Acute Triangulations of Trapezoids and Pentagons

Abstract

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).

1. 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 𝜋/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 [5–8]. 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∈𝒦, let 𝑓(𝐾) be the minimum size of an acute triangulation of 𝐾. Then, let 𝑓(𝒦) denote the maximum value of 𝑓(𝐾) for all 𝐾∈𝒦. The following results are known.

Theorem 1. (i) Reference [7]: let 𝒯 denote the family of all trapezoids, that is, quadrilaterals with at least one pair of parallel sides. Then, 𝑓(𝒯⧵ℛ)=7, where ℛ is the family of all rectangles (also including squares).
(ii) Reference [6]: let 𝒬 be the family of all quadrilaterals. Then, 𝑓(𝒬)=10.
(iii) Reference [5]: let 𝒬_𝑐 be the family of all convex quadrilaterals. Then, 𝑓(𝒬_𝑐)=8.
(iv) Reference [8]: let 𝒫 denote the family of all planar pentagons. Then, 𝑓(𝒫)≤54.

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 𝑃 of Γ is called an acute (resp., right) corner if the interior angle of Γ at 𝑃 is less than (resp., equal to) 𝜋/2; otherwise, 𝑃 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 𝑛, 𝑛_𝑖, and 𝑛_𝜕 be the number of vertices, interior vertices, and boundary vertices of 𝐓. Let 𝑚, 𝑚_𝑖, and 𝑚_𝜕 be the number of edges, interior edges, and boundary edges of 𝐓. Clearly, we have 𝑛=𝑛_𝑖+𝑛_𝜕, 𝑚=𝑚_𝑖+𝑚_𝜕, and 𝑛_𝜕=𝑚_𝜕. For each vertex 𝑃 in 𝐓, the number of edges incident to 𝑃 is called the degree of 𝑃, denoted by deg(𝑃). Let 𝜈_𝑘 denote the number of vertices in 𝐓 of degree 𝑘. Let 𝑓 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(1)𝑓=𝑚−𝑛+1=𝑛+𝑛_𝑖−2=𝑚_𝑖−𝑛_𝑖+1,(2)3𝑓=2𝑚−𝑚_𝜕=2𝑚_𝑖+𝑛_𝜕,(3)𝜈₁=0;𝜈₂+𝜈₃≤𝑛,(4)𝑛_𝜕=𝑚_𝜕≤𝑓+𝜈₂;4𝑓+𝜈₂≥2𝑚,(5)If a vertex 𝑃 of 𝐓 is an interior vertex, then deg(𝑃)≥5, if 𝑃 lies within a side of Γ, then deg(𝑃)≥4, and if 𝑃 is an obtuse or right corner of Γ, then deg(𝑃)≥3.

2. Characterizations of Trapezoids

2.1. Parallelograms

Let 𝐴𝐵𝐶𝐷 be a parallelogram with acute corners at 𝐵 and 𝐷 and |𝐴𝐵|≤|𝐵𝐶|. If the diagonal 𝐴𝐶 divides the angles at 𝐴 and 𝐶 into acute angles, then 𝐴𝐵𝐶𝐷 is triangulable with exactly two acute triangles. Otherwise, we have the following lemma that completes Theorem 2.1 in Section 2 of [7].

Lemma 3. Let 𝐴𝐵𝐶𝐷 be a parallelogram with acute corners at 𝐵 and 𝐷. Suppose that the diagonal 𝐴𝐶 does not divide the angle at 𝐴 (or 𝐶) into acute angles. Then, 𝐴𝐵𝐶𝐷 is triangulable with four acute triangles, and this bound is the best possible.

