# Polar Zonohedra Edge-Unfold to Nets

Joseph O’Rourke Departments of Computer Science and of Mathematics, Smith
College, Northampton, MA 01063, USA<EMAIL_ADDRESS>
###### Abstract

This note proves that every polar zonohedron has an edge-unfolding to a non-
overlapping net.

##  1 Introduction

What has become known as Dürer’s Problem [O’R13] asks whether or not every
convex polyhedron $P$ has an edge-unfolding—a spanning tree of the
$1$-skeleton which when cut, unfolds the surface of $P$ to a planar simple
polygon known as a _net_. A key requirement is that a net does not self-
overlap.111 Even though redundant, to block misinterpretation we sometimes
call it a _non-overlapping net_. Despite considerable effort since Shephard
first posed the question formally [She75], there are only finite classes of
polyhedra ($5$ Platonic solids, $13$ Archimedean solids, $92$ Johnson solids),
and a few infinite classes of polyhedra known to edge-unfold to a net: prisms,
domes [DO07, p. 323], prismoids [O’R01] [DO07, p. 325], nested prismatoids
[Rad21].

A _zonohedron_ is a centrally symmetric convex polyhedron that may be defined
as the Minkowski sum of a set of line segments. A _polar zonohedron_ is formed
when the generating segments constitute an equally spaced and equal (unit)
length star (or “umbrella”) at one of its two poles. It is symmetric about the
line through the two poles; we’ll assume this line is vertical. It is also
symmetric about a plane orthogonal to and through the midpoint of the poles
line. All of its edges have the same length; all of its faces are rhombs. A
polar zonohedron is determined by two parameters: $n$, the number of rhombs
incident to each pole, and ${\theta}\in(0,\pi/2)$, the angle that measures the
degree of closure at the poles. Thus there is a continuum of polar zonohedra.
Small ${\theta}$ produce pancake-like polyhedra, and ${\theta}$ near $\pi/2$
produce thin cigar-shaped polyhedra. See Fig. 1.

Figure 1: $n=8,{\theta}=50^{\circ}$. One zone $\mathcal{Z}$ of $(n-1)=7$
rhombs is marked in red.

The aim of this note is to prove this theorem:

###### Theorem 1

Every polar zonohedron has an edge-unfolding to a (non-overlapping) net.

The unfolding is illustrated in Fig. 1. It unfolds each of the $n$ zones, each
joined at the image $o$ of the north pole. A _zone_ of a polar zonohedron is a
collection of rhombs all sharing a set of parallel edges. There are $n$ such
congruent zones comprising the surface, each containing $(n-1)$ rhombs. The
edges cut between each pair of adjacent zones form what George Hart calls a
“surface helix.” Two further examples are shown in Figs. 2 and 3

I rely on Hart’s thorough exposition in [Har21], as well as employing his
Mathematica software. See also [Tow96].

Figure 2: $n=20$, ${\theta}=0.5\approx 28.6^{\circ}$, ${\alpha}\approx
15.8^{\circ}$. Figure 3: $n=12$, ${\theta}=\frac{1}{2}^{\circ}$.

##  2 Notation

Distinguish between a zone $\mathcal{Z}$ in $\mathbb{R}^{3}$ and that zone
unfolded, or developed in the plane as $Z$. Each $Z$ is composed of $(n-1)$
rhombs: $R_{1},R_{2},\ldots,R_{n-1}$. Let ${\alpha}$ be the angle of $R_{1}$
at the corner that is incident to the (north) pole point.

Let $o$ be the planar image of the north pole. We focus on two consecutive
unfolded zones, because if adjacent zones do not overlap, then there is no
overlap in the entire unfolding. All zones are congruent in $\mathbb{R}^{3}$
and therefore congruent in their planar unfoldings. Let $Z$ be one zone, and
$Z^{\prime}$ its counterclockwise (ccw) neighbor sharing an edge incident to
the pole $o$. See Fig. 4.

Figure 4: $Z$ and $Z^{\prime}$ for $n=16$. Left to right:
${\theta}=1^{\circ},20^{\circ},50^{\circ}$.

