ISRN Geometry

Volume 2013 (2013), Article ID 328095, 6 pages

http://dx.doi.org/10.1155/2013/328095

## On the Existence of a Point Subset with 3 or 6 Interior Points

Department of Mathematics and Statistics, Faculty of Science and Technology, Thammasat University, Pathumthani 12121, Thailand

Received 11 September 2013; Accepted 31 October 2013

Academic Editors: S. Hernández, A. Stipsicz, and S. Troubetzkoy

Copyright © 2013 Banyat Sroysang. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

For any finite planar point set in general position, an interior point of the set is a point of the set such that it is not on the boundary of the convex hull of the set . For any positive integer , let be the smallest integer such that every finite planar point set with no three collinear points and with at least interior points has a subset for which the interior of the convex hull of the set contains exactly or interior points of the set . In this paper, we prove that .

#### 1. Introduction

In this paper, we focus on finite planar point sets in general position; that is, no three points are collinear. In 1935, Erdős and Szekeres [1] posed a problem: for any integer , determine the smallest positive integer such that any finite point set of at least points has a subset of points whose convex hull contains exactly vertices. In 1961, they [2] showed that for all integer and then conjectured that for all integer . In 1974, Bonnice [3] proved that and . In 1970, Kalbfleisch et al. [4] showed that . In 2006, the computer solution for was presented by Szekeres and Peters [5]; that, .

In 2001, Avis et al. [6] posed an interior point problem: for any integer , determine the smallest positive integer such that any finite point set of at least points has a subset for which the interior of the convex hull of the set contains exactly points in the set . Moreover, they also showed the results that and . In 1974, Bonnice [3] showed that for all integer . In 2008, Wei and Ding [7] showed that for all integer . Moreover, in 2009, they [8] also showed that . In 2011, Sroysang [9] showed that for all integer . Moreover, in 2012, he [10] also showed that for all integer .

In 2001, Avis et al. [6] proved that 3 is the smallest positive integer such that any finite point set of at least 3 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 4 points in the set . Moreover, they [11] also proved that 7 is the smallest positive integer such that any finite point set of at least 7 interior points has a subset for which the interior of the convex hull of the set contains exactly 4 or 5 points in the set . In 2009, Wei and Ding [12] showed that any planar point set with 3 vertices and 9 interior points has a subset with 5 or 6 interior points of the set .

In 2010, Wei et al. [13] proved that 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 5 points in the set .

In 2012, Sroysang [14] proved that 7 is the smallest positive integer such that any finite point set of at least 7 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 7 points in the set .

In this paper, we pose an interior point problem: for any integer , determine the smallest positive integer such that any finite point set of at least points has a subset for which the interior of the convex hull of the set contains exactly or points in the set . We show that ; that is, 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 6 points in the set .

#### 2. Preliminaries

In this section, we list propositions and notations about the set , where is a finite planar point set such that no three points are collinear.

An *interior point* of the set is a point of the set such that it is not on the boundary of the convex hull of the set .

We denote notations as follows: the set of interior points of the set , the number of elements in the set , the convex hull of the set , the interior of the set , the set of vertices of the set , the number of elements in the set .

For , , the number of elements in the set .

For , the triangle with vertices , , and .

An *edge* of the set is an edge in . A subset of the set is called a -*int subset* if .

Note that there is such that .

Proposition 1 (see [8]). *9 is the smallest integer such that any finite point set of at least 9 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 points in the set .*

For any positive integer , we let be the smallest integer such that every planar point set with no three collinear points and with at least interior points has a subset for which the interior of the convex hull of the set contains exactly or points of the set .

For any positive integer ,

A finite planar point set is called a *deficient point set of type * and denoted by if , and for all .

An edge of the set is of *type * if there exists a subset of the set with such that the edge is an edge of the set .

Proposition 2 (see [8]). *Every edge of a deficient point set of type is of type 2.*

#### 3. Main Results

In this section, we will show that 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 6 points in the set .

Lemma 3. *. *

*Proof. *Let be a finite planar point set such that .

By Proposition 1, there is a subset of the set such that . Then . Hence, .

