1. Introduction
By a triangulation of a two-dimensional space, we mean a set of (full) triangles covering the space, in such a way that the intersection of any two triangles is either empty or consists of a vertex or of an edge. A triangle is called geodesic if all its edges are segments, i.e., shortest paths between the corresponding vertices. We are interested only in geodesic triangulations, all the members of which are, by definition, geodesic triangles. In rather general two-dimensional spaces, like Alexandrov surfaces, two geodesics starting at the same point determine a well defined angle.
An acute (non-obtuse) triangulation of a two-dimensional space is a geodesic triangulation such that the angles of all geodesic triangles are smaller (respectively, not greater) than
. The number of triangles in a triangulation is called its size.
The discussion of acute triangulations was firstly proposed in 1960 by Gardner (see [1] [2] [3] ). In the same year, independently, Burago and Zalgaller [4] proved the existence of acute triangulations of general two-dimensional polyhedral surfaces. However, their method could not give an estimate on the size of the existed acute triangulations. Acute triangulations of n-polygons (
) have been considered from 1980, such as acute triangulations of square [5], quadrilaterals [6] [7], trapezoids [8], pentagons [9] and arbitrary convex polygons [10] [11] [12].
Currently, some compact convex surfaces have also been triangulated, such as acute and non-obtuse triangulations of the surfaces of all Platonic solids in [13] [14] [15] [16], the surfaces of some Archimedean solids in [17] and [18], flat Möbius strips [19], flat tori [20]. Furthermore, other surfaces homeomorphic to the sphere have also been acutely triangulated, such as the double triangles [21], the double quadrilaterals [22] and double planar convex bodies in [23]. In 2018, Bau and Gagola III discussed acute decompositions (not triangulations) of closed orientable geometric surfaces in [24].
In 2009, acute triangulations of any polyhedral surface were considered again by Saraf [25], but there was still no estimate on the size of the existed acute triangulations. In 2011, H. Maehara [26] proved that every polyhedral surface admits a proper acute triangulation and got a upper bound for the size of the proper acute triangulation.
In this paper, we will consider the surface of any cone, and start with the surface of a bounded right circular cone. For the sake of convenience, let
be the surface of a bounded circular cone. Denote by T an acute triangulation of
and
a non-obtuse triangulation of
. Let
denote the size of T. We prove that
and
.
2. Non-Obtuse Triangulations
Theorem 1. The surface of any right circular cone can be triangulated into 8 non-obtuse triangles, and there is no non-obtuse triangulation with fewer triangles.
Proof. Consider the unfold figure of
. Denote by a the vertex of
, b the center of the bottom of
. Draw four segments from a to b such that the four segments divide equally the angles around a and b. Let
be the intersection points of the four segments with the circle of the bottom. Choose the points
,
,
,
such that
,
at c,
,
at e. Let
be the circle of the bottom,
,
,
,
.
Now we choose the suitable positions of c and e such that
, and
. Obviously,
.
We get a triangulation of
with 8 triangles:
It is not difficult to see that the four triangles
are congruent, and the four triangles
are also congruent. Therefore, we only need to consider the two triangles
and
. Since
, the sum of the interior angles of the triangle
is
, and the sum of the interior angles of the triangle
is
. By
,
, we have
. Then,
. Therefore, the above triangulation is a non-obtuse triangulation of
.
Suppose that there is a non-obtuse triangulation
of
with
, then
or 6. It implies that
has at least two vertices with degree 3. However, the total angle at any point except a on
is equal to 2π. Therefore,
has only one vertex possibly with degree 3, which is a contradiction.
The proof is complete. □
3. Acute Triangulations
Let o denote the vertex of the circular cone, and
denote the total angle around o on the surface
. Denote by
the center of the bottom of
.
Theorem 2. The surface of any right circular cone cannot be triangulated into less than 20 acute triangles.
Proof. Let T denote an acute triangulation of
. By the definition of triangulation, the degree of any vertex of T is at least 3. Notice that
, and the total angle at any point except o on
is equal to 2π, so in T, the vertex o can have degree less than 5, but any other vertex except o must have degree at least 5.
If
, then T has
edges and, by Euler’s formula,
vertices. Clearly, the sum of the degrees of all the vertices of T is
, so o must be a vertex of T. Furthermore, there are two cases about T: the vertex o with degree 4, any other ten vertices have degree 5; the vertex o with degree 3, one vertex with degree 6, any other nine vertices have degree 5. However, there is no triangulation with 18 triangles that can have these constructions.
If
, then we can immediately get that T has 24 edges and 10 vertices, and then the sum of the degrees of all the vertices of T is 48. Then there is only one case about T: the vertex o with degree 3, any other nine vertices have degree 5. However, there is no triangulation with 16 triangles can have this construction.