Arrange $Z$ so that the top edge of $R_{1}$ is horizontal. Then $Z^{\prime}$
is $Z$ rigidly rotated about $o$ by ${\alpha}$, because the zones are
congruent and angle ${\alpha}$ is incident to the pole $o$. In this
orientation all the $n$ unit edges of $Z$ are horizontal, all the unit edges
of $Z^{\prime}$ at angle ${\alpha}$.

##  3 Proof Plan

Because $Z^{\prime}$ is just a rigid rotation of $Z$ about $o$, it seems
almost obvious that they cannot overlap. But I have not found a simple proof.

However, the overall plan of the proof presented here is simple. First we
establish a condition on the shape of $Z$ that guarantees non-overlap. Let
$|\mathcal{C}(r)\cap Z|={\beta}(r)$ be the angular measure of the arc of the
circle $\mathcal{C}(r)$ centered on $o$ that intersects $Z$. (This definition
is revisited and clarified in the next section.) Then the shape condition is
that, if ${\beta}(r)\leq{\alpha}$ for all $r$, then $Z$ and the rotated
$Z^{\prime}$ do not overlap.

The remainder of the proof calculates ${\beta}(r)$ and verifies that it is
$\leq{\alpha}$, for all $n$ and all ${\theta}$. It is these calculations that
are not straightforward. The main open problem (Section 7) is to find a proof
that can establish ${\beta}\leq{\alpha}$ without these calculations.

##  4 Overlap Condition

We say that two regions $A$ and $B$ in the plane _overlap_ if there is a point
$p$ strictly interior to both $A$ and $B$.

Let $\mathcal{C}(r)$ be a circle of radius $r$ centered on $o$. Define
$|\mathcal{C}(r)\cap Z|={\beta}(r)$ as the angle subtended at $o$ by the
shortest arc of $\mathcal{C}(r)$ that covers $\mathcal{C}(r)\cap Z$. (We will
drop the $r$ in ${\beta}(r)$ when clear from the context.) Note here we are
allowing for the possibility that $\mathcal{C}(r)\cap Z$ might be
disconnected. And in fact, this is necessary, for in extreme situations
(${\theta}\to 0$), this can occur. See Fig. 5. We will revisit disconnected
arcs in Section 6.1.

Figure 5:  $|\mathcal{C}(r)\cap Z|$ is disconnected. $n=16$,
${\theta}=1^{\circ}$. $R_{n/2}=R_{8}$ is pink.

###### Lemma 1

If $Z$ and $Z^{\prime}$ overlap at point $p$ on circle $\mathcal{C}(r)$,
$r=|op|$, then $|\mathcal{C}(r)\cap Z|={\beta}(r)>{\alpha}$.

Proof: Suppose for contradiction that $Z$ and $Z^{\prime}$ overlap at $p$ but
$|\mathcal{C}(r)\cap Z|={\beta}\leq{\alpha}$. Because $Z^{\prime}$ is the
rotation of $Z$ about $o$ by ${\alpha}$, there is a gap of length
${\alpha}-{\beta}\geq 0$ along $\mathcal{C}(r)$, which separates
$\mathcal{C}(r)\cap Z$ to the left (cw) and $\mathcal{C}(r)\cap Z^{\prime}$ to
the right (ccw). Therefore, $Z$ and $Z^{\prime}$ are separated or just touch
along $\mathcal{C}(r$), and $p$ cannot exist.

Note that, because $\mathcal{C}(r)\cap Z$ could be disconnected, it is
possible that ${\beta}(r)>{\alpha}$ but there is no point $p$ witnessing
overlap. However, if there is a $p$, then it must be that
${\beta}(r)>{\alpha}$.

In Fig. 6, $\mathcal{C}(r)\cap Z$ is the red arc spanning
${\beta}(r)<{\alpha}$. Therefore when this arc is rotated by
${\alpha}>{\beta}(r)$ to the blue arc, the red and blue arcs cannot overlap.

