How to construct a chamfered dodecahedron with equilateral or coplanar faces?

Question

According to wikipedia, "The chamfered dodecahedron is a convex polyhedron with 80 vertices, 120 edges, and 42 faces: 30 hexagons and 12 pentagons. It is constructed as a chamfer (geometry) (edge-truncation) of a regular dodecahedron. The pentagons are reduced in size and new hexagonal faces are added in place of all the original edges." And according to this question, the steps to create a chamfered polyhedron are as follows:

1. shrink the given polyhedron's faces by a factor $r, 0<r<1$
2. translate the shrunken faces outward [I'll call this distance $s$], so that their edges are coplanar with the original polyhedron's edges
3. construct the required hexagons from the original polyhedron's vertices and the vertices of the shrunk faces.

I haven't found any other explicit details regarding the relationship between $r$ and the shape of the hexagons, and so my questions are these (note that the value of $s$ will be directly dependent on the shape of the original face, and so $s$ will need to be represented with some parameter(s) that represent the face's shape somehow):

1. What is the general relationship between $r$ and $s$ in order to get coplanar vertices? Does this relationship change depending on how many times the polyhedron is chamfered?
2. What is the general relationship between $r$ and $s$ in order to get equilateral vertices? Does this relationship change depending on how many times the polyhedron is chamfered?
3. For equilateral and coplanar vertices, what value does $r$ need to be so that the diameter of the resulting hexagon equals the original side length? (ie. given pentagons that share edge $AB$, the resulting hexagon between the two would have a diameter equal to the length of $AB$ - shown in the images below) I assume this relationship could be obtained by setting up the answers to the previous questions as a system of equations

Note that this is different than my question here as it takes a different approach in constructing the chamfered faces - scaling along the relative Z axis vs. translating.

Answer 1

I'll consider the case of faces that are simultaneously equilateral and coplanar, leveraging the work done by David McCooey in generating the coordinates of an equilateral chamfered dodecahedron (text link via dmccooey.com) with unit-length edges.

McCooey gives the coordinates in terms of these values: $$\begin{align} c_0 &:= \tfrac14\left(1 + \sqrt{5}\right) \\ c_1 &:= \tfrac1{10}\sqrt{10\left(5 + \sqrt{5}\right)} \\ c_2 &:= \tfrac1{10}\left( 5 + \sqrt{10\left(5 -\sqrt{5}\right)}\right) \\ c_3 &:= \tfrac1{20}\left( 5 + 5\sqrt{5} + 2\sqrt{10\left(5 + \sqrt{5}\right)}\right) \\ c_4 &:= \tfrac1{10}\left( 5 + 2\sqrt{ 5\left(5 + 2\sqrt{5}\right)}\right) \\ c_5 &:= \tfrac1{20}\left(15 + 5\sqrt{5} + 2\sqrt{10\left(5 + \sqrt{5}\right)}\right) \\ c_6 &:= \tfrac1{20}\left( 5 + 5\sqrt{5} + 4\sqrt{ 5\left(5 + 2\sqrt{5}\right)}\right) \\ c_7 &:= \tfrac1{20}\left(15 + 5\sqrt{5} + 4\sqrt{ 5\left(5 + 2\sqrt{5}\right)}\right) \\ \end{align} \tag{1}$$ Of the eighty vertices, we take two sets of five (where I'm assigning "$P$" and "$Q$" names to McCooey's "$V$"s):

$$\begin{align} P_0 := V_{0} = (0, c_2, c_7)\quad P_1 &:= V_{72} = (\phantom{-}c_3, c_3, c_3)\quad P_2 := V_{8} = (\phantom{-}c_2, c_7, 0)\\ P_3 &:= V_{10} = ( -c_2, c_7, 0)\quad P_4 := V_{76} = ( -c_3, c_3, c_3) \\[4pt] Q_0 := V_{60} = ( 0, c_4, c_5) \quad Q_1 &:= V_{52} = (\phantom{-}c_0, c_6, c_3) \quad Q_2 := V_{28} = (\phantom{-}\tfrac12, c_7, c_1) \\ Q_3 &:= V_{32} = (-\tfrac12, c_7, c_1) \quad Q_4 := V_{56} = ( -c_0, c_6, c_3) \end{align} \tag{2}$$ The $P$ vertices neighbor what would be a face of the dodecahedron in OP's images; such vertices lie on a sphere of approximate radius $2.87463$. The $Q$ vertices neighbor the corresponding pentagonal face of the chamfered dodecahedron; such vertices lie on a sphere of approximate radius $2.86094$. Define $P_\star := \frac15\sum P_i$ and $Q_\star:=\frac15\sum Q_i$, the centers of these faces.

