Simple Closed Quasigeodesics on Tetrahedra

Abstract

Pogorelov proved in 1949 that every convex polyhedron has at least three simple closed quasigeodesics. Whereas a geodesic has exactly a π surface angle to either side at each point, a quasigeodesic has at most a π surface angle to either side at each point. Pogorelov’s existence proof did not suggest a way to identify the three quasigeodesics, and it is only recently that a finite algorithm has been proposed. Here we identify three simple closed quasigeodesics on any tetrahedron: at least one through one vertex, at least one through two vertices, and at least one through three vertices. The only exception is that isosceles tetrahedra have simple closed geodesics but do not have a 1-vertex quasigeodesic. We also identify an infinite class of tetrahedra that each have at least 34 simple closed quasigeodesics.

Full text

Citation: O’Rourke, J.; Vîlcu, C.
Simple Closed Quasigeodesics on
Tetrahedra. Information 2022,13, 238.
https://doi.org/10.3390/
info13050238
Academic Editor: Giovanni Viglietta
Received: 25 March 2022
Accepted: 30 April 2022
Published: 5 May 2022
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information
Article
Simple Closed Quasigeodesics on Tetrahedra
Joseph O’Rourke 1,* and Costin Vîlcu 2
1Department of Computer Science, Smith College, Northampton, MA 01062, USA
2“Simion Stoilow” Institute of Mathematics of the Romanian Academy, 014700 Bucharest, Romania;
costin.vilcu@imar.ro
*Correspondence: jorourke@smith.edu
Abstract:
Pogorelov proved in 1949 that every convex polyhedron has at least three simple closed
quasigeodesics. Whereas a geodesic has exactly a
π
surface angle to either side at each point, a
quasigeodesic has at most a
π
surface angle to either side at each point. Pogorelov’s existence proof did
not suggest a way to identify the three quasigeodesics, and it is only recently that a finite algorithm
has been proposed. Here we identify three simple closed quasigeodesics on any tetrahedron: at least
one through one vertex, at least one through two vertices, and at least one through three vertices. The
only exception is that isosceles tetrahedra have simple closed geodesics but do not have a 1-vertex
quasigeodesic. We also identify an infinite class of tetrahedra that each have at least 34 simple
closed quasigeodesics.
Keywords: convex polyhedra; geodesics; quasigeodesics; tetrahedra
1. Introduction
It is well-known that every convex polyhedron has at least three simple closed quasi-
geodesics [
1
], a counterpart to the Lusternik–Schnirelmann theorem that every smooth
closed convex surface has at least three simple closed geodesics. Whereas a geodesic on a
convex polyhedron has exactly
π
surface angle to either side at each point, a quasigeodesic
has at most
π
surface angle to either side of any point. Unlike geodesics, quasigeodesics
can pass through vertices.
As Pogorelov’s result does not lead directly to an algorithm, it was posed as an
open problem to find a polynomial-time algorithm to construct at least one simple closed
quasigeodesic: Open Prob. 24.2 [
2
]. Even a finite algorithm was not known. Recently
there has been progress on this question [
3
], and an exponential-time algorithm has been
developed [4].
In this paper, we describe the three quasigeodesics guaranteed by Pogorelov, in the
particular case of tetrahedra.
In reference [
5
], we conjectured that every convex polyhedron has either a simple
closed geodesic, or a simple closed quasigeodesic through exactly one vertex. We proved
this conjecture for doubly-covered convex polygons [
5
], Chapter 17. Here we prove it for
all tetrahedra.
Theorem 1.
Every tetrahedron has a 2-vertex quasigeodesic, a 3-vertex quasigeodesic, and a simple
closed geodesic or a 1-vertex simple closed quasigeodesic.
Our result complements in some sense earlier work that determined closed geodesics
on simplices [6,7].
Many ellipsoids admit only three simple closed geodesics. (However, not all ellipsoids:
If sufficiently oblate, there are other simple closed geodesics [
8
].) Our second result estab-
lishes that many tetrahedra have an unexpected wealth of simple closed quasigeodesics.
Information 2022,13, 238. https://doi.org/10.3390/info13050238 https://www.mdpi.com/journal/information

Information 2022,13, 238 2 of 20
Theorem 2.
There exists an open set
O
in the space of all tetrahedra, each element of which has at
least 34 simple closed quasigeodesics.
All our proofs are constructive and lead to algorithms and constant-time in an ap-
propriate model of computation. See [
3
] for a discussion of models of computation for
quasigeodesics.
Alongside the above results, we also obtain a characterization of isosceles tetrahedra,
defined in Section 2. It complements results in [
9
,
10
]. A geodesic loop is a simple closed
curve that is a geodesic everywhere except at one point, the loop point.
Theorem 3. A tetrahedron is isosceles if and only if it admits no geodesic loop at a vertex.
After presenting our proofs, we conclude the paper with a short section of remarks
and open questions.
Notation
Here we list basic notation that we use throughout. More specialized notation and
preliminaries will be introduced where needed.
• Vertices of tetrahedron T:a,b,c,d;
• Face Ais opposite a; So: A=bdc,B=cda,C=adb,D=abc;
•
Face angles are specified by vertex and face. So the three face angles incident to vertex
aare: aB,aC,aD; etc. See Figure 1;
• Complete angle at a:θa=aB +aC +aD;
• Vertex curvature at a:ωa=2π−(aB +aC +aD).
For succinctness, we will often use the symbol
Qk
as shorthand for a “
k
-vertex simple
closed quasigeodesic”.
B
C
D
A
b
c
d
a
B
C
D
A
b
c
d
a
b
c
d
a
cBcD
cA
bC
bD
bA
dB
dC
dA
aB
aC
aD
Front
Back
Figure 1. A=bdc,B=cda,C=adb,D=abc.
2. Q0: Simple Closed Geodesics
We use the Gauss–Bonnet theorem in two forms:
1. The total curvature at the four vertices sums to 4π;
2. The turn τof a closed curve plus the curvature enclosed equals 2π:τ+ω=2π.
By the first form of Gauss–Bonnet, a simple closed geodesic
Q0
splits the vertex set of a
convex polyhedron into two subsets, the total curvature of each being 2
π
. Alexandrov [
11
]