Figure 6: ${\beta}\leq{\alpha}$ implies no overlap.

###### Corollary 1

If $|\mathcal{C}(r)\cap Z|={\beta}(r)\leq{\alpha}$ for all $r$, then $Z$ and
$Z^{\prime}$ do not overlap.

The remainder of the proof of Theorem 1 concentrates on proving
${\beta}\leq{\alpha}$.

##  5 Regular Polygon: ${\theta}=0$

There is a sense in which the situation when ${\theta}\to 0$ is the most
“dangerous,” for then there can be a flattened rhomb such as that illustrated
in Fig. 5. When ${\theta}=0$, the zonohedron degenerates to a doubly-covered
regular $n$-gon, with the only curvature at the vertices on the rim. (See the
earlier Fig. 3.) Then the unfolding of a zone $Z$ resembles mirrored $S$-shape
composed to two half-regular polygons. Fig. 7(a) shows an example with
${\theta}=1^{\circ}$, and (b) the ${\theta}=0$ counterpart.

Figure 7: $n=16$. (a) ${\theta}=1^{\circ}$. Two enlargements of rhomb $R_{8}$
are shown. (b) ${\theta}=0,{\alpha}=22.5^{\circ}$.

The $1^{\circ}$ example (a) shows that one rhomb $R_{8}$ is nearly flat; in
(b) that rhomb collapses to a line segment.

Whether one counts the doubly-covered regular $n$-gon as a “polyhedron,” it is
an interesting case, with regularities not present when ${\theta}>0$.

There is a slight difference for $n$ even or odd; compare Figs. 7 and Fig. 8.
We will concentrate on $n$ even, leaving $n$ odd to remarks.

Figure 8: $n=17$. (a) ${\theta}=1^{\circ}$. (b) ${\theta}=0,{\alpha}\approx
21.17^{\circ}$.

Let $\partial Z_{L}$ and $\partial Z_{R}$ be the right and left boundaries of
$Z$, with $Z$ in the previously described orientation. Let $Z^{+}$ and $Z^{-}$
be the upper and lower halves of the $S$-shape. Let $P_{n}$ be the regular
polygon of $n$ vertices that passes through the $n/2$ vertices of $\partial
Z^{+}_{R}$. Let $C_{R}$ be the circle circumscribing $P_{n}$.222 With unit
rhombs, the radius of $C_{R}$ is $1/(2\sin{\alpha}/2)$. In Fig. 7(b), $n=16$
and ${\alpha}=2\pi/n=22.5^{\circ}$.

Figure 9: $n=16$, ${\theta}=0$, ${\alpha}=22.5^{\circ}$. (a) Each unit segment
inscribed in $C_{R}$ spans ${\alpha}=2\pi/n$ from the center of $C_{R}$. (b)
Each segment spans ${\alpha}/2$ form a center on $C_{R}$.

We will prove that $|\mathcal{C}(r)\cap Z^{+}|$ is exactly ${\alpha}$, and
$|\mathcal{C}(r)\cap Z^{-}|$ is less than ${\alpha}/2$. Thus the rotation from
$Z$ to $Z^{\prime}$ leaves $Z^{+}$ and $Z^{\prime+}$ touching, with a large
gap between $Z^{-}$ and $Z^{\prime-}$. This is evident in Fig. 3 when
${\theta}=\frac{1}{2}^{\circ}$.

###### Lemma 2

$|\mathcal{C}(r)\cap Z^{+}|={\alpha}$ for all $r$ intersecting $Z^{+}$.

Proof: We refer to Figs. 7(b) and 9 throughout this proof.

Each side of a rhomb along $\partial Z^{+}_{R}$ is a unit-length chord of
$C_{R}$, and so spans angle ${\alpha}$ from the center of $C_{R}$; see Fig.
9(a). Therefore each edge spans ${\alpha}/2$ from $o=v_{0}$, which lies on
$C_{R}$; see Fig. 9(b). In Fig. 7(b), $v_{6}v_{7}$ is such an edge, of rhomb
$R_{6}$. A full rhomb of $Z^{+}$ spans a second edge, e.g., $v_{8}v_{9}$. This
determines a chord $c=v_{6}v_{8}$ (red in Fig. 7(b)), which therefore spans
${\alpha}$ from $o$. Thus $\mathcal{C}(r)$ with the appropriate $r$ intersects
a rhomb passing through diagonal endpoints, $R_{6}$ in our example.