Proof. Let 𝑓 be the smallest number of triangles in an acute triangulation 𝐓 of the parallelogram 𝐴𝐵𝐶𝐷. Some edge of 𝐓 incident with 𝐴 (resp., 𝐶) must meet the interior of 𝐴𝐵𝐶𝐷. Denote 𝐴𝑃 to be such an edge, and suppose that 𝑃 is an interior vertex of 𝐴𝐵𝐶𝐷, that is, 𝑛_𝑖≥1. Then, the vertices 𝐵 and 𝐷 have degree at least 2, the vertices 𝐴 and 𝐶 have degree at least 3, and the vertex 𝑃 has degree at least 5. Summing up, we have 2𝑚≥2×2+2×3+5=15, hence 2𝑚≥16. Since 𝜈₂≤2, by Lemma 2(4), we get 4𝑓+2≥16, so 𝑓≥4. Assume now that 𝑛_𝑖=0. Then, 𝑃 lies in the interior of the edge 𝐵𝐶 (or 𝐷𝐶). Otherwise, if 𝑃=𝐶, at least one of the vertices 𝐴 and 𝐶 must have degree ≥4 by hypothesis. This implies that there is a neighbor of it with degree ≥4, giving a contradiction as 𝑛_𝑖=0. Thus, 𝑛_𝜕=𝑛 and 𝑚_𝑖≥3 as the degree of 𝑃 is at least 4, and the degree of 𝐴 (resp., 𝐶) is at least 3. By Lemma 2(1), we get 𝑓=𝑚_𝑖−𝑛_𝑖+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 𝐴𝐵𝐶𝐷 be a trapezoid with parallel sides 𝐴𝐷 and 𝐵𝐶 with |𝐴𝐷|<|𝐵𝐶|. Let 𝐸 (resp., 𝐹) be the orthogonal projection of 𝐴 (resp., 𝐷) on the straight line ℓ_𝐵𝐶 containing 𝐵𝐶. Suppose that 𝐸 is interior to 𝐵𝐶 and 𝐹 exterior to 𝐵𝐶. If the diagonal 𝐴𝐶 divides the angles at 𝐴 and 𝐶 into acute angles, then 𝐴𝐵𝐶𝐷 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 𝐴𝐵𝐶𝐷 be a trapezoid with acute corners at 𝐵 and 𝐷, parallel sides 𝐴𝐷 and 𝐵𝐶 with an obtuse angle 𝐵𝐴^∧𝐶, and an acute angle 𝐶𝐴^∧𝐷. Then, 𝐴𝐵𝐶𝐷 is triangulable with four acute triangles, and this bound is the best possible.

Let 𝐴𝐵𝐶𝐷 be a trapezoid with two adjacent acute angles at 𝐵 and 𝐶, parallel sides 𝐴𝐷 and 𝐵𝐶, and, consequently, |𝐴𝐷|<|𝐵𝐶|. If there exists an interior point 𝑃 on 𝐵𝐶 such that the triangles 𝐵𝐴𝑃, 𝐴𝑃𝐷, and 𝑃𝐷𝐶 are acute, then 𝐴𝐵𝐶𝐷 is triangulable by 3 acute triangles. Otherwise, we have the following lemma.

Lemma 5. Let 𝐴𝐵𝐶𝐷 be a trapezoid with two adjacent acute angles at 𝐵 and 𝐶 and parallel sides 𝐴𝐷 and 𝐵𝐶. Suppose that there is no interior point 𝑃 on 𝐵𝐶 such that 𝐵𝐴𝑃, 𝐴𝑃𝐷, and 𝑃𝐷𝐶 are acute triangles. Then, 𝐴𝐵𝐶𝐷 is triangulable with five acute triangles, and this bound is the best possible.