OP's chamfering process scales a dodecahedral face by a factor of $r$ (taking $P_i$ to $P_\star+r(P_i-P_\star)$, then translates it by $z$. It's not clear if OP's "$z$" is a scalar or a vector, so I'll introduce a scalar $s$ and say that the translation is by the vector $(s-1)P_\star$. Then we have

$$Q_i = sP_\star + r (P_i - P_\star) \tag{3}$$

The scaling values are simply $$\begin{align} r &= \frac{|Q_0Q_1|}{|P_0P_1|} = \frac{1}{\sqrt{(c_2-c_3)^2 + c_3^2 +(c_3 -c_7)^2}} \approx 0.487457 \tag{4}\\[8pt] s &= \frac{|Q_\star|}{|P_\star|} = \sqrt{ \frac{ (2c_1+2c_3+c_5)^2 + (c_4+2c_6+2c_7)^2 }{(c_2+2c_3+2c_7)^2 + (2 c_3 + c_7)^2}} \approx 1.19577 \tag{5} \end{align}$$

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Of course, the above numerics are specific to the first iteration of the subdivision process OP is seeking to define in a previous question. What to do after that isn't entirely clear, although it should be "obvious" that every iteration changes the geometry so that particular scale factors don't simply carry-over.

The first iteration is "easy" because all faces are equivalent, as are edges. At the second iteration, we have two types of faces (regular pentagons and non-regular hexagons) and two types of edges (bordering two hexagons, and bordering a hexagon and a pentagon). From there, the types of hexagons and the types of edges will only proliferate, adding complexity.

Consideration of strategies for later iterations would seem to go beyond the scope of this question, so I'll end my discussion here.

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Appendix. Here's a bit more of a derivational approach that lends some insight into later chamfering iterations. In what follows, I'll be using $\lambda$ as the scale factor for edges, and $p$ (and also $q$) as the distance by which a scaled face is translated above its forebear.

The equilaterality condition.

Consider an edge $\overline{AB}$ of length $2m$, and let $C$ be on a neighboring face such that $C$ serves as the apex of isosceles $\triangle ABC$ with height $c$. Let $A_\lambda$ be the dilation of $A$ in $C$ by factor $\lambda$; likewise define $B_\lambda$. (Necessarily, $|A_\lambda B_\lambda| = \lambda|AB|=2\lambda m$.) Then, let $A_p$, $B_p$, $C$ be the translates of $A_\lambda$, $B_\lambda$, $C$ by distance $p$ perpendicular to the plane of $|\triangle ABC|$.

Equilaterality of the chamfer requires that $|A_pB_p|=|AA_p|=|BB_p|$. The second equality follows from $\triangle ABC$ being isosceles, so we need only consider the first. A couple of appeals to Pythagoras yields $$4\lambda^2m^2 = p^2 + (1-\lambda)^2|AC|^2 = p^2 + (1-\lambda)^2 (c^2+m^2) \tag{A.1}$$ Repeating the argument for isosceles $\triangle ABD$ of height $d$ on the other neighboring face of $\overline{AB}$, with $q$ the corresponding translation distance, gives for the same scale factor $\lambda$ $$4\lambda^2m^2 = q^2 + (1-\lambda)^2 (d^2+m^2) \tag{A.2}$$

The coplanarity condition.