Information 2022,13, 238 3 of 20
(pp. 377–378) observed that such a condition is uncommon among all convex polyhedra.
This fact was further refined by Gruber [
12
] (as a preliminary step of his general result
proof) and by Gal’perin [
13
] (for polyhedra homeomorphic to the sphere). This led to a
proof that, for a fixed number of vertices, the set of convex polyhedra having a simple
closed geodesic is closed and has measure zero in the space of all convex polyhedra.
Particularizing to our framework, there is a special class of tetrahedra which do have
many simple closed geodesics. An isosceles tetrahedron (also called an isotetrahedron, a
tetramonohedron, or an isohedral tetrahedron) is a tetrahedron whose four vertices each have
curvature π, or, equivalently, all four faces are congruent acute triangles.
It is a beautiful result that isosceles tetrahedra are the only convex surfaces that have
arbitrarily long simple closed geodesics [
6
,
14
]. Consequently, they have infinitely many such
geodesics. This wealth of Q0s is balanced in some sense by the non-existence of Q1s.
Lemma 1. No isosceles tetrahedron has a 1-vertex quasigeodesic.
This lemma complements the remark in [
9
] and was reproved in [
15
], that a regular
tetrahedron has no geodesic loop. See also [
10
] for a characterization of isosceles tetrahedra
as the only tetrahedra having three distinct minimal loops through any point on the face.
Proof.
Refer to Figure 2. Here we use the second form of Gauss–Bonnet. Let
Q
be a 1-vertex
quasigeodesic through vertex
d
, with
a
and
b
strictly to
Q
’s left, and
c
strictly to
Q
’s right.
Since
ωa+ωb=
2
π
,
Q
must have no turn,
τ=
0, to its left at
d
, and turn
τ=π
to its right
at
d
. Having no turn to its left means the total angle of
π
is to the left of
Q
at
d
. Turning
π
to the right means that
Q
turns around completely, folding back on itself, which then forces
Q
to contain vertex
c
. Thus,
c
does not lie strictly to
Q
’s right:
Q
is a 2-vertex quasigeodesic,
not a 1-vertex quasigeodesic.
a b
c
d
d
d
D
D
C
A
B
B
C
A
ab
cd d
d
(a)
(b)
Figure 2.
(
a
,
b
) Unfoldings of two isosceles tetrahedra. Quasigeodesic
Q
(red) is a degenerate doubling
of edge dc.
With a different argument, briefly presented next, one can prove a stronger result. No
isosceles tetrahedron has a geodesic loop at a vertex.
To see this, assume the isosceles tetrahedron
T
admits a geodesic loop
Λ
at its vertex
a
. Star-unfold
T
with respect to
a
, resulting in an acute triangle
¯
T
. The star-unfolding is
discussed in Section 4.2. It cuts the three edges incident to
a
. Tile the plane with
¯
T
. In that
tiling,
Λ
corresponds to a segment
˜
Λ
joining two images of
a
, say
aj
and
ak
. However,
aj
and
ak
are determining either a side or a diagonal apexed at images of
a
. In both cases,
˜
Λ
would contain the image of another vertex of
T
, showing that
Λ
necessarily passes through
the respective vertex.
3. Q1: 1-Vertex Quasigeodesics
We have just seen in Section 2that isosceles tetrahedra have no 1-vertex quasigeodesic,
but do have simple closed geodesics. The goal of this section is to prove this theorem:

Information 2022,13, 238 4 of 20
Theorem 4.
Every non-isosceles tetrahedron has at least one 1-vertex simple closed quasigeodesic.
Notice that Theorem 3follows immediately from the above result and the remark
ending the previous section.
In the remainder of this section, we assume all tetrahedra are not isosceles.
Properties of Q1.
A quasigeodesic
Q1
through exactly one vertex
v
on a tetrahedron
T
must satisfy these
conditions.
(1) Qmust form a geodesic loop with loop point v.
(2)
To satisfy the Gauss–Bonnet theorem,
Q
must partition the other three vertices two
to one side and one to the other, such that the total curvatures to each side of
Q
are at
most 2
π
(and not both sides equal to 2
π
, for then there is no curvature at
v
and it is
not a vertex).
Of course the quasigeodesic angle criterion must be satisfied at v.
Sketch of Proof for Theorem 4.
The proof follows a case analysis based first on how many curvatures are greater than
π
, and second on the distances from low-curvature vertices to high-curvature vertices. The
curvatures greater than
π
lead to convex vertices in unfoldings, and which vertices are
closest to these high-curvature vertices permits concluding that particular geodesic loops
are inside certain disks and so live on
T
. Then the angles to either side at the geodesic loop
vertex must be verified to be at most πto conclude it is a quasigeodesic.
3.1. Case 1
For Case 1, assume exactly one vertex has curvature exceeding
π
:
ωa>π
. Let
d
be
the closest vertex to
a
among
b
,
c
,
d
. Then star-unfold
T
with respect to
d
, as illustrated in
Figure 3: Faces
C
,
D
,
B
are incident to
a
, and face
A
is attached to face
D
along edge
bc
.
Label the three images of das d1,d2,d3as illustrated.
We claim that
Q=d1d2
(red in the figure) is a simple closed quasigeodesic containing
just the vertex
d
. It will help to view
Q
as directed from
d1
to
d2
. Note that, because
ωa>π
,
θa<π.
First note that, because
|ad|
is shorter than or equal to
|ab|
and
|ac|
,
Q
separates
b
,
c
from
a
:
Q
is a chord of a circle of radius
|ad|
centered on
a
, and
b
and
c
lie on or outside
that circle.
Next,
Q
is a straight segment between two images of vertex
d
in this unfolding, and so
a geodesic loop on
T
includes
d
. It remains to show that the angle to either side of
Q
at
d
is ≤π.
Let
α1
and
α2
be the angles of
4d1d2a
above
Q
, as illustrated in the figures. Then it is
immediate that α1+α2<π.
Let
β1
and
β2
be the angles
∠d2d1b
and
∠d1d2c
below
Q
, as illustrated. The angle of
Q
to the right side of
d
we seek to bound is
β1+β2+dA
. The reason angle
dA
is to the
right of Qis: (a) d A is incident to vertex d, and (b) face 4bcd3is right of d1d2.
Now note that the external angles at
b
and
c
in the unfolding are
ωb
and
ωc
respectively.
Because
ωb
,
ωc<π
, the triangle
4d1d2d3
includes face
A
and so includes
b
and
c
. Therefore,
β1+β2+dA
must be smaller than
π
because those three angles are each smaller than the
corresponding angles of 4d1d2d3.
Therefore, we have proved that the angle to the right of
Q
at
a
is less than
π
, and so
Q
is a simple closed quasigeodesic as claimed.

