Acute Triangulations of Trapezoids and Pentagons

An acute triangulation of a polygon is a triangulation whose triangles have all their angles less than π/2. The number of triangles in a triangulation is called the size of it. In this paper, we investigate acute triangulations of trapezoids and convex pentagons and prove new results about such triangulations with minimum size. This completes and improves in some cases the results obtained in two papers of Yuan (2010).


Introduction and Preliminaries
A triangulation of a planar polygon is a finite set of nonoverlapping triangles covering the polygon in such a way that any two distinct triangles are either disjoint or intersect in a single common vertex or edge.An acute (resp., nonobtuse) triangulation of a polygon is a triangulation whose triangles have all their angles less (resp., not larger) than /2.The number of triangles in a triangulation is called the size.Burago and Zalgaller [1] and, independently, Goldberg and Manheimer [2] proved that every obtuse triangle can be triangulated into seven acute triangles and this bound is the best possible.Cassidy and Lord [3] showed that every square can be triangulated into eight acute triangles and eight is the minimum number.This remains true for any rectangle as proved by Hangan et al. in [4].Acute triangulations of trapezoids, quadrilaterals, and pentagons were investigated in [5][6][7][8].Further information, historical notes, and problems about acute triangulations of polygons and surfaces can be found in the survey paper [9].Let K denote a family of planar polygons, and for  ∈ K, let () be the minimum size of an acute triangulation of .Then, let (K) denote the maximum value of () for all  ∈ K.The following results are known.[7]: let T denote the family of all trapezoids, that is, quadrilaterals with at least one pair of parallel sides.Then, (T \ R) = 7, where R is the family of all rectangles (also including squares).
In this paper, we discuss acute triangulations of trapezoids and convex pentagons and prove new results of such triangulations with minimum size.For example, we get the following characterization of the right trapezoids: they are the only trapezoids needing exactly six triangles and one interior vertex for an acute triangulation of minimum size.For the family of convex pentagons, we show that the bound stated in Theorem 1(iv) can be improved under some additional conditions.
Let Γ be a convex planar polygon.A vertex  of Γ is called an acute (resp., right) corner if the interior angle of Γ at  is less than (resp., equal to) /2; otherwise,  is called an obtuse corner.Let T be an acute triangulation of Γ.A vertex or edge of T is called a boundary (resp., interior) vertex or edge if it lies on the boundary of Γ (resp., lies inside Γ).Let ,   , and   be the number of vertices, interior vertices, and boundary vertices of T. Let ,   , and   be the number of edges, interior edges, and boundary edges of T. Clearly, we have  =   +   ,  =   +   , and   =   .For each vertex  in T, the number of edges incident to  is called the degree of , denoted by deg().Let   denote the number of vertices in T of degree .Let  denote the number of triangles in T. The following lemma is easily verified (cf.[6, Lemma 1]).Lemma 2. Let T be an acute triangulation of a planar convex n-gon Γ.Then, one has  Let  be a trapezoid with two adjacent acute angles at  and , parallel sides  and , and, consequently, || < ||.If there exists an interior point  on  such that the triangles , , and  are acute, then  is triangulable by 3 acute triangles.Otherwise, we have the following lemma.

Characterizations of Trapezoids
Lemma 5. Let  be a trapezoid with two adjacent acute angles at  and  and parallel sides  and .Suppose that there is no interior point  on  such that , , and  are acute triangles.Then,  is triangulable with five acute triangles, and this bound is the best possible.
Proof.Let  be the smallest number of triangles in an acute triangulation T of any trapezoid as in the statement.Some edge  must meet the interior of .Suppose that  is an interior point of , that is,   ≥ 1.Then, it immediately follows that deg () ≥ 5 and so  ≥ 5. Suppose that there is no interior vertex in , that is,   = 0 and  is a boundary vertex.Some edge  must meet the interior of , and  is also a boundary vertex.We can assume that  ̸ =  since at least one of the triangles , , and  is not acute by hypothesis.This implies that   ≥ 4. By Lemma 2(1),  =   −   + 1 ≥ 5. Now, for a trapezoid as in the statement, an acute triangulation with 5 triangles is given in [7, Section 3].
The following result gives a characterization of the right trapezoids.