Viewing along $\overline{AB}$ (with midpoint $M$), the edge visually collapses to a point and the neighboring faces to segments $\overline{CM}$ and $\overline{DM}$ of lengths $c$ and $d$. Extending $\overline{CC_p}$ to meet the plane of $\square ABB_pC_p$ at $C'$ (and likewise obtaining $D'$), the dihedral angles between the chamfer plane(s) and the faces of the polyhedron become $\gamma := \angle CMC'$ and $\delta := \angle DMD'$.

Writing $\theta$ as the polyhedron's dihedral angle along $\overline{AB}$, we have a planar hexagon $AA_pB_pBB_qA_qA$ when $\gamma+\delta+\theta=180^\circ$. Consequently, $$\begin{align} \tan\theta = -\tan(\gamma+\delta) & = -\frac{\tan\gamma+\tan\delta}{1-\tan\gamma\tan\delta} \tag{A.3}\\[4pt] &= -\frac{\dfrac{p}{(1-\lambda)c}+\dfrac{q}{(1-\lambda)d}}{1-\dfrac{pq}{(1-\lambda)^2cd}} \tag{A.4} \end{align}$$ Thus, $$(pq)-(1-\lambda)^2 c d) \tan\theta = (1-\lambda) (cq+dp) \tag{A.5}$$

Equilaterality and coplanarity.

Equations $(A.1)$, $(A.2)$, $(A.5)$ comprise a system of three equations in five unknowns: $\lambda$, $c$, $d$, $p$, $q$. In the specific case of chamfering the dodecahedron, we have $c=d=m\tan54^\circ$ and $p=q$, leaving just the unknowns $\lambda$ and $s$. Taking $m=1$, we can solve to obtain $$\begin{align} \lambda &= -5 - 2 \sqrt{5} + \sqrt{2 (25 + 11 \sqrt{5})} \approx 0.487457 \\ p &\approx 0.435995 \end{align} \tag{A.6}$$ Scale factor $\lambda$ here matches $r$ in $(4)$. Distance $p$ is consistent with $s$ in $(5)$ in that $(p+i)/i=s$, for the dodecahedral inradius $i= \frac{m}{10}\sqrt{250+110\sqrt{5}}\approx 2.22703$.

More generally, we don't have enough information to solve for unique values of our five parameters. But our equations only think locally along one edge of the polyhedron being chamfered. Every edge gives rise to a corresponding system; edges sharing a common face share the translation parameter (the analogue of $p$ and $q$); and, for universal equilaterality, the systems share the same scale factor $\lambda$. So, globally, we "should" obtain enough equations to determine unique parameters (or else find that no parameters can do everything we ask of them).

Those considerations are beyond the scope of this discussion, but I will note that we (and by "we", I mean "Mathematica") can eliminate $c$ and $d$ from the single-edge system to get a relation between the common scale factor $\lambda$ and the translation distances $p$ and $q$ for neighboring faces.

$$(1 + \lambda) (3\lambda-1) m^2 \sin^2\theta = p^2 + q^2 \mp 2 p q \cos\theta \tag{A.7}$$ (where I suspect "$\pm$" is always "$-$" (it is for the regular dodecahedron) but haven't verified this). Interestingly, this is exactly a Law of Cosines statement identifying $p$, $q$, and $m\sin\theta\sqrt{(1+\lambda)(3\lambda-1)}$ as sides of a triangle, with $p$ and $q$ surrounding angle $\theta$ (or its supplement). This suggests there might be a clean geometric argument behind all of this. (It also shows that we must have $\lambda \geq 1/3$.) Substituting back into the system, we find

$$ c = \frac{|p\cos\theta\mp q|}{(1-\lambda)\sin\theta} \qquad d = \frac{|p\mp q\cos\theta|}{(1-\lambda)\sin\theta} \tag{A.8} $$

That's enough typing for now.