Information 2022,13, 238 5 of 20
λ2
λ1
ρ1
b
c
a
D
C
B
A
Q
dA
d2
d1
d
3
ρ2b
ca
d
A
D
B
Q
(a) (b)
Figure 3.
Case 1. (
a
) Unfolding of a tetrahedron when
ωa>π
and the other three curvatures are
<π
.
d
is closest to
a
. Curvatures at
a
,
b
,
c
,
d
are 282
◦
, 140
◦
, 173
◦
, 124
◦
. Auxiliary yellow triangles added.
(b) The quasigeodesic in 3D.
3.2. Case 2
For Case 2, assume
T
has at least two vertices with curvatures more than
π
:
ωa≥
ωb≥π.
The reader may find it easier to follow the proof on the particular case of
T
, a doubly
covered quadrilateral. The next argument, however, is valid for the general situation.
We will consider geodesic loops
Qv
at vertex
v
, with
v∈ {a
,
b
,
c
,
d}
suitably chosen, as
constructed in Case 1.
Consider a closest vertex vto aamong b,c,d.
Case 2.1:
v=b
:
|ab|≤|ac|
,
|ad|
:
b
is closest to
a
. Then there exists a geodesic loop
Qb
at
b
which
separates
a
from
c
,
d
. We now justify this claim. As illustrated in Figure 4, the segment
Qb=b1b2
cannot be blocked by vertex
a
because
θa≤π
, and cannot be blocked by vertices
cor d, because they fall outside the circle centered at aof radius |ab|.
An equivalent but different way to view this is as follows. View
a
as the apex of
T
,
and remove the base
bcd
. Extend the faces
B
,
C
,
D
incident to
a
to a cone
C
. Then on
C
there
are geodesic loops based on each of b,c,d. Because bis closest to a, the loop Qblives on T.
Because
ωb≥π
, the complete angle
θb
at
b
is at most
π
, hence
Qb
has less than
π
to each side, and so is a simple closed quasigeodesic. This happens irrespective of
ωc
,
ωd
being larger or smaller than π.
d
c
a
b2
b1
b3
D
C
BA
Figure 4. Case 2.1: bis closest to a. Curvatures at a,b,c,dare 196◦, 190◦, 159 ◦, 175◦.
Case 2.2:
v=c; i.e., c(or equivalently d) is closest to a:|ac|≤|ab|,|ad|.
Consider now a closest vertex wto bamong a,c,d.
Case 2.2.1:

Information 2022,13, 238 6 of 20
This is handled by Case 2.1 with the roles of aand breversed.
Case 2.2.2:
|bd|≤|ba|,|bc|.|ac|≤|ab|,|ad|:cis closest to aand dis closest to b.
As illustrated in Figure 5a,b, the geodesic loop
Qc
at
c
separates
a
from
b
,
d
because
c
is closest to
a
and so
b
,
d
cannot interfere, and the geodesic loop
Qd
at
d
separates
b
from
a,c, because dis closest to band so a,ccannot interfere.
Now we argue that one or the other of
Qc
,
Qd
is a quasigeodesic. Note first that,
because
ωa
,
ωb≥π
, the angle at
c
toward
a
(left in the figure), and the angle at
d
toward
b
(right in the figure), are
<π
in
4ac1c2
and
4bd1d2
respectively. Next we examine the
angles to the other side of Qc,Qd.
With
a
,
b
separated by
Qc
and
Qd
, those two geodesic loops bound a vertex-free region
R
on the surface of
T
, homeomorphic to a cylinder. As Figure 5c illustrates, cut
R
is
isometric to a planar quadrilateral
c1c2d1d2
. Because the quadrilateral angles sum to 2
π
, it
cannot be that the two angles at
c
and the two angles at
d
both exceed
π
. At least one must
be
≤π
. The respective geodesic loop is therefore a quasigeodesic. Figure 6shows the two
geodesic loops in 3D.
d
b
a
c2
c1
c3
d
b
a
c2
c1
c3
b
c
a
d2
d1
d3
D
C
BA
D
C
BA
b
a
c2
c1d2
d1
D
C
B
A
(a) (b)
(c)
Figure 5.
Case 2.2.2:
c
is closest to
a
and
d
is closest to
b
. (
a
)
Qc
red. (
b
)
Qd
green. Curvatures at
a,b,c,dare 226◦, 205◦, 138◦, 151◦. (c) The two segments c1c2and d1d2are slightly non-parallel.
b
a
c
d
D
C
B
QcQd
Figure 6. Case 2.2.2: Qcand Qdon the 3D tetrahedron unfolded in Figure 5.Qcis a quasigeodesic.
Case 2.2.3:
w=c; i.e., cis closer than dto both aand b:|bc|≤|ba|,|bd|.|ac|≤|ab|,|ad|.