If
, then T has 21 edges and 9 vertices, and thus, the sum of the degrees of all the vertices of T is 42. Then the degree of the vertex o must be 2, a contradiction.
If
, it is not difficult to see that there are at least two vertices of T with degree at most 4, which implies impossible.
Theorem 3. The surface of any right circular cone can be triangulated into 20 acute triangles.
Proof. Let
be five points in the relative interior of the flank of the circular cone such that
, and
. Clearly, these five points
on
form a regular pentagon with center o. Draw two segments from
on the outside of the triangle
such that the three angles at
are equal to
respectively.
Now we respectively choose the suitable distances from
to o such that the intersection points of the above ten segments, denoted by
, are located in the bottom of
. Hence the pentagon
is regular, and with center
.
The ten segments
,
,
,
,
,
,
,
,
,
divide the bottom of
into ten arcs, denote by
,
,
,
,
,
,
,
,
,
. Furthermore,
, and
. As we respectively slide
in direction
,
,
,
and
, the distances from
to
are monotone decreasing. Choose the suitable positions of
such that
.
We obtain a triangulation T of
with 20 triangles:
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
.
It is clear that the triangles
,
,
,
,
are planar congruent isosceles ones with vertex angle
and the triangles
,
,
,
,
are also planar congruent isosceles ones with vertex angle 2π/5.
The remaining ten triangles are isosceles. Furthermore, the five triangles
,
,
,
,
are congruent, and the other five triangles
,
,
,
,
are also congruent. Hence we only need to consider the triangles
and
.
It is not difficult to see that the sum of curvature on the arc
is
, then the sum of curvature of
is
, and the sum of curvature of
is
. By the Gauß-Bonnet formula, the sum of the interior angles of the triangle
is
. Since
, we have
. Similarly, the sum of the interior angles of the triangle
is
. Because of
, then
. □
4. Acute Triangulations with Degree of o Less Than 5
The total angle at each point of
except for o is
. According to the total angle of the vertex o, can we find a triangulation T of
with the number of triangles as few as possible such that the degree of o in T is less than 5?
If
, it is not difficult to see that the curvature of o is
, hence we can get a triangle on
with all the three interior angles being equal to
such that o is located in the interior of this triangle. If
, we can find an triangulation of
such that the degree of o is 4. If
, we prove that there is an triangulation of
such that the degree of o is 3.
Theorem 4. If
, then the surface of the right circular cone admits an acute triangulation with 20 triangles such that the degree of o is 0.
Proof. Let
denote three points in the relative interior of the flank of
such that
and
. Let
denote the midpoint of
,
,
respectively. By the Gauß-Bonnet formula, the sums of the interior angles of the two triangle
and
are both
. Outside of the triangle
, draw three segments from
respectively such that the four angles at
are equal to
.
Let
,
,
. Consider the segments
,
,
,
,
,
. Clearly,
,
,
. Let
denote the midpoint of
,
,
respectively. It is not difficult to see that
belongs to the middle segment starting from
respectively, and
,
,
.
Therefore, we obtain a triangulation
of
with 20 triangles:
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
.
Now choose the suitable positions of
such that the twelve segments
,
,
,
,
,
,
,
,
,
,
and
equally divide the circle of the bottom.
It is not difficult to see that the four triangles
,
,
,
are congruent equilateral triangles with angle
. It is clear that both the triangle
and
are equilateral triangles. Notice that the sums of the interior angles of the two triangles
and
are both
, thus,
, and then
. Hence, the three triangles
,
and
are both acute ones.
By the symmetry, the six triangles
,
,
,
,
and
are congruent right triangles. Hence we only need to consider the triangle
. In the triangle
, the sum of the interior angles is
. Notice that
,
, we can get
.
Similarly, the six triangles
,
,
,
,
and
are also congruent right triangles, then we only consider the triangle
. In the triangle
, the sum of the interior angles is
. Since
,
, we have
.
Now, we will slightly change the positions of some vertices of
to get an acute triangulation of
. During all the steps, the original acute angles remain acute.
Slide
slightly in direction
,
,
,
,
,
respectively, then the twelve right angles
,
,
,
,
,
,
,
,
,
,
and
become acute. Then all the triangles become acute.
The proof is complete. □
Theorem 5. If
, then the surface of the right circular cone cannot be triangulated into less than 22 acute triangles such that the degree of o is less than 5.
Proof. By Theorem 2, we know that
. If
, then T has
edges and, by Euler’s formula,
vertices. Clearly, the sum of the degrees of all the vertices of T is
. Since
, the vertex o of the cone cannot be in the interior of a triangle of T. That is to say, if o has degree less than 5, then o must be a vertex of T.