Answer 2

tl;dr If $y$ represents the side length of a face, $b$ the distance between a face's center and $y$'s midpoint, $l$ the distance between the face's center and a vertex on $y$, and $\theta_0$ the angle between the face and the adjacent face across $y$, then the expressions for $r$ and $s$ to get equilateral (only for the first two chamfers) and coplanar faces are:

$ \begin{align} \theta=\frac {180-\theta_0}{2},\quad u&=b\,tan(\theta),\quad y=2\,\sqrt {l^2-b^2} \\ r&=1-\frac {y}{\sqrt {l^2+u^2} + y} \\ s&=\frac {u\,y}{\sqrt {l^2+u^2} + y} \end{align} \tag {0}$

So I managed to figure out the answer to each of the questions - what was hindering my progress was trying to figure it all out without the second image below. Once I figured that out, everything else just fell into place! The following image shows all the variables used (note that $\theta_0$ is slightly misleading this this image: the second image below is more accurate):

1. Relationship between $r$ and $s$ so that the vertices are coplanar

In order for the vertices to be coplanar, they all need to be on the same plane that connects the two new pentagons together, which is shown in the first image below. This was the main thing that stumped me, as I didn't know how to easily represent "coplanar vertices" without a whole lot of unknowns with vectors. This drastically simplified the workflow, and the relationship between $s$ and $r$ was quickly found (note that $\theta_0$ is the angle between the two original faces, and half of the remaining angle - $180-\theta_0$ - gets divided between them, which is why $\theta=\frac {180-\theta_0}{2}$):

$ \begin{align} \theta&=\frac {180-\theta_0}{2} \\ c&=\frac{s}{tan(\theta)} \\ r&=1-\frac{c}{b} \\ r&=1-\frac{s}{b\,tan(\theta)} \\ s&=(1-r)\,b\,tan(\theta) \end{align} \tag {1}$

2. Relationship between $r$ and $s$ so that the vertices are equilateral

This is was notably more challenging than the previous solution, and it even tricked me into thinking I had solved it a couple times - but to make things easier, I conjured up $y$ to represent the side length of the original face (not shown in the images unfortunately: this is the line perpendicular to $b$). The solution for $s$ isn't that bad, but getting $r$ was certainly not as pretty as the one before (note that only two variables from $b$, $l$, and $y$ are required, as they are connected by the expression $l=\sqrt {b^2+\frac {y^2}{4}}$)...

$ \begin{align} r&=1-\frac {d}{l} \\ d&=\sqrt {x^2-s^2} \\ r&=1-\frac {\sqrt {x^2-s^2}}{l} \\ x&=r\,y \\ r&=1-\frac {\sqrt {r^2\,y^2-s^2}}{l} \\ s&=\sqrt {r^2\,y^2 - l^2\,(1-r)^2} \\ 0&=r^2\,(y^2-l^2)+r\,(2\,l^2)+(-l^2-s^2) \\ r&=\frac {-l^2+\sqrt {l^4-(y^2-l^2)(-l^2-s^2)}}{y^2-l^2} \end{align} \tag {2.1}$

Alternatively, if we replace $r$ with $1-\frac {c}{b}=1-\frac {s}{b\,tan(\theta)}$ (instead of just leaving it in and then solving for a quadratic later), a less complex solution can be found for $r$:

$ \begin{align} r&=1-\frac {s}{b\,tan(\theta)} \\ r&=1-\frac {\sqrt {r^2\,y^2-s^2}}{l} \\ r&=1-\frac {\sqrt {y^2\,(1-\frac {s}{b\,tan(\theta)})^2-s^2}}{l} \end{align} \tag {2.2}$

3. Relationship between $r$ and $s$ so that the vertices are coplanar and equilateral

Given that I need both coplanar and equilateral faces here, I simply used a system of equations using both the previous solutions and solved for $r$ and $s$ independently. There was a lot of simplifying done and the process was not pretty in the least, so for this answer I just laid out the groundwork and then skipped to the solution - let me know if you wish to see the entire process, or even if you find an easier/better way to do this! For the sake of clarity, I replaced all instances of $b\,tan(\theta)$ with $u$:

Solving for r (using the upper root):

$ \begin{align} s&=(1-r)\,u \\ s&=\sqrt {r^2\,y^2-l^2\,(1-r)^2} \\ (1-r)\,u&=\sqrt {r^2\,y^2-l^2\,(1-r)^2} \\ (1-r)^2\,u^2&=r^2\,y^2-l^2\,(1-r)^2 \\ &... \\ r&=\frac {-(l^2+u^2)+\sqrt {(l^2+u^2)^2-(y^2-l^2-u^2)\,(-l^2-u^2)}}{y^2-l^2-u^2} \end{align} \tag {3.1}$