Now we argue for the same conclusion when $\mathcal{C}(r)$ passes through
$Z^{+}$ at an arbitrary location, rather than spanning a diagonal of some
specific rhomb $R_{i}$. Let $\mathcal{C}(r)$ enter $Z^{+}$ at $x$ and exit at
$y$; see Fig. 10. We now argue that the length of the arc $\mathcal{C}(r)\cap
Z^{+}$ is exactly ${\alpha}$, just as it is for the diagonal of each $R_{i}$.
First note that the arc can cross at most two rhombi, i.e., it can cross at
most one horizontal segment of $Z^{+}$. Second, note that the arc can be
partitioned into two parts, each part is symmetric with respect to a diagonal.
In the figure, the diagonals are $v_{8}o$ and $v_{7}o$. Then the left part of
the arc (green) has length $2a$ and the right part (orange) has length $2b$.
Now notice that $a+b$ is the length of the arc spanned by two adjacent
diagonals. But we know that is ${\alpha}/2$. So we have established that
$|\mathcal{C}(r)\cap Z^{+}|={\alpha}$.

Thus we proved that $|\mathcal{C}(r)\cap Z^{+}|={\alpha}$ for all $r$ crossing
$Z^{+}$.

Figure 10: $|\mathcal{C}(r)\cap Z^{+}|=2a+2b$.

Now we turn to the claim for $|\mathcal{C}(r)\cap Z^{-}|$. Even though this
arc is small, I have not found a simple proof that it is $\leq{\alpha}$. In
fact, it is strictly less than half of ${\alpha}$. Because my proof is
technical, and only applies to the degenerate polar zonohedron when
${\theta}=0$, it is relegated to the Appendix.

###### Lemma 3

$|\mathcal{C}(r)\cap Z^{-}|<{\alpha}/2$ for all $r$ crossing $Z^{-}$.

Proof: See the Appendix, Section 8.  .

There remains the situation where $\mathcal{C}(r)$ intersects both $Z^{+}$ and
$Z^{-}$. We analyze this case separately in Lemma 5 in Section 6.1, which
establishes that arc $\mathcal{C}(r)\cap Z$ subtends exactly ${\alpha}$ from
$o$.

##  6 Positive ${\theta}$

In comparison to ${\theta}=0$, $Z$ for ${\theta}>0$ is “stretched out” in that
the turn angle of $\partial Z_{L}$ between adjacent rhombs is less than
${\alpha}$. (This turn angle is not constant rhomb-to-rhomb but it is strictly
less than ${\alpha}$ when ${\theta}>0$.) This means that $\mathcal{C}(r)$
crosses $Z$ closer to the horizontal, intersecting rhombs nearer to unit
length. Thus the intuition is that these arcs are short, and become shorter as
$r$ increases and pushes the chords further from $o$.

Figure 11: $n=16$, ${\theta}=20^{\circ}$. (a) Contrast with ${\theta}=0$.
Diagonal midpoint perpendicular of $R_{5}$ falls below $o$. (b) All diagonal
perpendiculars fall below $o$.

###### Lemma 4

$|\mathcal{C}(r)\cap Z|\leq{\alpha}$ for all $r$ crossing $Z$.

Proof: The proof has four parts:

  1. (1)

The chord $c$ determined by $\mathcal{C}(r)\cap Z$ is more shallow (more
horizontal) than rhomb diagonals.

  2. (2)

If the chord $c$ includes a corner of rhomb $R_{i}$, then it remains in
$R_{i}$: $c\subset R_{i}$.

  3. (3)

If $c$ does not include a corner of a rhomb, then it fits inside the rhomb
immediately above or below.