Therefore, there are three cases about T: the vertex o with degree 4, one vertex with degree 6, any other ten vertices have degree 5; the vertex o with degree 3, one vertex with degree 7, any other ten vertices have degree 5; the vertex o with degree 3, two vertices with degree 6, any other nine vertices have degree 5. However, there is no triangulation with 20 triangles can have these constructions. □
Theorem 6. If
, then the surface of the right circular cone admits an acute triangulation with 22 triangles such that the degree of o is less than 5.
Proof. Case 1.
Let
denote three points in the relative interior of the flank of
such that
and
. Let
be the midpoint of the segment
,
,
,
,
,
respectively. Draw two segments from
on the outside of the triangle
such that the three angles around
are equal to
respctively. Let the points
be on the above segments such that
,
,
. Choose the suitable positions of
such that the nine segments
,
,
,
,
,
,
,
and
equally divide the circle of the bottom. Obviously,
,
,
.
Therefore, we obtain a triangulation
of
with 22 triangles:
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
.
Clearly, the triangle
are equilateral triangle with angle
. It is clear that the twelve triangles
,
,
,
,
,
,
,
,
,
,
and
are congruent isosceles triangles with vertex angle
, and hence the other two angles are both
.
By the Gauß-Bonnet formula, the sums of the interior angles of all the nine triangles
,
,
,
,
,
,
,
and
are
. Furthermore, the six right triangles
,
,
,
,
and
are congruent and the three triangles
,
and
are also congruent. Thus, we only consider the two triangles
and
.
In the triangle
,
,
, then
. In the triangle
,
. It is easy to see that
.
Now slide the vertices
of
slightly in direction
,
,
respectively. Then the six right angles
,
,
,
,
and
become less than
. During all the steps, the original acute angles remain acute.
Case 2.
Let
denote two points in the relative interior of the flank of
such that
and
. Clearly, there are two segments from a to b. Let
be the midpoints of the two segments.
Let
denote the points in the relative interior of the bottom such that the segments
and
are orthogonal to the segment
which contains c, and the segments
and
are orthogonal to the segment
which contains d. Let
,
,
,
such that
,
. Extend the segments
and
, and then let i be the intersection point. Extend
and
, then let j be the intersection point. Choose the suitable positions of
such that the ten segments
,
,
,
,
,
,
,
,
,
equally divide the circle of the bottom.
We get a triangulation
of
with 22 triangles as follows:
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
.
The four triangles
,
,
,
are congruent right triangles, so we only consider the triangle
. It is clear that
,
and
. Obviously, the four triangles
,
,
,
are congruent right triangles. We only consider the triangle
. Without loss of generality, suppose the radius of the flank of
is unit. Clearly,
,
,
, that is,
. Hence
and
. The two triangles
and
are congruent isosceles triangles, so we only consider the triangle
. It is easy to see that
, and then the other two angles
.
The two triangles
and
are congruent, so we only consider the triangle
. Clearly,
. By the Gauß-Bonnet formula, the sum of the interior angles of the triangles
is
, then
. Since the four triangles
,
,
,
are congruent right triangles, we only consider the triangle
. Clearly,
,
. Notice that the sum of the interior angles of the triangle
is
, thus
.
The four triangles
,
,
,
are congruent, so we only consider the triangle
. Let
,
,
, then we have
,
. Obviously,
. It is not difficult to check that
, that is,
. Then
,
, and hence
.
The two triangles
and
are congruent isosceles triangles, so we only consider the triangle
. It is clear that
, and then the other two angles
.
Now, we will slightly change the positions of some vertices of
to get an acute triangulation of
. During all the slidings of vertices, once an angle has been acute, then the next steps will be performed so gently that the angle remains acute.
Step 1. Slide
slightly in direction
,
respectively, then the eight right triangles
,
,
,
,
,
,
,
become less than
.
Step 2. Slide
slightly in direction
,
respectively, then the four right angles
,
,
,
become less than
.
Step 3. Slide
slightly in direction
,
,
,
respectively. Meanwhile, maintain that the ten segments
,
,
,
,
,
,
,
,
,
still equally divide the circle, then the four right angles
,
,
,
become less than
. In this step, don’t change the values of the four angles
,
,
and
.
Step 4. Slide
slightly in direction
,
respectively, then the four right angles
,
,
,
become less than
.
During all the steps, the original acute angles remain acute. Then all the triangles become acute.
The proof is completed. □
Acknowledgements
The first author gratefully acknowledges financial supports from Hebei Science Foundation (A2019106041) and Shijiazhuang University (18BS003), the third author was supported by Hebei University of Business and Economics (2017KYQ02).