Proposition 6. Every trapezoid with exactly two right angles is triangulable with six acute triangles and one interior vertex, and this bound is the best possible.
Proof.Let  be the smallest number of triangles in an acute triangulation T of any trapezoid  with exactly two right corners at  and .Let  be an acute corner, thus || ≤ ||.Some edge  must meet the interior of .Suppose that  is an interior vertex of  and   = 1.Then,  is an end vertex of five interior edges.At least one neighbour of  is interior to a side of the trapezoid and is therefore incident of a further interior edge.So,   ≥ 6,   = 1 and  =   −   + 1 ≥ 6. Suppose that  is an interior vertex and   ≥ 2.Then, there are at least two interior vertices with degree ≥ 5, the degree of  (resp.,  and ) is ≥ 3, the degree of  is ≥ 2, and there is at least one vertex of degree ≥ 4.Then, we have 2 ≥ 3×3+2×5+4+2 ≥ 25, so 2 ≥ 26.Since  2 ≤ 1 by Lemma 2(4), we get 4 + 1 ≥ 26, so  > 6. Suppose that   = 0 and   = .Some edge of  must meet the interior of .Then  cannot lie on the edge ; otherwise, some edge incident to  must meet the interior of , and we get an interior vertex against the fact that   = 0. So,  must be in the interior of .If the angles at  are not right, we get a contradiction since  is a trapezoid with only two right angles which admits an acute triangulation of at least size .But this contrasts with the minimum size  of T. If the angles at  are right, then  is a square.But any acute triangulation of a square must have at least an interior vertex.So, we get again a contradiction as   = 0. Now, for a trapezoid as in the statement, an acute triangulation with 6 triangles and one interior vertex are given in [7, Section 3].

Lemma 7.
Let  be a trapezoid with acute corners at  and , parallel sides  and , and  and  both exterior to .Then,  is triangulable with seven acute triangles.This bound is the best possible among the acute triangulations of such a trapezoid which have at least one interior vertex.
Proof.Let T be an acute triangulation with at least one interior vertex for a trapezoid as in the statement.Let  denote the size of T. Suppose that   = 2.Then, there are two interior vertices  and  in T, and at least two neighbours of  and/or  have degree at least 4. Since   ≥ 9 and   = 2, Lemma 2(1) gives  =   −   + 1 ≥ 8. Suppose that   ≥ 3.Then, there are at least three interior vertices , , and  with degree ≥ 5.The degrees of  and  are ≥ 3, and those of  and  are ≥ 2. There are at least two neighbours of ,  and/or  with degree ≥ 4. Summing up, we have 2 ≥ 3 × 5 + 2 × 3 + 2 × 2 + 2 × 4 = 33, hence 2 ≥ 34.Since  2 ≤ 2, by Lemma 2(4), we get 4 + 2 ≥ 34, hence  ≥ 8. So, we can assume that   = 1.The interior vertex  cannot be connected to all the vertices of .Otherwise, there is a neighbour of  which is interior to a side of the trapezoid.It is therefore adjacent to a further interior vertex of the trapezoid, that is,   ≥ 2. This contradicts   = 1.If  is connected to exactly three vertices of the trapezoid, say , , and , there are two neighbours of  which have degree ≥ 4. Further, at least one of the obtuse corners  and  must have degree ≥ 4.Then, we have   ≥ 7,   =   = −1 and +−1 =   +  ≥ 7+−1, hence  ≥ 7. If  is joined to exactly two vertices of , then there are three neighbours of  with degree ≥ 4. The degree of the two vertices of  joined with  is ≥ 3, and the remaining two vertices have degree ≥ 2. Summing up, we get 2 ≥ 5 + 2 × 2 + 2 × 3 + 3 × 4 = 27, hence 2 ≥ 28.Since  2 ≤ 2, by Lemma 2(4), we have 4 + 2 ≥ 28, hence  ≥ 7. If  is joined to exactly one vertex of , then four neighbours of  have degree ≥ 4. Thus,   ≥ 7, and, by Lemma 2(1),  =   −   + 1 ≥ 7. Now, for a trapezoid as in statement, an acute triangulation with 7 acute triangles was described in [7, Section 3].