  4. (4)

The most broad presentation of a rhomb to $o$ is the nearly flat rhomb, with
diagonal near $2$, which subtends strictly less than
$2\cdot{\alpha}/2={\alpha}$ from $o$.

#### (1).

Let $c=xy=\mathcal{C}(r)\cap Z$ be the chord of $\mathcal{C}$. It must be that
the perpendicular bisector of $xy$ passes through $o$, the center of
$\mathcal{C}(r)$. Recall that the diagonals in the regular-polygon
case—${\theta}=0$, Fig. 9—extend directly through $o$. The two diagonals of a
rhomb meet orthogonally at their midpoints, so if $xy$ is one diagonal, the
other aims toward $o$.

Fig. 11(a) contrasts the ${\theta}>0$ situation with ${\theta}=0$. There, for
example, the diagonal of $R_{5}$, because of the turn-angle
reduction/straightening, cannot any longer aim toward $o$ (red), as it did
when ${\theta}=0$ (dashed). Extending this to all rhomb diagonals leads to
Fig. 11(b), which displays all these potential diagonal chords, and shows that
$c$ cannot be a diagonal of $R_{i}$ for $i\geq 2$: for each possible rhomb
diagonal, the chord perpendicular falls below $o$.333 Just barely below for
$R_{2}$, by $10^{-6}$.

#### (2).

This shows that the diagonals are too vertically steep (in our orientation of
$Z$) to serve as the chord $c$: $c$ must be more nearly horizontal. This
implies that $c$ would stay inside $R_{i}$ were it to pass through a corner of
$R_{i}$: $c\subset R_{i}$.

Figure 12: The red arc $\mathcal{C}(r)\cap Z$ can be translated up or down to
fit into an adjacent rhomb. The convex corner at an endpoint of the blue
translated arc is marked in each case.

#### (3).

First we note that if an arc of $\mathcal{C}(r)$ intersects three or more
rhombs, then with one exception, its chord $c$ is steeper than the diagonal of
a middle rhomb. The exception concerns the nearly flat rhomb $R_{n/2}$
illustrated earlier in Fig. 5. Because the argument for this exception is not
straightforward, we address that in a separate lemma, Lemma 5 in Section 6.1
below. So henceforth assume $c=xy$ crosses a horizontal and intersects two
rhombs, $R_{i}$ and $R_{i+1}$.

Fig. 12 shows that the arc fits in either the adjacent upper or lower rhomb,
by sliding an arc endpoint toward the convex vertex of $\partial Z$. One
endpoint of the horizontal $R_{i}\cap R_{i+1}$ is a convex vertex and the
other endpoint a reflex vertex. Sliding/translating the arc endpoint $x$ along
the side of the rhomb toward the convex vertex results in the other end of the
arc $y$ to lie strictly internal to either $R_{i}$ or $R_{i+1}$. This follows
because the reflex vertex implies that the extension of the incident $R_{i}$
rhomb side enters $R_{i+1}$, and the extension of the incident $R_{i+1}$ rhomb
side enters $R_{i}$. The only exception is when there is zero turn angle
between the rhombs (which occurs for example with $R_{8}$ and $R_{9}$ in Fig.
8), when the translated arc has both endpoints on rhomb sides. So in all
cases, the arc fits inside a rhomb. In the next step it does not matter in
which rhomb it fits, for we use an upperbound over all rhombs.

#### (4).

Now that we know $\mathcal{C}(r)\cap Z$ fits inside some rhomb $R_{i}$ of $Z$,
we look at the angle subtended at $o$ by the full rhomb $R_{i}$. This is an
upperbound on the measure of the arc inside (often a considerably generous
upper bound).

Now, the broadest extent of a rhomb from the point of view of $o$ is the
nearly flat rhomb whose long diagonal is nearly $2$ units. For example, this
is the middle rhomb $R_{8}$ illustrated previously in Fig. 7(a). As we know
from Fig. 9, each unit side of a rhomb subtends ${\alpha}/2$ in $Z^{+}$ when
${\theta}=0$. For ${\theta}>0$, each unit side subtends $<{\alpha}/2$, not
only in $Z^{+}$ but for all unit sides in a rhomb of $Z$. Therefore a whole
rhomb subtends $<{\alpha}$.

