values for φ\varphi and ξ\xi:

In [26]:
phi = .5 + .5*5.**.5; phi
Out[26]:
1.61803398874989\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]:
0.9431512592438817\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]:
0.4497506184100908\renewcommand{\Bold}[1]{\mathbf{#1}}0.4497506184100908

For the numerical results it seems more natural to have an edge length equal to 11. For the remainder of this section we will therefore divide lengths by ll, areas by l2l^2 and volumes by l3l^3.

The circumradius:

In [30]:
A.dot_product(A)**.5/l
Out[30]:
2.155837375115645\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]:
0.299802068545933\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]:
1.13604037325491\renewcommand{\Bold}[1]{\mathbf{#1}}1.13604037325491

The total volume:

In [33]:
80.*V3 + 12.*V5
Out[33]:
37.6166499627336\renewcommand{\Bold}[1]{\mathbf{#1}}37.6166499627336

The triangular faces inradius:

In [34]:
C3 = 1./3.*sum([R3**i*A for i in range(3)]) C3.dot_product(C3)**.5/l
Out[34]:
2.07708965974321\renewcommand{\Bold}[1]{\mathbf{#1}}2.07708965974321

The pentagonal faces inradius:

In [35]:
C5 = 1./5.*sum([R5**i*A for i in range(5)]) C5.dot_product(C5)**.5/l
Out[35]:
1.98091594728185\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]:
164.1753660560339\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]:
152.929920275835\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, η\eta:

In [42]:
eta = phi/xi

It satisfies η32ηφ=0\eta^3 -2\eta - \varphi = 0:

In [43]:
eta^3 - 2*eta - phi
Out[43]:
0\renewcommand{\Bold}[1]{\mathbf{#1}}0

Now write Adams expression as 10φ3A+φ22B\begin{equation} \frac{10\varphi}{3}\sqrt{A} + \frac{\varphi^2}{2}\sqrt{B} \end{equation} (write 5=2φ1\sqrt{5}= 2\varphi-1):

In [44]:
A = (phi^2 + 3*eta*(phi + eta)); A
Out[44]:
3ξ2+9ξ+ϕ+7\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]:
(5ϕ+10)ξ2+(15ϕ+30)ξ+10ϕ+25\renewcommand{\Bold}[1]{\mathbf{#1}}\left(5 \phi + 10\right) \xi^{2} + \left(15 \phi + 30\right) \xi + 10 \phi + 25

Now put A=(3ξ2ϕ2)AA^\prime = (3\xi^2 - \phi^2)A and B=(3ξ2ϕ2)BB^\prime = (3\xi^2 - \phi^2)B:

In [46]:
A1 = (3*xi^2 - phi^2)*A; A1
Out[46]:
1\renewcommand{\Bold}[1]{\mathbf{#1}}1
In [47]:
B1 = (3*xi^2 - phi^2)*B; B1
Out[47]:
20ϕξ2+15ϕ+10\renewcommand{\Bold}[1]{\mathbf{#1}}-20 \phi \xi^{2} + 15 \phi + 10

Now B=(4φ2)ξ2+(4φ2)ξ3φ1\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]:
0\renewcommand{\Bold}[1]{\mathbf{#1}}0

Now Adams expression can be written as

(10φ3A+φ22B)/3ξ2φ2\begin{equation} (\frac{10\varphi}{3}\sqrt{A^\prime} + \frac{\varphi^2}{2}\sqrt{B^\prime})/\sqrt{3\xi^2-\varphi^2} \end{equation}

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]:
(3ϕ+1)ξ2+(3ϕ+1)ξ16ϕ2\renewcommand{\Bold}[1]{\mathbf{#1}}\left(3 \phi + 1\right) \xi^{2} + \left(3 \phi + 1\right) \xi - \frac{1}{6} \phi - 2

Snub_Dodecahedron

laevo  and  dextro\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. IteratetofindCircumradiusC0=2.3Cn+1=f(Cn)f(Cn)f(x)=256(35)x8128(1325)x6+32(3535)x416(195)x2+(295)f(x)=2048(35)x7768(1325)x5+128(3535)x332(195)xVolume=(203C213+10(5+25)C25((7+35)2)\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 S. Adams

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 2φ\frac{2}{\color{DarkGoldenrod}{\varphi}}, Adams' solution inscribes a Snub Dodecahedron on to a regular icosahedron with an edge length of 11.

Figure_1._for_solution_to_Snub_Dodecahedron.png

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

gd2D24=gf2+(abafeb)2+ed2D24=0 \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…

D23+112D3cosα+sin2β12Dcosα23[2Dcosα313+12Dcosα23]D24=0 \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

D24Dcosα+1sin2β(12Dcosα)2=0(Eq.1) \color{red}{ D^2 \,-\, 4D\,\cos\alpha \,+\, 1 \,-\, \sin^2\beta \,(1 \,-\, 2\,D\,\cos\alpha)^2 \,=\, 0 \quad(Eq.1)}

ParseError: KaTeX parse error: Invalid color: 'blue ' at position 377: …}{b}} } \color{̲b̲l̲u̲e̲ ̲}̲{)^2} \,+\, \c…

ParseError: KaTeX parse error: Invalid color: 'blue ' at position 377: …overline{\color{̲b̲l̲u̲e̲ ̲}̲{a}\color{green…