Proof. Let 𝑓 be the smallest number of triangles in an acute triangulation 𝐓 of any trapezoid as in the statement. Some edge 𝐴𝑃 must meet the interior of 𝐴𝐵𝐶𝐷. Suppose that 𝑃 is an interior point of 𝐴𝐵𝐶𝐷, that is, 𝑛_𝑖≥1. Then, it immediately follows that deg(𝑃)≥5 and so 𝑓≥5. Suppose that there is no interior vertex in 𝐴𝐵𝐶𝐷, that is, 𝑛_𝑖=0 and 𝑃 is a boundary vertex. Some edge 𝐷𝑄 must meet the interior of 𝐴𝐵𝐶𝐷, and 𝑄 is also a boundary vertex. We can assume that 𝑃≠𝑄 since at least one of the triangles 𝐵𝐴𝑃, 𝐴𝑃𝐷, and 𝑃𝐷𝐶 is not acute by hypothesis. This implies that 𝑚_𝑖≥4. By Lemma 2(1), 𝑓=𝑚_𝑖−𝑛_𝑖+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 𝑓 be the smallest number of triangles in an acute triangulation 𝐓 of any trapezoid 𝐴𝐵𝐶𝐷 with exactly two right corners at 𝐷 and 𝐶. Let 𝐵 be an acute corner, thus |𝐴𝐵|≤|𝐵𝐶|. Some edge 𝐴𝑃 must meet the interior of 𝐴𝐵𝐶𝐷. Suppose that 𝑃 is an interior vertex of 𝐴𝐵𝐶𝐷 and 𝑛_𝑖=1. Then, 𝑃 is an end vertex of five interior edges. At least one neighbour of 𝑃 is interior to a side of the trapezoid and is therefore incident of a further interior edge. So, 𝑚_𝑖≥6, 𝑛_𝑖=1 and 𝑓=𝑚_𝑖−𝑛_𝑖+1≥6. Suppose that 𝑃 is an interior vertex and 𝑛_𝑖≥2. Then, there are at least two interior vertices with degree ≥5, the degree of 𝐴 (resp., 𝐶 and 𝐷) is ≥3, the degree of 𝐵 is ≥2, and there is at least one vertex of degree ≥4. Then, we have 2𝑚≥3×3+2×5+4+2≥25, so 2𝑚≥26. Since 𝜈₂≤1 by Lemma 2(4), we get 4𝑓+1≥26, so 𝑓>6. Suppose that 𝑛_𝑖=0 and 𝑛_𝜕=𝑛. Some edge of 𝐴𝑃 must meet the interior of 𝐴𝐵𝐶𝐷. Then 𝑃 cannot lie on the edge 𝐷𝐶; otherwise, some edge incident to 𝐷 must meet the interior of 𝐴𝐵𝐶𝐷, and we get an interior vertex against the fact that 𝑛_𝑖=0. So, 𝑃 must be in the interior of 𝐵𝐶. If the angles at 𝑃 are not right, we get a contradiction since 𝐴𝑃𝐶𝐷 is a trapezoid with only two right angles which admits an acute triangulation of at least size 𝑓. But this contrasts with the minimum size 𝑓 of 𝐓. If the angles at 𝑃 are right, then 𝐴𝑃𝐶𝐷 is a square. But any acute triangulation