Fig. 13 illustrates the actual ${\beta}_{i}$ subtended angles for $R_{i}$,
$i=1,\ldots,n-1$, in two examples. As claimed, ${\beta}<{\alpha}$, with
equality only for $i=1$. (The small upticks visible at $R_{12}$ when $n=16$,
and at $R_{18}$ when $n=24$, reflect the lower tail of the $S$-shape, where
the rhombs become more flattened.)

Equality to ${\alpha}$ only occurs when $\mathcal{C}(r)$ crosses the first
rhomb $R_{1}$. For all larger $r$, the inequality is strict.

Figure 13: ${\beta}_{i}$ is the angle (in degrees) subtended from $o$ by rhomb
$R_{i}$: for $n=16$, ${\theta}=20^{\circ}$ when ${\alpha}=22.5^{\circ}$; and
for $n=24$, ${\theta}=40^{\circ}$, when ${\alpha}\approx 11.48^{\circ}$.

###  6.1 Nearly Flat Rhomb $R_{n/2}$

We encountered a disconnected arc $\mathcal{C}(r)\cap Z$ earlier in Fig. 5.
This is also an example where the arc crosses three rhombs. This nearly flat
transitional rhomb only occurs when $n$ is even. Even as ${\theta}\to 0$, no
rhomb for $n$ odd becomes arbitrarily close to flat; see Fig. 8.

In order to complete step (4) of Lemma 4—to show the angle subtended at $o$ by
the full rhomb $R_{i}$ is $<2$—we analyze this exceptional case in the
following lemma.

Figure 14: $n=16$. (a) ${\theta}=0$: $|xy|=|ab|$. (b) ${\theta}=1^{\circ}$:
$|xy|<2$.

###### Lemma 5

If $\mathcal{C}(r)$ intersects $R_{n/2}$, $n$ even, then the chord $c$ of arc
$\mathcal{C}(r)\cap Z$ has length $<2$.

Proof: To simplify notation, let $R_{0}=R_{n/2}$, and let $R_{-}$ and $R_{+}$
be the congruent rhombs above and below $R_{0}$. Refer to Fig. 14 throughout.

We first analyze the situation when ${\theta}=0$ when $R_{0}$ reduces to a
segment. If $\mathcal{C}(r)$ passes through the upper corner $a$ of $R_{+}$,
then it also passes through the lower corner $b$ of $R_{-}$. See the blue arc
in (a) of the figure. If $r^{\prime}<r$ is a bit smaller, then still the span
of $\mathcal{C}(r^{\prime})\cap Z=|xy|$ is exactly $2$. This follows because
the portion of the arc in $R_{-}$ is symmetric about the diagonal line $bo$,
so $|ax|=|by|$. Note that $\mathcal{C}(r^{\prime})\cap Z$ (red) is
disconnected, and intersects all three rhombs.

Now consider the situation when ${\theta}$ is small, (b) of the figure. (This
is an annotated version of the earlier Fig. 5.) $R_{0}$ is a nearly flat rhomb
between $R_{-}$ and $R_{+}$. If $\mathcal{C}(r)$ passes through the upper
corner $a$ of $R_{+}$, it falls below the lower corner $b$ of $R_{-}$, because
the small intervening height of $R_{0}$ lifts up $b$. See the blue arc in (b)
of the figure. If $r^{\prime}<r$ is a bit smaller, then the right end of the
$\mathcal{C}(r^{\prime})$ arc (red) crosses $R_{-}$, but because $b$ is
lifted, the span $\mathcal{C}(r^{\prime})\cap R_{-}$ is smaller than in the
${\theta}=0$ case. The result is that the span $|xy|$ is strictly less than
$2$.