Information 2022,13, 238 7 of 20
As illustrated in Figure 7a, just as in Case 1, because
c
is closest to both
a
and
b
, there
exist geodesic loops
Qc
,
Q0
c
at
c
such that
Qc
separates
a
from
b
,
d
and
Q0
c
separates
b
from
a,d: vertex dcannot interfere with either Qc=c1c2nor Q0
c=c1c3.
However, although one can construct two geodesic loops at
d
,
Qd
on the cone apexed
at
a
and
Q0
d
on the cone apexed at
b
, they may not stay inside
T
. We now argue that at least
one of Qd,Q0
dlives on T.
Figure 7b illustrates the situation when one, in this case
Qd=d1d2
, falls outside
T
.
It should be clear that
d1
can see
c
, because
ωa
,
ωb≥π
, so the two faces
C∪D
form a
convex quadrilateral. The exterior angle gap at
c
can only block visibility from one of
d2
,
d3
.
Another way to view this is that, if
Qd=d1d2
does not live on
T
, then it separates
a
,
c
from
b
. However, then
Q0
d=d1d3
even more so separates
a
,
c
from
b
, and lives on (remains
inside the surface of) T.
d
c2
c1
b
a
c3
b
c
d2
d1
d3
C
D
C
BA
aD
BA
(a) (b)
Figure 7.
Case 2.2.3:
c
is closer than
d
to both
a
and
b
.
Qc
,
Q0
c
red (
a
),
Qd
,
Q0
d
green (
b
). Curvatures at
a,b,c,dare 284◦, 200◦, 58◦, 178◦.
So now we have two geodesic loops, say,
Qc=c1c2
and
Qd=d1d2
. Then, just as in
Case 2.2.2, we have identified a vertex-free region
R
, with one geodesic excluding
a
to the
left, the other bto the right. Following the same logic as in Case 2.2.2, we conclude that at
least one of the angles toward
R
at
c
or
d
must be
≤π
. It is straightforward that the angles
to the other side are ≤π:4ac1c2and 4bc1c2and 4bd1d3.
This completes the proof of Theorem 4. There is a sense in which this theorem cannot
be strengthened, because there are tetrahedra that have only one such
Q1
. (That more
complex “spiraling” geodesic loops are not possible is a consequence of Lemma 8 in [
4
]).
We claim that Figure 8is an example of a tetrahedron with only one simple Q1.
bcb
a
b
d
b
c
d
a
(a) (b)
Figure 8.
Tetrahedron with just one
Q1
. (
a
) Unfolded. (
b
) Coordinates of
a
,
b
,
c
,
d
:
(−0.65, 0, 1.56),(0, 0, 0),(1, 0, 0),(0.89, 0.25, 0).

Information 2022,13, 238 8 of 20
4. Q2: 2-Vertex Quasigeodesics
The goal of this section is to prove this theorem:
Theorem 5. Every tetrahedron has a 2-vertex simple closed quasigeodesic.
4.1. Degenerate 2-Vertex Quasigeodesics
If a tetrahedron has at least two vertices with curvature of at least
π
, then the complete
angle incident to those two vertices is each
≤π
. So the double edge connecting them
constitutes a degenerate simple closed quasigeodesic: at each endpoint, the angle to one
side is 0, and to the other side at most
π
. We call such a degenerate quasigeodesic an
edge-loop.
Define a tetrahedron as pointed if it has just one vertex with curvature exceeding
π
.
We will consistently use the label afor that vertex, so it is pointed at a.
The remainder of this section concentrates on pointed tetrahedra. First, we review
some tools used in the proof.
4.2. Star-Unfolding and Cut Locus
Ageodesic segment on a Pis a shortest path between its extremities.
The cut locus
C(x)
of the point
x
on
P
is the set of endpoints (different from
x
) of all
nonextendable geodesic segments (on the surface P) starting at x.
The star-unfolding of
T
with respect to its vertex
v
is obtained by cutting along the
edges incident to vand unfolding to the plane.
We need one property of the star-unfolding that derives from [
16
] and is stated as
Lemma 3.3 in [
17
]. To avoid introducing notation not needed here, we specialize this lemma
to our situation:
Lemma 2.
Let
Sv
be the star-unfolding of a tetrahedron from vertex
v
. Then the cut locus
C(v)
is
the restriction to
Sv
of the Voronoi diagram of the images of
v
. Moreover,
C(v)
lies entirely in the
face opposite
v
. In particular, the degree-3ramification point
y
lies in that face, and is connected by
segments to that face’s three vertices.
One can see this intuitively: If
y
were interior to a face incident to
v
, then there would
be three paths from
v
to
y
: one straight in 3D, but the other two with some 3D aspect, a
contradiction.
Sketch of Proof for Theorem 5.
The proof first establishes a visibility relation in the star-unfolding that yields an edge-
loop. Second, the quasigeodesic condition is proved to hold at both ends of the geodesic,
thereby establishing a 2-vertex simple closed quasigeodesic.
4.3. Visibility
Let
T
be a tetrahedron pointed at
a
, and
Sa
the star-unfolding of
T
with respect to
a
.
Label the three images of
a
as
ab
,
ac
,
ad
, with
ab
opposite
b
, and similarly for
ac
and
ad
. Say
that two vertices
v
,
u∈Sa
are visible to one another if the segment
vu
is nowhere exterior to
Sa, and touches ∂Saonly at uand v. So uv is a vertex-to-vertex diagonal of Sa.
Lemma 3.
If
T
is pointed at
a
, then at least one of
b
,
c
,
d
is visible in
Sa
to
ab
,
ac
,
ad
respectively.
Sometimes there is only one such visibility relation.
Proof. The tightness claim of the lemma is established by Figure 9.
Because
T
is pointed at
a
,
Sa
is reflex at
b
,
c
,
d
. Partition the plane into six regions
by extending the three edges of
4bcd
. Call the cone regions
Cb
,
Cc
,
Cd
incident to
b
,
c
,
d
respectively. If
ab∈Cd
or
ab∈Cc
, then
ab
cannot see
b
, and similarly for
c
and
d
. So, for
contradiction, we will show that if we have two visibility segments
abb
,
acc
blocked, then
adcan see d.