Acute Triangulations of Pentagons
The following proposition follows directly from the results proved in [5].Proposition 8. Every convex quadrilateral admits an acute triangulation of size at most eight, such that there are at most two new vertices introduced on each side.
It was shown in [8,Lemma 3.1] that every pentagon with at least one acute corner can be triangulated into at most 32 acute triangles.Under the hypothesis of convexity, we have the following.

Proposition 9. Every convex pentagon with at least one acute corner can be triangulated into at most 25 acute triangles.
Proof.Let Γ =  (in the anticlockwise order) be a convex pentagon with at least one acute corner, say .We distinguish some cases.
Case 1.The triangle  is acute.By Proposition 8, the convex quadrilateral  has an acute triangulation with size ≤ 8 such that there are at most 2 side vertices on .

Subcase 1.1.
There is no side vertex on .Then, Γ admits an acute triangulation with at most 9 triangles.

Subcase 1.2.
There is precisely one side vertex, say , on .By Lemma 4 of [6], since  is an acute triangle, for any point  on the side , there are two points  on  and  on  such that the line segments , , and  divide  into four acute triangles.Then, we get an acute triangulation of Γ into at most 12 triangles.Subcase 1.3.There are exactly two side vertices, say  and , on .In this case, the convex quadrilateral  has an acute triangulation of size 7, as shown in [5].Suppose that  is an interior point of .Let  be the point on  such that  is parallel to .Then, the triangle  is acute.By Lemma 4 of [6], the triangle  can be triangulated into four acute triangles with  as the only side vertex on  and two new vertices  and  on the edges  and , respectively.Let  and  be the orthogonal projections of  and  on the edge .The segments  and  divide the right trapezoid  into three right triangles.By [7, Section 3], there is an acute triangulation of the right trapezoid  of size 6 without new vertices introduced on the sides  and .Then, we can slightly slide  and  in direction from  to  such that the triangles , , and  become acute.This gives an acute triangulation of Γ whose size is at most 20.
Case 2. The triangle  is nonacute, that is, , for example, is a nonacute corner.

Subcase 2.1.
There is no side vertex on .Then, there exists an acute triangle, say , which belongs to an acute triangulation of  with size ≤ 8. Let  be the orthogonal projection of  on the side .By Theorem 2 of [6], since  is a triangle with nonacute corner , for any point  on the side , there is an acute triangulation of  with size 7 such that  is the only side vertex lying on .Such an acute triangulation of  has new vertices  (resp.,  and ) introduced on the side  (resp., ).Finally, we can slightly slide  away from  in direction perpendicular to  such that the triangles  and  become acute.This gives an acute triangulation of Γ into at most 16 acute triangles.

Subcase 2.2.
There is precisely one side vertex  on .As in the previous subcase, by Theorem 2 of [6], for any point  on the side , there is an acute triangulation of the triangle  with size 7 such that  is the only side vertex lying on .This gives an acute triangulation of Γ into at most 15 acute triangles.

Subcase 2.3.
There are exactly two side vertices, say  and , on .In this case,  has an acute triangulation of size 7 by [5].Suppose that  is an interior point of .Let  be the point on  such that  is parallel to .Then, the triangle  has a nonacute corner .By Theorem 2 of [6],  can be triangulated into 7 acute triangles with  as the only side vertex on  and new vertices , respectively,  and  on the edges , respectively, .Let , , and  be the orthogonal projections of , , and  on the edge , respectively.The line segments , , , and  divide the right trapezoid  into 5 right triangles.By [7, Section 3], there is an acute triangulation of the right trapezoid  of size 6 without new vertices introduced on the sides  and .Then, we can slightly slide , , and  in direction from  to  such that the triangles , , , , and  become acute.This gives an acute triangulation of Γ whose size is at most 25.