of a square must have at least an interior vertex. So, we get again a contradiction as 𝑛_𝑖=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 𝐴𝐵𝐶𝐷 be a trapezoid with acute corners at 𝐵 and 𝐷, parallel sides 𝐴𝐷 and 𝐵𝐶, and 𝐸 and 𝐹 both exterior to 𝐵𝐶. Then, 𝐴𝐵𝐶𝐷 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 𝑓 denote the size of 𝐓. Suppose that 𝑛_𝑖=2. Then, there are two interior vertices 𝑃 and 𝑆 in 𝐓, and at least two neighbours of 𝑃 and/or 𝑆 have degree at least 4. Since 𝑚_𝑖≥9 and 𝑛_𝑖=2, Lemma 2(1) gives 𝑓=𝑚_𝑖−𝑛_𝑖+1≥8. Suppose that 𝑛_𝑖≥3. Then, there are at least three interior vertices 𝑃, 𝑅, and 𝑆 with degree ≥5. The degrees of 𝐴 and 𝐶 are ≥3, and those of 𝐵 and 𝐷 are ≥2. There are at least two neighbours of 𝑃, 𝑅 and/or 𝑆 with degree ≥4. Summing up, we have 2𝑚≥3×5+2×3+2×2+2×4=33, hence 2𝑚≥34. Since 𝜈₂≤2, by Lemma 2(4), we get 4𝑓+2≥34, hence 𝑓≥8. So, we can assume that 𝑛_𝑖=1. The interior vertex 𝑃 cannot be connected to all the vertices of 𝐴𝐵𝐶𝐷. Otherwise, there is a neighbour of 𝑃 which is interior to a side of the trapezoid. It is therefore adjacent to a further interior vertex of the trapezoid, that is, 𝑛_𝑖≥2. This contradicts 𝑛_𝑖=1. If 𝑃 is connected to exactly three vertices of the trapezoid, say 𝐴, 𝐵, and 𝐶, there are two neighbours of 𝑃 which have degree ≥4. Further, at least one of the obtuse corners 𝐴 and 𝐶 must have degree ≥4. Then, we have 𝑚_𝑖≥7, 𝑚_𝜕=𝑛_𝜕=𝑛−1 and 𝑛+𝑓−1=𝑚_𝑖+𝑚_𝜕≥7+𝑛−1, hence 𝑓≥7. If 𝑃 is joined to exactly two vertices of 𝐴𝐵𝐶𝐷, then there are three neighbours of 𝑃 with degree ≥4. The degree of the two vertices of 𝐴𝐵𝐶𝐷 joined with 𝑃 is ≥3, and the remaining two vertices have degree ≥2. Summing up, we get 2𝑚≥5+2×2+2×3+3×4=27, hence 2𝑚≥28. Since 𝜈₂≤2, by Lemma 2(4), we have 4𝑓+2≥28, hence 𝑓≥7. If 𝑃 is joined to exactly one vertex of 𝐴𝐵𝐶𝐷, then four neighbours of 𝑃 have degree ≥4. Thus, 𝑚_𝑖≥7, and, by Lemma 2(1), 𝑓=𝑚_𝑖−𝑛_𝑖+1≥7. Now, for a trapezoid as in statement, an acute triangulation with 7 acute triangles was described in [7, Section 3].