Information 2022,13, 238 9 of 20
Let H(x,y)be the halfplane to the left of the directed segment xy.
c
d
ab
ac
ad
b
Figure 9. Pointed Twith just one visible segment, add. Both aband acare blocked.
Case 1: Two a-images in one cone.
Let
ab
,
ac∈Cd
, with no loss of generality. To contradict the claim of the lemma, we
would need
ad
in either
Cc
or
Cb
. Again with no loss of generality, let
ad∈Cc
. This requires
ad∈H(d
,
c)
, for the boundary of
H(d
,
c)
is the (lower) boundary of
Cc
. See Figure 10. At
the same time, it must be that
ad∈H(b
,
ac)
in order for
Sa
to be reflex at
b
. However, it is
not possible for
ad
to be in both of those halfplanes below
bc
, for the intersection is only
non-empty above cd.
d
bc
ac
ad
ab
Cd
Cc
Cb
H(d,c)
H(b,ac)
Figure 10. Case 1. adcannot be in both Ccand H(b,ac).
Case 2: Two a-images in two different cones.
Let
ab∈Cc
and
ac∈Cd
, with no loss of generality. Then we seek to show that
ad
can
see
d
. If
ad∈Cc
or
ad∈Cb
, then it is blocked from seeing
d
. If
ad∈Cc
, then both
ad
and
ab
are in the same cone, already handled by Case 1. So we must have
ad∈Cb
, which requires
ad∈H(b,d), as the boundary of H(b,d)is the lower boundary of Cb. See Figure 11.

Information 2022,13, 238 10 of 20
By Lemma 2, the cut locus is entirely inside
4bcd
. Thus, each of
b
,
c
,
d
connects by a
segment to the ramification point
y
. Each of these segments of the cut locus is a subsegment
of a bisector. In particular, the bisector of aband adlies along the segment cy.
Let
β
,
γ
,
δ
be the angles of
4bcd
. Now, in order for the bisector to penetrate into
4bcd
at
c
,
ad
must lie in the halfplane
H(c
,
x)
, where the ray
cx
makes an angle of
γ
with respect
to the lower boundary of cone
Cc
. In the example shown in Figure 11, it is clear that it
is impossible for
ad
to be in both
H(c
,
x)
and in
H(b
,
d)
, because those halfplanes only
intersect above bc.
However, if
γ
is larger, then the two halfplanes could intersect below
bc
, and thus
there is a possible placement of
ad
in
Cb
satisfying the bisector condition. It is easy to see
that the critical inequality is that we need 2γ>π−βfor a placement to be available.
However, now a similar inequality is needed for the placements of
ac
and
ab
, for each
of those to both lie in the appropriate cones, and lead to bisectors that penetrate into
4bcd
at band drespectively. Thus these three inequalities must hold:
2γ>π−β
2β>π−δ
2δ>π−γ.
The reason that the inequalities are strict is that equality implies that the boundaries
of
H(b
,
d)
and
H(c
,
x)
are parallel, and there is no spot at which to locate
ad
(and similarly
for acand ab). Summing the three inequalities leads to 2π>2π, a contradiction.
Therefore, it is not possible to have all three of
ab
,
ac
,
ad
located in three different cones.
Since the two Cases cover all possibilities, at least one image of
a
must lie in a region
between cones, and so can see the corresponding vertex of 4bcd.
d
bc
ac
ad
ab
Cd
Cc
Cb
β
δ
γ
γ
γ
H(b,d)
H(c,x)
x
bisector
Figure 11. Case 2: admust lie in H(c,x)for the bisector of adand abto penetrate 4bcd at c.

Information 2022,13, 238 11 of 20
As a consequence of Lemma 3, we have established that every pointed tetrahedron
has an edge-loop. Our next goal is to show that this edge-loop is in fact a simple closed
quasigeodesic, which requires at most πangle at the endpoints of the visibility segment.
Lemma 4.
Every pointed tetrahedron has a non-degenerate edge-loop forming a simple closed
quasigeodesic.
Proof.
Let the visibility segment be
add
without loss of generality. First, because the
curvature at
a
is
>π
, the total angle there is
<π
, and so the quasigeodesic condition is
satisfied at that end. Now we turn to the other end at d.
Consider the segments incident to
ad
. Segment
adb
is left of and
adc
is right of
add
, just
by the counterclockwise labeling convention for the base
bcd
. Now, because
ωb<π
and
ωc<π
, the angles at
b
and at
c
in the unfolding
Sa
are both reflex. See Figure 12. Therefore
the segment endpoints incident to
ad
follow the counterclockwise order
ab
,
c
,
d
,
b
,
ac
. There-
fore, the angle at
d
right of
add
is the apex of the triangle
4abdad
and the angle to the left is
the apex of
4acdad
. Because both angles are
<π
, we have established that the edge-loop is
in fact a simple closed quasigeodesic.
c
d
ab
ac
ad
b
Figure 12. 4abdadand 4acdad: triangle angles at dare <π.
Note Lemma 4holds for every visibility segment: although there is always at least
one, there can be as many as three.
Together with the degenerate 2-vertex quasigeodesics on non-pointed tetrahedra, we
have established Theorem 5.
4.4. A Geometrical Interpretation of Edge Loops
In this subsection, we provide a geometrical interpretation of edge loops of tetrahedra.
Our construction is based on Alexandrov’s Gluing Theorem and the technique of vertex
merging; see [5] for a description and other applications of these tools.
We are still in the case of tetrahedra
T
pointed at
a
, with an edge loop based on the
edge ad, see Figure 12. Then we have θa+θd<2π.
Denote by
n
the mid-point of the edge
ad
. Cut
T
along
ad
and glue back differently,
via Alexandrov’s Gluing Theorem. Precisely, we identify
a
and
d
, and for the two banks of

