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Research Article | Open Access

Volume 2013 |Article ID 747128 | 5 pages | https://doi.org/10.1155/2013/747128

# Acute Triangulations of Trapezoids and Pentagons

Accepted06 Feb 2013
Published18 Mar 2013

#### Abstract

An acute triangulation of a polygon is a triangulation whose triangles have all their angles less than . 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 . The number of triangles in a triangulation is called the size. Burago and Zalgaller  and, independently, Goldberg and Manheimer  proved that every obtuse triangle can be triangulated into seven acute triangles and this bound is the best possible. Cassidy and Lord  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 . Acute triangulations of trapezoids, quadrilaterals, and pentagons were investigated in . Further information, historical notes, and problems about acute triangulations of polygons and surfaces can be found in the survey paper . Let denote a family of planar polygons, and for , let be the minimum size of an acute triangulation of . Then, let denote the maximum value of for all . The following results are known.

Theorem 1. (i) Reference : let denote the family of all trapezoids, that is, quadrilaterals with at least one pair of parallel sides. Then, , where is the family of all rectangles (also including squares).
(ii) Reference : let be the family of all quadrilaterals. Then, .
(iii) Reference : let be the family of all convex quadrilaterals. Then, .
(iv) Reference : let denote the family of all planar pentagons. Then, .

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 be a convex planar polygon. A vertex of is called an acute (resp., right) corner if the interior angle of at is less than (resp., equal to) ; otherwise, is called an obtuse corner. Let be an acute triangulation of . A vertex or edge of is called a boundary (resp., interior) vertex or edge if it lies on the boundary of (resp., lies inside ). Let , , and be the number of vertices, interior vertices, and boundary vertices of . Let , , and be the number of edges, interior edges, and boundary edges of . Clearly, we have , , and . For each vertex in , the number of edges incident to is called the degree of , denoted by . Let denote the number of vertices in of degree . Let denote the number of triangles in . The following lemma is easily verified (cf. [6, Lemma 1]).

Lemma 2. Let be an acute triangulation of a planar convex n-gon . Then, one has(1),(2),(3),(4),(5)If a vertex of is an interior vertex, then , if lies within a side of , then , and if is an obtuse or right corner of , then .

#### 2. Characterizations of Trapezoids

##### 2.1. Parallelograms

Let be a parallelogram with acute corners at and and . 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 that completes Theorem 2.1 in Section 2 of .

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

Proof. Let be the smallest number of triangles in an acute triangulation of the parallelogram . Some edge of incident with (resp., ) must meet the interior of . Denote to be such an edge, and suppose that is an interior vertex of , that is, . Then, the vertices and have degree at least 2, the vertices and have degree at least 3, and the vertex has degree at least 5. Summing up, we have , hence . Since , by Lemma 2(4), we get , so . Assume now that . Then, lies in the interior of the edge (or ). Otherwise, if , at least one of the vertices and must have degree by hypothesis. This implies that there is a neighbor of it with degree , giving a contradiction as . Thus, and as the degree of is at least 4, and the degree of (resp., ) is at least 3. By Lemma 2(1), we get . Now, for a parallelogram as in the statement, an acute triangulation with 4 triangles is given in [7, Section 2].

##### 2.2. Trapezoids

Following , 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 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.

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.

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 of any trapezoid as in the statement. Some edge must meet the interior of . Suppose that is an interior point of , that is, . Then, it immediately follows that and so . Suppose that there is no interior vertex in , that is, 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 . By Lemma 2(1), . 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 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 . 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, , and . Suppose that is an interior vertex and . Then, there are at least two interior vertices with degree , the degree of (resp., and ) is , the degree of is , and there is at least one vertex of degree . Then, we have , so . Since by Lemma 2(4), we get , so . Suppose that 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 . 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 . 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 . 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 be an acute triangulation with at least one interior vertex for a trapezoid as in the statement. Let denote the size of . Suppose that . Then, there are two interior vertices and in , and at least two neighbours of and/or have degree at least 4. Since and , Lemma 2(1) gives . Suppose that . Then, there are at least three interior vertices , , and with degree . The degrees of and are , and those of and are . There are at least two neighbours of , and/or with degree . Summing up, we have , hence . Since , by Lemma 2(4), we get , hence . So, we can assume that . 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, . This contradicts . If is connected to exactly three vertices of the trapezoid, say , , and , there are two neighbours of which have degree . Further, at least one of the obtuse corners and must have degree . Then, we have , and , hence . If is joined to exactly two vertices of , then there are three neighbours of with degree . The degree of the two vertices of joined with is , and the remaining two vertices have degree . Summing up, we get , hence . Since , by Lemma 2(4), we have , hence . If is joined to exactly one vertex of , then four neighbours of have degree . Thus, , and, by Lemma 2(1), . 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 .

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 (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 such that there are at most 2 side vertices on .
Subcase  1.1. There is no side vertex on . Then, admits an acute triangulation with at most 9 triangles.
Subcase  1.2. There is precisely one side vertex, say , on . By Lemma 4 of , 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 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 . 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 , 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 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 . Let be the orthogonal projection of on the side . By Theorem 2  of , 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 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 , 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 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 . 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 , 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 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 , 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 be a convex pentagon which has at least one acute corner and two parallel sides, nonincident to it. Then, can be triangulated into at most 14 acute triangles.

Proof. Let (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 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, 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, 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, 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 , 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, 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 of . Let be the orthogonal projection of on the edge . By Theorem 2 of , there is an acute triangulation of with size 7 such that is the only side vertex on . Then, 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 with size .
Subcase  2.3. Let be as in Subcase 1.4. By Theorem 2 of  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 with size .

#### Acknowledgment

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

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