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=𝐴𝐵𝐶𝐷𝐸 (in the anticlockwise order) be a convex pentagon with at least one acute corner, say 𝐴. We distinguish some cases.
Case  1. The triangle 𝐴𝐵𝐸 is acute. By Proposition 8, the convex quadrilateral 𝐵𝐶𝐷𝐸 has an acute triangulation with size ≤8 such that there are at most 2 side vertices on 𝐵𝐸.
Subcase  1.1. There is no side vertex on 𝐵𝐸. Then, Γ admits an acute triangulation with at most 9 triangles.
Subcase  1.2. There is precisely one side vertex, say 𝑃, on 𝐵𝐸. By Lemma 4 of [6], since 𝐴𝐵𝐸 is an acute triangle, for any point 𝑃 on the side 𝐵𝐸, there are two points 𝑅 on 𝐴𝐸 and 𝑆 on 𝐴𝐵 such that the line segments 𝑃𝑅, 𝑅𝑆, and 𝑃𝑆 divide 𝐴𝐵𝐸 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 𝑀 and 𝑁, on 𝐵𝐸. In this case, the convex quadrilateral 𝐵𝐶𝐷𝐸 has an acute triangulation of size 7, as shown in [5]. Suppose that 𝑀 is an interior point of 𝐸𝑁. Let 𝑃 be the point on 𝐴𝐸 such that 𝑃𝑁 is parallel to 𝐴𝐵. Then, the triangle 𝑃𝑁𝐸 is acute. By Lemma 4 of [6], the triangle 𝑃𝑁𝐸 can be triangulated into four acute triangles with 𝑀 as the only side vertex on 𝐸𝑁 and two new vertices 𝑅 and 𝑆 on the edges 𝐸𝑃 and 𝑁𝑃, respectively. Let 𝐻 and 𝐾 be the orthogonal projections of 𝑆 and 𝑁 on the edge 𝐴𝐵. The segments 𝑁𝐾 and 𝑆𝐾 divide the right trapezoid 𝐻𝐵𝑁𝑆 into three right triangles. By [7, Section 3], there is an acute triangulation of the right trapezoid 𝐴𝐻𝑆𝑃 of size 6 without new vertices introduced on the sides 𝑃𝑆 and 𝐻𝑆. Then, we can slightly slide 𝐾 and 𝐻 in direction from 𝐵 to 𝐴 such that the triangles 𝑆𝐻𝐾, 𝑆𝐾𝑁, and 𝐵𝐾𝑁 become acute. This gives an acute triangulation of Γ whose size is at most 20.
Case  2. The triangle 𝐴𝐵𝐸 is nonacute, that is, 𝐸, for example, is a nonacute corner.
Subcase  2.1. There is no side vertex on 𝐵𝐸. Then, there exists an acute triangle, say 𝐸𝐵𝑃, which belongs to an acute triangulation of 𝐵𝐶𝐷𝐸 with size ≤8. Let 𝐻 be the orthogonal projection of 𝑃 on the side 𝐵𝐸. By Theorem 2  of [6], since 𝐴𝐵𝐸 is a triangle with nonacute corner 𝐸, for any point 𝐻 on the side 𝐵𝐸, there is an acute triangulation of 𝐴𝐵𝐸 with size 7 such that 𝐻 is the only side vertex lying on 𝐵𝐸. Such an acute triangulation of 𝐴𝐵𝐸 has new vertices 𝑅 (resp., 𝑆 and 𝑇) introduced on the side 𝐴𝐸 (resp., 𝐴𝐵). Finally, we can slightly slide 𝐻 away from 𝐸𝐵𝑃 in direction perpendicular to 𝐻𝑃 such that the triangles 𝐸𝐻𝑃 and 𝐵𝐻𝑃 become acute. This gives an acute triangulation of Γ into at most 16 acute triangles.
Subcase  2.2. There is precisely one side vertex 𝑃 on 𝐵𝐸. As in the previous subcase, by Theorem 2 of [6], for any point 𝑃 on the side 𝐵𝐸, there is an acute triangulation of the triangle 𝐴𝐵𝐸 with size 7 such that 𝑃 is the only side vertex lying on 𝐵𝐸. This gives an acute triangulation of Γ into at most 15 acute triangles.