Information 2022,13, 238 12 of 20
the cut,
an
with
dn
. Because
θa+θd<
2
π
, the result after gluing is a convex polyhedron
F
with 5 vertices:
b
,
c
,
n1
and
n2
obtained from
n
, and
w=ab
obtained from both
a
and
b
. By
construction, the curvatures at n1,n2are precisely π.
On
F
, we can merge the vertices
c
and
n1
, and respectively
d
and
n2
, to obtain a new
convex polyhedron
∆
with three vertices:
w
,
u
obtained from
c
and
n1
, and
v
obtained from
dand n2. Therefore, ∆is a doubly covered (obtuse) triangle. See Figure 13.
The edge loop of
T
based on
ad
corresponds to the geodesic loop at
w
on
∆
obtained
by doubling the height hfrom w.
a
n1
143
an2
d
n1
a
n1
c
a
234
B
d
D
b
c
d
a
v
n2
A
B
CD
b
c
d
a
aa
A
B
CD
(a)
(b) u
Figure 13.
(
a
) Unfolding of a pointed tetrahedron, with edge-loop red. (
b
) One side of
∆
, an obtuse
doubly-covered triangle with edge-loop along the altitude. (n1is on the back side.)
5. Q3: 3-Vertex Quasigeodesics
The goal of this section (a revision of [18]) is to prove this theorem:
Theorem 6.
Every tetrahedron
T
has at least one face
F
whose boundary
∂F
is a 3-vertex simple
closed quasigeodesic Q3.
We first establish two preliminary lemmas that will be used in the proof.
5.1. Preliminary Lemmas
Lemma 5.
Let
α1
,
α2
,
α3
be the face angles incident to vertex
v
of a tetrahedron
T
. Then the angles
satisfy the triangle inequality:
α1<α2+α3
, and similarly
α2<α1+α3
, and
α3<α1+α2
. The
inequalities are strict unless T is flat.
Proof. See Lemma 2.8 in [5].
To simplify the calculations, angles will be represented in inequalities in units of
π
:
1
≡π
, 2
≡
2
π
, etc. Thus, under this convention, each of the 12 face angles of a tetrahedron
lies in (0, 1).
We say that “face
F
fails at vertex
v
” if the two angles incident to
v
not in
F
exceed
π
.
So, for face
A
to fail on vertex
b
, then among the three face angles
bA
,
bC
,
bD
incident to
b
,
the two angles not in
A
satisfy
bC +bD >
1. This means that
∂A
is not a quasigeodesic,
because to one side—the other side from bA—the angle exceeds π.

Information 2022,13, 238 13 of 20
Example.
Figure 14 shows a tetrahedron with
∂C
a quasigeodesic, but none of the other face
boundaries is a quasigeodesic. Thus the “at least one” claim of Theorem 6cannot be
strengthened. Its vertex coordinates are:
a,b,c,d= (−3.54, 1.98, 4.58),(0, 0, 0),(1, 0, 0),(4.91, 3.24, 0).
For example, back face
C
does not fail at vertex
b
:
bD +bA =
125
◦+
33
◦=
159
◦<π
.
Front face Bfails at vertex c:cD +c A =48◦+140◦=188◦>π.
A
b
c
d
a
B
C
D
Figure 14.
The (red) boundary of shaded back face
C=abd
is a quasigeodesic, but none of
∂A
,
∂B
,
∂D
are quasigeodesics.
Lemma 6. If a face A fails at a vertex b, then ω(b)<1.
Proof. Since face Afails at b, by definition, bC +bD >1. Therefore
ω(b) = 2−(bA +bC +bD)
ω(b) = 2−(bC +bD)−bA
ω(b)<1−bA
ω(b)<1
This establishes the claim of the lemma.
5.2. Case Analysis
We now undertake a case analysis to show that it is not possible for all four faces
of tetrahedron
T
to fail at vertices. The cases, illustrated in Figure 15, distinguish first
the number of distinct vertices among the four face-failures, and second, the pattern of
the failures.

Information 2022,13, 238 14 of 20
A
B
C
D
a
d
c
b
A
B
C
D
a
d
c
b
A
B
C
D
a
d
c
b
A
B
C
D
a
d
c
b
A
B
C
D
a
d
c
b
Case 1
Case 2a Case 2b
Case 3a Case 3b
Figure 15. Failures. Case 1: at 4 vertices. Case 2: at 2 vertices. Case 3: at 3 vertices.
The proof analyzes the 12 face angles of
T
, and shows that the set of solutions in
(
0, 1
)12
is empty (under the convention that each angle is in
(
0, 1
)
). So we are representing
tetrahedra by their 12 face angles. The four faces each have a total of
π
angle, which
reduces the dimension of the tetrahedron configuration space from 12 to 8. It is known
that in fact the configuration space is 5-dimensional, not 8-dimensional [
19
], but the proof
to follow works without including the various additional trigonometric relations that
tetrahedron angles must satisfy. It suffices to use linear equalities and inequalities among
the 12 face angles.
Case 1: 4 vertices.
Suppose first that each of the four faces
A
,
B
,
C
,
D
fail on four distinct vertices. Then
Lemma 6shows that
ω(v)<
1 for each vertex
v
. But then
∑ω(v)<
4, contradicting the
Gauss–Bonnet theorem.
Case 2a: 2 vertices, 3 +1.
Suppose now that the four faces fail on a total of two vertices. This can occur in two
distinct ways: three faces fail on one vertex, which we call Case 2a, or two faces fail each on
two vertices, Case 2b. Say that bis the vertex at which three faces fail. We then have:

Information 2022,13, 238 15 of 20
Afails at b:bC +bD >1
Bfails at a:aC +aD >1
Cfails at b
Dfails at b.
It turns out that we do not need to use the fact that
C
and
D
fail at some vertices, so
the implied inequalities are suppressed. Summing the failure inequalities above leads to
a contradiction:
(aC +bC) + (aD +bD)>2
(1−dC) + (1−cD)>2
2>2+ (dC +cD)
0>dC +cD.
This is a contradiction because all angles have a positive measure.
Case 2b: 2 vertices, 2 +2.
This follows the exact same proof, as again
C
and
D
failures are not needed to reach a
contradiction.
Case 3a: 3 vertices, double outside.
The three vertices at which faces fail bound a face, say
A
. One vertex of
A
, say
b
,
is “doubled” in the sense that two faces fail at
b
. Case 3a is distinguished in that neither
face failing on
b
is the three-vertex face
A
(Swapping
B
to fail on
c
and
A
to fail of
d
is
symmetrically equivalent to the case illustrated).
We again do not need all failures, in particular, we only need those for faces
B
and
D
:
Afails at c
Bfails at d:dA +dC >1
Cfails at b
Dfails at b:bA +bC >1.
Adding these inequalities leads to the same contradiction:
(bA +d A) + (bC +dC)>2
(1−cA) + (1−aC)>2
0>aC +cA;
again a contradiction.
Case 3b: 3 vertices, double inside.
In contrast to Case 3a, in this case, one of the faces that fail on
b
is the three-vertex face
A. (Swapping Bto fail on c,Dto fail on b, and Cto fail on d, is symmetrically equivalent.)
This is the only difficult case, and the only case in which the triangle inequalities guaranteed
by Lemma 5are needed.
The angles of face
A
satisfy
bA +c A +dA =
1. Assume without loss of generality that
bA ≤c A ≤d A. Three faces, B,C,Dfail at the three vertices of face A:d,b,crespectively.
To build intuition, we first run through the proof for specific A-face angles:

Information 2022,13, 238 16 of 20
(bA,cA,dA)=(0.1, 0.3, 0.6)
Afails at b
Bfails at d:dA +dC >1 : dC >0.4
Cfails at b:bA +bD >1 : bD >0.9
Dfails at c:cA +cB >1 : cB >0.7
Note 0.4
+
0.9
+
0.7
=
2; this holds for arbitrary
A
angles. Now apply the triangle
inequality to each of dB,cB,dC:
bD <bA +bC :bC >bD −bA :bC >0.8
cB <cA +cD :cD >cB −cA :cD >0.4
dC <dA +dB :dB >dC −dA :dB >−0.2
Note 0.8 +0.4 −0.2 =1; this again holds for arbitrary Aangles.
Triangle face Dsatisfies: bD +cD +aD =1.
bD >0.9
cD >0.4
bD +cD >1.3
bD +cD +aD >1.3 >1,
which contradicts bD +cD +aD =1.
Without specific angles assigned to
(bA
,
cA
,
dA)
, the argument is less transparent.
Again assume that bA ≤cA ≤dA.
Afails at b
Bfails at d:dA +dC >1 : dC >1−d A
Cfails at b:bA +bD >1 : bD >1−bA
Dfails at c:cA +cB >1 : cB >1−cA.
Note the sum of the above three right-hand sides is 3
−(dA +b A +c A) =
2. Now
apply the triangle inequality to dB,cB,dC:
bD <bA +bC :bC >bD −bA :bC >1−2·b A
cB <cA +cD :cD >cB −cA :cD >1−2·cA
dC <dA +dB :dB >dC −dA :dB >1−2·dA.
Note the sum of the above three right-hand sides is 3
−
2
(dA +b A +cA) =
1. Face
D
’s angles satisfy
bD +cD +aD =
1. Now we reach a contradiction using the inequali-
ties above.
bD >1−bA
cD >1−2·cA
bD +cD >2−(bA +2·cA).
We have
(bA +
2
·cA)≤
1 because
bA +c A +dA =
1 and
cA ≤d A
. Of course every
angle is positive, so aD >0. So we have:
bD +cD >1
bD +cD +aD >1,

Information 2022,13, 238 17 of 20
which contradicts bD +cD +aD =1.
That the inequalities for each of the above cases cannot be simultaneously satisfied has
been verified by Mathematica’s
FindInstance[]
function, which uses Linear Programming
over the rationals to conclude that the set of solutions in R12 is empty.
Replacing the triangle inequalities with equalities when the tetrahedron is flat (e.g.,
aB =aC +aD instead of aB <aC +aD) again leads to the same contradiction.
We have thus established Theorem 6and, together with remarks in Section 2,
Theorems 4–6establish Theorem 1.
6. Tetrahedra with Many Q1,2,3
As mentioned in the Introduction (Section 1), one cannot expect there to exist more
than three simple closed quasigeodesics on a general convex surface, so one could expect
that the same fact also holds for general tetrahedra.
In this section we provide an open subset of the space of tetrahedra, each tetrahedron
of which has (unexpectedly) many such quasigeodesics.
Let
T
be the space of all tetrahedra in
R3
, with the topology induced by the usual
Pompeiu–Hausdorff metric. Two polyhedra in
T
are then close to each other if and only if
they have close respective vertices.
The goal of this section is to prove the theorem previously stated in the Introduction:
Theorem 2.
There exists an open set
O
of tetrahedra, each element of which has at least 34 simple
closed quasigeodesics.
We call a tetrahedron f-acute if all its faces are acute triangles.
Lemma 7. The set Fof f-acute tetrahedra is open in T.
Proof.
The face angles at the vertices of
T
depend continuously on the vertex positions in
R3. Once they are <π, they remain so in a neighborhood.
We further restrict our study to a special open subset
O
of
F
, of tetrahedra near a
regular tetrahedron, all having three vertices of curvature
>π
. The introduction of these
tetrahedra is justified by the considerations in Section 4.
We start with a regular tetrahedron of apex
a
and horizontal base
bcd
and move
a
downward a short distance along a vertical line. The new tetrahedron
N
has base
vertices of curvatures slightly larger than
π
and top vertex
a
of curvature slightly less
than
π
. Moreover, all faces of
N
remain acute triangles. So we consider
O
to be a small
neighborhood of N.
The next lemma can be proved with an argument similar to Lemma 7’s proof.
Lemma 8. All tetrahedra in Oare f-acute and have three vertices of curvature >π.
The following is a particular case of Lemma 17.2 in [5].
Lemma 9.
Assume the tetrahedron
T
has a simple closed quasigeodesic
Qk
through
k≥
1vertices,
such that its left and right angles at each of the
k
vertices are all strictly less than
π
. Then, all
tetrahedra sufficiently close to T in Thave such a quasigeodesic.
In view of Lemmas 8and 9, it suffices to count the simple closed quasigeodesics on
our reference tetrahedron N∈ O.
• We saw that a general tetrahedron has no Q0;
•
There exists at least one
Q1
on every tetrahedron. In fact,
N
has
6
such quasigeodesics.
To see this, consider the four star-unfoldings of
N
with respect to its vertices. Because
of the symmetry of
N
, three of the unfoldings from the base vertices
b
,
c
,
d
are isometric.
See Figure 16. One can then check that through each base vertex pass two
Q1
s, as