Corollary 10. Every convex pentagon with at least two nonadjacent acute corners can be triangulated into at most 16 acute triangles.
Proof.By the hypothesis and the results from [5], we can avoid subcases 1.3 and 2.3 in the above proof.The remaining cases give the requested bound.
The following proposition follows directly from the results proved in [3,4,7].
Proposition 11.Every trapezoid (resp., rectangle) admits an acute triangulation of size at most 7 (resp., 8) such that there are at most one new vertex introduced on each side.Proposition 12. Let Γ be a convex pentagon which has at least one acute corner and two parallel sides, nonincident to it.Then, Γ can be triangulated into at most 14 acute triangles.
Proof.Let Γ =  (in the anticlockwise order) be a convex pentagon with at least one acute corner, say , and two parallel sides  and  with || ≤ ||.We distinguish some cases.
Case 1.The triangle  is acute.Subcase 1.1.Let  and  be the orthogonal projections of  and , respectively, on the straight line ℓ  containing the edge .Suppose that  (resp., ) is interior (resp., exterior) of .By [7, Section 2], the trapezoid  admits an acute triangulation of size at most 4 such that there are no new vertices on the side .Then, Γ has an acute triangulation of size at most 5. Subcase 1.2.Suppose that the above orthogonal projections  and  are interior to the edge .By [7, Section 2], the trapezoid  admits an acute triangulation of size at most 5 such that no new vertices are introduced on .Then, Γ has an acute triangulation of size at most 6.Subcase 1.3.Suppose that  is a right trapezoid (this implies that the triangle  is acute).By [7, Section 3],  can be triangulated into 6 acute triangles such that there are no new vertices on .Then, Γ has an acute triangulation of size at most 7.
Subcase 1.4.Suppose that the above orthogonal projections  and  are exterior to .By [7, Section 3],  admits an acute triangulation of size at most 7 such that there is only one vertex, say , on the side .By Lemma 4 of [6], for any point  in the side , there are two points  on  and  on  such that the line segments , , and  divide  into 4 acute triangles.Then, Γ can be triangulated into at most 11 acute triangles.
Case 2. The triangle  is nonacute, that is, the corner , for example, is nonacute.Subcase 2.1.Let  be as in subcase 1.1.There is an acute triangle, say , which belongs to the triangulation of size ≤ 4 of .Let  be the orthogonal projection of  on the edge .By Theorem 2 of [6], there is an acute triangulation of  with size 7 such that  is the only side vertex on .Then, Γ has an acute triangulation of size at most 12. Subcase 2.2.Let  be as in subcase 1.2.Reasoning as in the previous subcase gives an acute triangulation of Γ with size ≤ 13.Subcase 2.3.Let  be as in Subcase 1.4.By Theorem 2 of [6] the triangle  can be triangulated into at most 7 acute triangles such that the only side vertex on  is .This gives an acute triangulation of Γ with size ≤ 14.
5) If a vertex  of T is an interior vertex, then deg() ≥ 5, if  lies within a side of Γ, then deg() ≥ 4, and if  is an obtuse or right corner of Γ, then deg() ≥ 3.
Assume now that   = 0.Then,  lies in the interior of the edge  (or ).Otherwise, if  = , at least one of the vertices  and  must have degree ≥ 4 by hypothesis.This implies that there is a neighbor of it with degree ≥ 4, giving a contradiction as [7].Thus,   =  and   ≥ 3 as the degree of  is at least 4, and the degree of  (resp., ) is at least 3.By Lemma 2(1), we get  =   −   + 1 ≥ 4. Now, for a parallelogram as in the statement, an acute triangulation with 4 triangles is given in [7, Section 2].2.2.Trapezoids.Following[7], we say that a trapezoid is a quadrilateral with at least one pair of parallel sides.Let  be a trapezoid with parallel sides  and  with || < ||.Let  (resp., ) be the orthogonal projection of  (resp., ) on the straight line ℓ  containing .Suppose that  is interior to  and  exterior to .If the diagonal  divides the angles at  and  into acute angles, then  is triangulable with exactly two acute triangles.Otherwise, we have the following lemma which can be proved in the same manner as Lemma 3. Lemma 4. Let  be a trapezoid with acute corners at  and , parallel sides  and  with an obtuse angle  ∧  , and an acute angle  ∧  .Then,  is triangulable with four acute triangles, and this bound is the best possible.