Subcase  2.3. There are exactly two side vertices, say 𝑀 and 𝑁, on 𝐵𝐸. In this case, 𝐵𝐶𝐷𝐸 has an acute triangulation of size 7 by [5]. Suppose that 𝑀 is an interior point of 𝐸𝑁. Let 𝑃 be the point on 𝐴𝐸 such that 𝑃𝑁 is parallel to 𝐴𝐵. Then, the triangle 𝐸𝑁𝑃 has a nonacute corner 𝐸. By Theorem 2 of [6], 𝐸𝑁𝑃 can be triangulated into 7 acute triangles with 𝑀 as the only side vertex on 𝐸𝑁 and new vertices 𝑅, respectively, 𝑆 and 𝑇 on the edges 𝐸𝑃, respectively, 𝑁𝑃. Let 𝐻, 𝐾, and 𝐿 be the orthogonal projections of 𝑆, 𝑇, and 𝑁 on the edge 𝐴𝐵, respectively. The line segments 𝑁𝐿, 𝑇𝐿, 𝑇𝐾, and 𝑆𝐾 divide the right trapezoid 𝐻𝐵𝑁𝑆 into 5 right triangles. By [7, Section 3], there is an acute triangulation of the right trapezoid 𝐴𝐻𝑆𝑃 of size 6 without new vertices introduced on the sides 𝑃𝑆 and 𝐻𝑆. Then, we can slightly slide 𝐿, 𝐾, and 𝐻 in direction from 𝐵 to 𝐴 such that the triangles 𝐵𝐿𝑁, 𝑇𝑁𝐿, 𝐾𝑇𝐿, 𝐾𝑇𝑆, and 𝑆𝐻𝐾 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=𝐴𝐵𝐶𝐷𝐸 (in the anticlockwise order) be a convex pentagon with at least one acute corner, say 𝐷, and two parallel sides 𝐴𝐸 and 𝐵𝐶 with |𝐴𝐸|≤|𝐵𝐶|. We distinguish some cases.
Case  1. The triangle 𝐶𝐷𝐸 is acute.
Subcase  1.1. Let 𝐻 and 𝐾 be the orthogonal projections of 𝐴 and 𝐸, respectively, on the straight line ℓ_𝐵𝐶 containing the edge 𝐵𝐶. Suppose that 𝐻 (resp., 𝐾) is interior (resp., exterior) of 𝐵𝐶. By [7, Section 2], the trapezoid 𝐴𝐵𝐶𝐸 admits an acute triangulation of size at most 4 such that there are no new vertices on the side 𝐶𝐸. Then, Γ has an acute triangulation of size at most 5.
Subcase  1.2. Suppose that the above orthogonal projections 𝐻 and 𝐾 are interior to the edge 𝐵𝐶. By [7, Section 2], the trapezoid 𝐴𝐵𝐶𝐸 admits an acute triangulation of size at most 5 such that no new vertices are introduced on 𝐶𝐸. Then, Γ has an acute triangulation of size at most 6.
Subcase  1.3. Suppose that 𝐴𝐵𝐶𝐸 is a right trapezoid (this implies that the triangle 𝐶𝐷𝐸 is acute). By [7, Section 3], 𝐴𝐵𝐶𝐸 can be triangulated into 6 acute triangles such that there are no new vertices on 𝐶𝐸. Then, Γ has an acute triangulation of size at most 7.
Subcase  1.4. Suppose that the above orthogonal projections 𝐻 and 𝐾 are exterior to 𝐵𝐶. By [7, Section 3], 𝐴𝐵𝐶𝐸 admits an acute triangulation of size at most 7 such that there is only one vertex, say 𝑃, on the side 𝐶𝐸. By Lemma 4 of [6], for any point 𝑃 in the side 𝐶𝐸, there are two points 𝑅 on 𝐷𝐸 and 𝑆 on 𝐶𝐷 such that the line segments 𝑃𝑅, 𝑃𝑆, and 𝑅𝑆 divide 𝐶𝐷𝐸 into 4 acute triangles. Then, Γ can be triangulated into at most 11 acute triangles.
Case  2. The triangle 𝐶𝐷𝐸 is nonacute, that is, the corner 𝐸, for example, is nonacute.
Subcase  2.1. Let 𝐴𝐵𝐶𝐸 be as in subcase 1.1. There is an acute triangle, say 𝐶𝐸𝑃, which belongs to the triangulation of size ≤4 of 𝐴𝐵𝐶𝐸. Let 𝑅 be the orthogonal projection of 𝑃 on the edge 𝐶𝐸. By Theorem 2 of [6], there is an acute triangulation of 𝐶𝐷𝐸 with size 7 such that 𝑅 is the only side vertex on 𝐶𝐸. Then, Γ has an acute triangulation of size at most 12.
Subcase  2.2. Let 𝐴𝐵𝐶𝐸 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 𝐴𝐵𝐶𝐸 be as in Subcase 1.4. By Theorem 2 of [6] the triangle 𝐶𝐷𝐸 can be triangulated into at most 7 acute triangles such that the only side vertex on 𝐶𝐸 is 𝑃. This gives an acute triangulation of Γ with size ≤14.

Acknowledgment

The author would like to thank the anonymous referees for their careful reading and very useful comments and suggestions.