Information 2022,13, 238 18 of 20
represented in Figure 16b. On the other hand, the three geodesic loops through apex
a
,
represented in Figure 16a, are not quasigeodesics;
•
Because
N
is chosen sufficiently close to a regular tetrahedron, Lemma 9shows that
every edge of
N
provides three non-degenerate
Q2
s, as in Figure 17. The three vertices
of curvatures >πprovide three more degenerate Q2’s. They all sum up to 21 Q2s;
•
The boundary of every face of
N
is a
Q3
, because
N
is f-acute, hence there are
4
such
Q3quasigeodesics;
•
Every partition of the face set of
N
into two faces provides a
Q4
, again because
N
is
f-acute, hence there are 3Q4s, namely corresponding to AB :CD,AC :BD,AD :BC.
Thus we have found a tetrahedron
N
in whose neighborhood
O
, every tetrahedron
has at least 34 quasigeodesics, verifying Theorem 2.
bc
d
a
a a
b
c
d
d
d
(a) (b)
a
Figure 16.
Two star unfoldings of a near-regular tetrahedron. In this example,
ωa=
142
◦
and
ωb=ωc=ωd=
193
◦
. Geodesic loops are solid red segments. In (
a
), each loop has angle 12
◦
to one
side of aand 206◦to the other side. In (b), θd=167◦, so the loops are quasigeodesics.
b
c
d
a
Figure 17.
Three non-degenerate and one degenerate (red)
Q2
on a regular tetrahedron. The blue,
green, and purple geodesic segments each connect to ab.
7. Remarks and Open Problems
Our work leaves open several questions of various natures.
Open Problem 1.
The 2-vertex quasigeodesics that we identified in Section 4are all
edge-loops, i.e., they contain the edge joining the respective vertices. Is this necessarily
the case?
According to Theorem 2, some tetrahedra have at least 34 simple closed quasigeodesics,
and this happens on an open subset of the space Tof tetrahedra.

Information 2022,13, 238 19 of 20
Open Problem 2.
Does there exist an upper bound on the number of simple closed
quasigeodesics a tetrahedron can have? Of course, this is not the case for pure simple closed
geodesics, see Section 2.
Open Problem 3.
Find examples of tetrahedra with
k≥
3 simple closed quasi-
geodesics, for as many values of
k
as possible. For example, is there any tetrahedron that
has only the
k=
3 simple closed quasigeodesics that Pogorelov guarantees and we describe
in Theorem 1? Such a tetrahedron would be a polyhedral counterpart of an ellipsoid with
exactly three simple closed geodesics.
Our quest for simple closed quasigeodesics on tetrahedra lead us to investigate the
acuteness of face angles incident to a vertex. In this direction, we established the next
elementary result of some independent interest.
Proposition 3.
Let
¯
e
be a longest edge of the tetrahedron
T
. Then at least one extremity of
¯
e
has all
incident face angles acute, or T is a doubly covered rectangle.
Notice that the face angles incident to a vertex
v
of
ωv>π
are all acute, directly from
Lemma 5.
Proof.
Assume
T=abcd
and
¯
e=ab
. Then
ab
is in particular a longest edge in the triangle
faces Cand D, hence the angles aC,aD,bC,bD are all acute.
Assume now that the statement does not hold, hence
aB ≥π
,
bA ≥π
. Unfold the
union of faces
A∪B
in the plane, to a quadrilateral
a0b0c0d0
. Clearly, the triangles
a0c0d0
and
acd
are congruent, as are
b0c0d0
and
bcd
. However,
|a0b0|≥|ab|
. The angle conditions
aB ≥π
,
bA ≥π
imply, via an elementary geometry result, that the points
a0
and
b0
lie on,
or in the interior of, the circle of diameter
cd
. Therefore, we get
|a0b0|≤|cd|≤|ab|≤|a0b0|
,
impossible unless we have equalities everywhere. In this case,
T
is a doubly covered
rectangle, and the conclusion holds.
Our proofs involve the vertex of Tof largest curvature.
Open Problem 4.
Is the longest edge of a tetrahedron always incident to the vertex of
largest curvature? This is indeed the case for degenerate tetrahedra, which correspond to
planar quadrilaterals.
Author Contributions:
J.O. and C.V. contributed equally to this work. All authors have read and
agreed to the published version of the manuscript.
Funding:
The second author was partially supported by Unitatea Executiva pentru Finantarea
Invatamantului Superior, a Cercetarii, Dezvoltarii si Inovarii (UEFISCDI), project no. PN-III-P4-ID-
PCE- 2020-0794.
Data Availability Statement: Not applicable.
Acknowledgments: We benefitted from corrections and suggestions from the referees.
Conflicts of Interest: The authors declare no conflict of interest.
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