D212+112D12(cosα+3sinα)+sin2β12Dcosα23[D(cosα+3sinα)2313+12Dcosα23]D24=0 \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

2D2D(cosα+3sinα)+1sin2β(12Dcosα)[1+D(cosα+3sinα)]=0(Eq.2) \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)}

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

γ3tanαΓ3cosα3sinα \color{red}{\gamma \,\equiv\,\sqrt{3}\,\tan\alpha} \quad \color{blue}{\Gamma \,\equiv\,3\cos\alpha-\sqrt{3}\,\sin\alpha}

cos2α+sin2α=1=cos2α(1+γ23)orcos2α=11+γ23 \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}} }}

Γcosα=(cosα3sinα)cosα=(3γ)cos2α=3γ1+γ23 \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}} }}

3(1+γ23)[Γcosα3γ1+γ23]=Γcosαa  γ2+3b  γ+3(Γcosα3)c=0 \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

PositiveRoot:  γ=b+b24ac2a=3+912Γcosα(Γcosα3)2Γcosα(Eq.3) 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) }

Γ2=Γcosα(3γ)=3Γcosα12[3+912Γcosα(Γcosα3)](Eq.4) \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)}

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

GoldenRatiophi:  φ1+52Icosasymmetry:  sin2β=13φ2 \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}}

Combine:3φ2(Eq.2)  +  3φ2(Eq.1)y=0 \color{blue}{Combine:}\quad\color{blue}{ 3\varphi^2(Eq.2)} \;+\; \color{red}{3\varphi^2(Eq.1) \color{green}{y}} \,=\,0

[3φ2(y2)+2((12y)cosα3sinα)cosα]iD2((4y+1)cosα+3sinα)φ4jD  +(y+1)φ4k=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

Definemuandlambda:  j24ik(μcosα+λ3sinα)2  =  μ2cos2α+(μλ)23cosαsinα+λ23sin2α \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}

[φ2(4y+1)24φ2(y+1)(3φ2(y2)+2(12y))]μ2cos2α+[φ8(4y+1)+4φ4(y+1)](μλ)23cosαsinα+[φ84φ6(y+1)(y2)]λ23sin2α \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}

Sumcomponents(μλ)2μ2λ2=0using  identitiesφn+1φn1=φnandφn+2+φn2=3φn \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}}

φ16  1616  88  11φ14  64  32  144  80144  32144φ12  48144    960    32+128288  40200  8184  144φ10  64  32  192  32  64φ8  16  32  16÷144φ12y42y2  φ2y    φ \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}

Firstrootofy:  1y3y2p=1yq=1φr=φa=p23+q  =  131  =  43(y+1)  y42y2φ2yφb=2p227pq3+ry4+y3b=22713φ  =  49275540y3y2Secondrootofy:0y2yp3b2+b24+a3273b2b24+a3273  =  η2130φyφeta:ηφ2+12φ5273  +  φ212φ527300 \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}

First Golden Circle:  If  a  istakenas  23,  then  y  becomesan  OrigintoonethirdpowersoftheGolden Ratio. \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}.}

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

[2(3cosα3sinα)cosα9φ2]D  +  φ4(3cosα3sinα)  =  [2Γcosα9φ2]D  +  φ4Γ  =  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

D  =  φ4Γ9φ22Γcosα(Eq.5) \color{green}{D\;=\;\frac{ \varphi^4\Gamma}{9\varphi^2 - 2\Gamma\cos\alpha} \quad(Eq.5)}

Γ(9φ22Γcosα)3φ2D(Eq.1)  =  0(substituteDwithEq.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)}

Γ(9φ22Γcosα)3φ2[(3φ24cos2α)(φ4Γ9φ22Γcosα)4φ4cosα+φ4(9φ22Γcosαφ4Γ)]  =  0 \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

4(Γcosα)236φ2Γcosα+27φ2+φ4Γ2  =  0(substituteΓ2withEq.4) \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)}

4(Γcosα)2+3φ2(φ412)Γcosα+32φ2(φ2+18)  =  φ42912Γcosα(Γcosα3) \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)}}

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

16(Γcosα)4+24φ2(φ212)(Γcosα)3+36φ2(21φ2+11)(Γcosα)2+54φ4(φ236)Γcosα+81φ4(φ2+9)  =  0 \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

Define  x:x23Γcosα and divide by 81: \color{green}{ Define\;x:\quad x\equiv\frac{2}{3}\Gamma\cos\alpha \quad \text{ and divide by 81:}}

x4+φ2(φ212)x3+φ2(21φ2+11)x2+φ4(φ236)x+φ4(φ2+9)  =  0 \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

x39φ2p=9φ2x2+φ2(21φ2+2)q=φ2(21φ2+2)xφ4(φ2+9)r=φ4(φ2+9)(x1)  x4+φ2(φ212)x3+φ2(21φ2+11)x2+φ4(φ236)x+φ4(φ2+9)x4x309φ2x3+9φ2x20φ2(21φ2+2)x2φ2(21φ2+2)x0φ4(φ2+9)xφ4(φ2+9)00 \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}

Firstrootofx:  1a=p23+q  =  27φ4+φ2(21φ2+2)  =  2φ6b=2p227pq3+rb=54φ6+3φ4(21φ2+2)φ4(φ2+9)  =  φ10Secondrootofx:p3b2+b24+a3273b2b24+a3273  =  3φ2φ3η \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}