Lemma 4. *. *

*Proof. *This suffices to show the existence of a deficient point set of type . We construct a deficient point set of type as shown in Figure 1. Hence, .

Lemma 5. *Let be a finite planar point set. Assume that and . Then the set has a 3-int or 6-int subset.*

*Proof. *Suppose that each subset of a planar point set is not a 3-int subset. In [8], we have only three different configurations of the type as shown in Figures 2, 3, and 4. However, each configuration has a subset for which the interior of the convex hull of the set contains exactly 6 points of the set . Hence, the set has a 3-int or 6-int subset.

Lemma 6. *Let be a finite planar point set. Assume that and . Then the set has a 3-int or 6-int subset.*

*Proof. *Let be such that vertices are put into counterclockwise positions, respectively (see Figure 5).

Suppose that each subset of the planar point set is not a 6-int subset. Then the sets , , , and are not 6-int subsets.

If is a 2-int subset, then the set is a 6-int subset. Then the set is not a 2-int subset. Similarly, the sets , , and are not 2-int subsets.

Let .

To show that the set has a 3-int subset, we divide into seven cases.*Case 1*. There is an element in the set such that the set is a 3-int subset.

In this case, the set has a 3-int subset.*Case 2*. There is an element in the set such that the set is a 5-int subset.

Without loss of generality, we assume that . Then the set is a 3-int subset. Thus, the set has a 3-int subset.*Case 3*. There is an element in the set such that the set is a 7-int subset.

Without loss of generality, we assume that . Then the set is a 1-int subset. If the set has a 3-int subset, then the set has a 3-int subset. Assume that the set is a deficient point set of type . By Proposition 2, there is a subset of the set with such that the edge is an edge of the set . Let . Then . Thus, the set has a 3-int subset.*Case 4*. There is an element in the set such that the set is a 1-int subset.

Without loss of generality, we assume that . Then the set is a 7-int subset. Similar to Case 3, the set has a 3-int subset.*Case 5*. There is an element in the set such that the set is an 8-int subset.

By Lemma 5, the set has a 3-int subset.*Case 6*. There is an element in the set such that the set is a 0-int subset.

Without loss of generality, we assume that . Then the set is an 8-int subset. By Lemma 5, the set has a 3-int subset.*Case 7*. The sets , , , and are 4-int subsets.

If one of them has a 3-int subset, then the set has a 3-int subset. Assume that they are deficient point sets without a 3-int subset. If the edge of the set is of type 2, then we obtain that . It follows that, the edge of the set is of type 0 or type 1. If the edge of the set is of type 0, then the edge of the set is of type 3, so the set has a 3-int subset. Next, we will assume that the edge of the set is of type 1. Similarly, it suffices to assume that the edge of the set is only of type 1, the edge of the set is only of type 1, and the edge of the set is only of type 1. Hence, we obtain only one possible configuration as shown in Figure 6.

However, there is a subset of such that , as shown in Figure 7. Thus, the set has a 3-int subset.

Therefore, the set has a 3-int or 6-int subset. This proof is completed.

Lemma 7. *Let be a finite planar point set. Assume that and . Then the set has a 3-int or 6-int subset.*

*Proof. *Let and be such that vertices are put into counterclockwise positions, respectively (see in Figure 8).

Suppose that each subset of the set is not a 6-int subset. Then the set is not a 6-int subset for all .

Let .

To show that the set has a 3-int subset, we divide into six cases.*Case 1.* There is an element in the set such that the set is a 3-int subset.

In this case, the set has a 3-int subset.*Case 2*. There is an element in the set such that the set is a 7-int subset.

Then the set is a 1-int subset for some . Without loss of generality, we assume for some . If the set has a 3-int subset, then the set has a 3-int subset. Assume that the set is a deficient point set of type . By Proposition 2, there is a subset of the set with such that the edge is an edge of the set . Let . Then . Thus, the set has a 3-int subset.*Case 3.* There is an element in the set such that the set is a 5-int subset.

We divide into three subcases.*Subcase 3.1.* There is an element in the set such that the set is a 3-int subset.