We can quantify the lifting as follows. Symbolic calculation show that the
angle of the central rhomb $R_{0}$ is exactly $2{\theta}$; so $2^{\circ}$ in
Fig. 5(b). Let $d=|ax|\leq 1$ be the horizontal distance from $a$ to $x$. Then
the entry of $\mathcal{C}(r^{\prime})$ at $x$ into $R_{0}$ is lifted by
$d\sin{\theta}$, whereas the lifting of $b$ is at least twice that:
$2\sin{\theta}$.

Although we have ignored the slight deviation from the horizontal of the
bottom and top edges of $R_{-}$ and $R_{+}$ when ${\theta}$ is positive, that
deviation can be removed by a slight rotation, leaving the key inequality
$|xy|<2$ intact.

The slight deviation from the horizontal noted above implies that the nearly
vertical diagonal of $R_{0}$ does not aim through $o$, and so the diagonal
argument used in Lemma 4 still holds for $R_{n/2}$. This is discernible in the
enlargement of $R_{n/2}$ shown in Fig. 7(a).

With Lemma 5 completing Lemma 4, we have now proved Theorem 1 via Corollary 1
for all polar zonohedra for ${\theta}>0$. And Lemmas 2 and 3 settle it for the
degenerate polyhedron when ${\theta}=0$.

##  7 Open Problem

That the zone-by-zone unfolding avoids overlap seems almost obvious, yet the
proof presented of Theorem 1 feels labored. Surely there is a simpler proof.

#### Acknowledgements.

My unfolding software employs code from Hart’s supplement to [Har21]. I
benefitted from discussions with Boris Aronov, Richard Mabry, and Joseph
Malkevitch.

##  8 Appendix: Lower Regular $n$-gon

In order to prove Lemma 3, we first abstract the $S$-shaped regular polygon to
circles, surrounding $Z^{+}$ and $Z^{-}$, passing through $\partial Z_{L}$ and
$\partial Z_{R}$. We call these circles $C^{\pm}_{L}$ and $C^{\pm}_{R}$. We
compute arc lengths through these circles. These arc lengths are clearly an
upperbound on the arcs intersecting $Z$ sandwiched between the circles.

Secondly, for computational convenience, we reflect $Z^{-}$ across the
vertical so that all arcs are to the right of the vertical. The key result is
a “crescent lemma” that is perhaps known under another name, as it is pure
elementary geometry.

###  8.1 The Crescent Lemma

Refer to Fig. 15 throughout the proof below.

###### Lemma 6

Let $\mathcal{C}(r)$ centered on $o$ intersect $C^{-}_{L}$ and $C^{-}_{R}$,
the two circles circumscribed and inscribed about $Z^{-}$. Let $S$ be the
crescent shape between $C^{-}_{L}$ and $C^{-}_{R}$ (shaded blue in the
figure). Then $|\mathcal{C}(r)\cap S|$, the angle subtended from $o$ across
$S$, is a constant independent of $r$.

Proof: Incident to $o$ we label four rays $A,B,C,D$. The arc
$\mathcal{C}(r)\cap S$ subtends an angle ${\beta}$ at $o$, between $B$ and
$D$. To simplify notation, we will use $\angle BD$ to indicate that angle, and
similarly for the other labeled rays.

We first define three of the four rays.

  * •

$A$ is the vertical ray through $o$ and $C^{-}_{L}\cap C^{-}_{R}$.

  * •

$B$ is the ray through the point where $\mathcal{C}(r)$ crosses through
$C^{-}_{L}$.

  * •

$D$ is the ray through the point where $\mathcal{C}(r)$ exits through
$C^{-}_{R}$.

Let ${\beta}=\angle BD$ and ${\gamma}=\angle AB$. Define the fourth ray $C$ so
that $\angle CD={\gamma}$.

Rotate $BD$ rigidly clockwise about $o$ by ${\gamma}$. By definition, this
maps $B$ to $A$, and maps $D$ to $C$. So now we see that ${\beta}=\angle AC$.