Solving for s (using the lower root):

$ \begin{align} r&=1-\frac{s}{u} \\ r&=1-\frac {\sqrt {y^2\,(1-\frac {s}{u})^2-s^2}}{l} \\ 1-\frac{s}{u}&=1-\frac {\sqrt {y^2\,(1-\frac {s}{u})^2-s^2}}{l} \\ \frac{s^2}{u^2}&=\frac {y^2\,(1-\frac {s}{u})^2-s^2}{l^2} \\ &... \\ s&=\frac {u\,y^2-\sqrt {u^2\,y^4-(y^2-l^2-u^2)\,(u^2\,y^2)}}{y^2-l^2-u^2} \end{align} \tag {3.2}$

Thanks to WolframAlpha, a much simpler solution could be produced if all the values are assumed to be positive (given that we will never have negative distance or angles, this is a valid assumption):

$ \begin{align} s&=\frac {u\,y}{\sqrt {l^2+u^2} + y},\;u=b\,tan(\theta),\;y=2\,\sqrt {l^2-b^2} \end{align} \tag {3.3}$

And given that $r=1-\frac {s}{u}$, this gives us a nice solution for $r$ as well:

$ \begin{align} r&=1-\frac {y}{\sqrt {l^2+u^2} + y},\;u=b\,tan(\theta),\;y=2\,\sqrt {l^2-b^2} \end{align} \tag {3.4}$

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In applying this to my dodecahedron (the magnitude of each vertex being $1$), I can now show that the solutions I discovered are correct (credit for the values for $r$ and $s$ goes to Blue's answer - note that these values are all approximations, rounded to the fourth digit-place):

Given: $b=0.4911,\quad l=0.6071,\quad y=0.7136,\quad \theta=\frac {180-116.5650}{2}=31.7175,\quad u=b\,tan(\theta)=0.3035$

Unknown: $r=0.4875,\quad s=0.1556$

1. Relationship between $r$ and $s$ so that the vertices are coplanar

$ \begin{align} r&=1-\frac{s}{b\,tan(\theta)}&=1-\frac{0.1556}{0.4911\,tan(31.7175)}&=.4873 \\ s&=(1-r)\,b\,tan(\theta)&=(1-0.4875)\,0.4911\,tan(31.7175)&=.1556 \end{align}$

2. Relationship between $r$ and $s$ so that the vertices are equilateral

$ \begin{align} r&=1-\frac {\sqrt {y^2\,(1-\frac {s}{b\,tan(\theta)})^2-s^2}}{l} &=1-\frac {\sqrt {0.7136^2\,(1-\frac {0.1556}{0.4911\,tan(31.7175)})^2-0.1556^2}}{0.6071}&=.4877 \\ s&=\sqrt {r^2\,y^2 - l^2\,(1-r)^2}&=\sqrt {0.4875^2\,0.7136^2 - 0.6071^2\,(1-0.4875)^2}&=.1556 \end{align}$

3. Relationship between $r$ and $s$ so that the vertices are coplanar and equilateral

$ \begin{align} r&=1-\frac {y}{\sqrt {l^2+u^2} + y}&=1-\frac {0.7136}{\sqrt {0.6071^2+0.3035^2} + 0.7136}&=.4875 \\ s&=\frac {u\,y}{\sqrt {l^2+u^2} + y}&=\frac {(0.3035)\,(0.7136)}{\sqrt {0.6071^2+0.3035^2} + 0.7136}&=.1555 \end{align}$

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Update

These equations may work for future subdivisions/chamfering as well, however faces will not continue to be equilateral (while still being coplanar). This is to be expected, however an issue arises of how to uniformly scale and translate the vertices of a face if it is not equilateral (given that $b$, $l$, and $y$ may vary per vertex). As such, my chamfering process would need to be updated so that it works with such faces.