In this subcase, the set has a 3-int subset.*Subcase 3.2.* There exist elements , , and in the set such that the sets , , and are 1-int subsets.

It follows that , , , and , where . Without loss of generality, we assume that . Then the set is a 6-int subset. This is impossible.*Subcase 3.3.* There exist elements , in the set such that the set is a 1-int subset and the set is a 2-int subset.

It follows that , and where . If is not between and , then the set is a 6-int subset which is impossible. Thus, is between and . We choose . Then . Hence, the set has a 3-int subset.*Case 4.* There is an element in the set such that the set is a 4-int subset.

We divide into five subcases.*Subcase 4.1.* There is an element in the set such that the set is a 3-int subset.

In this subcase, the set has a 3-int subset.*Subcase 4.2.* There exist elements , , , and in the set such that the sets , , , and are 1-int subsets.

It follows that , , , , and , where , , , , . Without loss of generality, we can assume that . If , then the set is a 3-int subset. If , then the set is a 3-int subset. Thus, the set has a 3-int subset if or . Next, we will show that the statement “” is impossible. We suppose that . Then the set is a 6-int subset which is a contradiction.*Subcase 4.3.* There exist elements and in the set such that the sets and are 2-int subsets.

It follows that , and where . Without loss of generality, we assume . Then the set is a 6-int subset. This is impossible.*Subcase 4.4.* There exist elements , , and in the set such that the sets and are 1-int subsets and the set is a 2-int subset.

It follows that , , , and , where . Without loss of generality, we can assume that . Let and . If , then the set is a 6-int subset. If , then the set is a 6-int subset. Thus, we obtain that . Then the set is a 3-int subset. Hence, the set has a 3-int subset.*Subcase 4.5.* There is an element in the set such that the set is a 4-int subset.

Without loss of generality, we assume that and , where . Let and . Then the sets and are not 6-int subsets. If the set is an 8-int subset or the set is an 8-int subset then, by Lemma 6, the set has a 3-int subset. If the set is a 7-int subset, then the set is a 3-int subset, so the set has a 3-int subset. If the set is a 7-int subset, then the set is a 3-int subset, so the set has a 3-int subset. If the set is a 5-int subset, then the set is a 3-int subset, so the set has a 3-int subset. If the set is a 5-int subset, then the set is a 3-int subset, so the set has a 3-int subset. Next, we assume that the sets and are 4-int subsets. Then is a 0-int subset. Then the set is an 8-int subset. By Lemma 6, the set has a 3-int subset.*Case 5.* There is an element in the set such that the set is an 8-int subset.

By Lemma 5, the set has a 3-int subset.*Case 6.* We have for all .

If there exist elements , , , and in the set such that the sets , , , and are 2-int subsets where the sets , , , and put into anticlockwise positions, then the set is a 6-int subset. Thus, we obtain that there is an element in the set such that it is a 1-int subset. It is easy to see that has a 3-int subset.

Therefore, the set has a 3-int or 6-int subset. This proof is completed.

Theorem 8. *One has .*

*Proof. *By Lemmas 3 and 4, it follows that . By Lemmas 5, 6, and 7, we obtain that . Hence, .

#### 4. Conclusion and Discussion

In [6], 3 is the smallest positive integer such that any finite point set of at least 3 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 4 points in the set .

In [13], 7 is the smallest positive integer such that any finite point set of at least 7 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 5 points in the set .

In this paper, 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 6 points in the set .

In [14], 7 is the smallest positive integer such that any finite point set of at least 7 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 7 points in the set .

In [15], 8 is the smallest positive integer such that any finite point set of at least 8 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or 8 points in the set . Moreover, 9 is the smallest positive integer such that any finite point set of at least 9 interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or points in the set , where .

For any positive integer , we let be the smallest integer such that every planar point set with no three collinear points and with at least interior points has a subset for which the interior of the convex hull of the set contains exactly 3 or points of the set .

For any positive integer ,

Thus, we have the following formulas:

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgment

The author would like to thank the referees for their useful comments and suggestions.

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