Now consider any other $r^{\prime}$ with $\mathcal{C}(r^{\prime})$
intersecting $A$ and $C$. Then the arc between $A$ and $C$ for this
$r^{\prime}$ is exactly ${\beta}$, for this arc is just an arc of the circle
$\mathcal{C}(r^{\prime})$, concentric with $\mathcal{C}(r)$, delimited by
$\angle AC$. Therefore, $|\mathcal{C}(r)\cap S|={\beta}$ for all $r$ crossing
the crescent.

Figure 15: $Z^{-}$ lies in the crescent between $C^{-}_{L}$ and $C^{-}_{R}$.
$\mathcal{C}(r)$ intersects the crescent in an arc of angle ${\beta}$
independent of $r$.

We label the next result as a Proposition rather than a Lemma, because we rely
on a bound calculated from an explicit equation.

###### Proposition 1

The angle ${\beta}=\angle BD$ defined in Lemma 6 is strictly less than
${\alpha}/2$.

Proof: Because all the arc $|\mathcal{C}(r)\cap S|$ are the same length
${\beta}$, it suffice to measure the length of one, say the vertically lowest
arc shown in Fig. 15. We would like to compare this ${\beta}$ to ${\alpha}$,
which is the counterpart in $Z^{+}$. These angles depend on $n$, which
determines the length of each rhomb side. (Here we use unit radius circles
$C^{-}_{L}$ and $C^{-}_{R}$, so the rhomb sides have length $L$ as a function
of $n$.) Even though $n$ is a natural number, in the context of circles
replacing the discrete rhombs, we can calculate that

$\displaystyle{\alpha}$ $\displaystyle=2\pi/n$ $\displaystyle L$
$\displaystyle=2\sin({\alpha}/2)$ $\displaystyle n$
$\displaystyle=\pi/\arcsin(L/2)$

The last equation turns $n$ into a continuous variable.

Using this, I calculated the ratio ${\beta}/{\alpha}$ as a function of this
continuous $n$. The result is shown in Fig. 16.

Figure 16: The ratio ${\beta}/{\alpha}$ for different values of $n$.

For all $n\geq 3$, ${\beta}/{\alpha}<\frac{1}{2}$. It is evident that as
$n\to\infty$, the fraction diminishes to $\frac{1}{3}$, and as $n$ gets small,
the fraction increases but does not reach $\frac{1}{2}$ by $n=3$, the smallest
$n$.444For $n=3$, the polar zonohedron is a combinatorial cube.

A formal proof that ${\beta}/{\alpha}<\frac{1}{2}$ is left for future work,
but in light of this plot and the equations below, there can be little doubt
it holds.

We have established Lemma 3: $|\mathcal{C}(r)\cap Z^{-}|={\beta}<{\alpha}/2$.

###  8.2 Equations

Using $L$ to represent the length of a rhomb side when the radii of
$C^{-}_{L}$ and $C^{-}_{R}$ are both $1$, the equations that lead to the plot
in Fig. 16 are explicitly:

$\tan{\alpha}=\frac{L\sqrt{4-L^{2}}}{2-L^{2}}\;.$ (1)

$\tan{\beta}=\frac{3\left(L^{4}-4L^{2}-\sqrt{4-L^{2}}\sqrt{L^{4}\left(4-L^{2}\right)}\right)}{L\left(9\sqrt{4-L^{2}}L^{2}-36\sqrt{4-L^{2}}+\sqrt{L^{4}\left(4-L^{2}\right)}\right)}\;.$
(2)

${\beta}/{\alpha}=\frac{\arctan(\tan{\beta})}{\arctan(\tan{\alpha})}\;.$ (3)

Calculations with these equations verify that the graph shown in Fig. 16 is
accurate. In particular, a Taylor-series expansion of the ratio
$\tan{\beta}/\tan{\alpha}$ results in an expression that verifies that, as
$n\to\infty$ (and so $L\to 0$), the ratio approaches ${\beta}/{\alpha}$ and
its limit is indeed $1/3$:

$\lim_{L\to 0}\frac{\sqrt{L^{4}}+L^{2}}{6L^{2}}=\frac{1}{3}\;.$ (4)

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