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

Abstract

For any finite planar point set 𝑃 in general position, an interior point of the set 𝑃 is a point of the set 𝑃 such that it is not on the boundary of the convex hull of the set 𝑃. For any positive integer 𝑘≥3, let ℎ(𝑘) be the smallest integer such that every finite planar point set 𝑃 with no three collinear points and with at least ℎ(𝑘) interior points has a subset 𝑄 for which the interior of the convex hull of the set 𝑄 contains exactly 𝑘 or 𝑘+3 interior points of the set 𝑃. In this paper, we prove that ℎ(3)=8.

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 𝑘≥3, determine the smallest positive integer 𝑓(𝑘) such that any finite point set of at least 𝑓(𝑘) points has a subset of 𝑘 points whose convex hull contains exactly 𝑘 vertices. In 1961, they [2] showed that 𝑓(𝑘)≥2^𝑘−2+1 for all integer 𝑘≥3 and then conjectured that 𝑓(𝑘)≥2^𝑘−2+1 for all integer 𝑘≥3. In 1974, Bonnice [3] proved that 𝑓(3)=3 and 𝑓(4)=5. In 1970, Kalbfleisch et al. [4] showed that 𝑓(5)=9. In 2006, the computer solution for 𝑘=6 was presented by Szekeres and Peters [5]; that, 𝑓(6)=17.

In 2001, Avis et al. [6] posed an interior point problem: for any integer 𝑘≥1, determine the smallest positive integer 𝑔(𝑘) such that any finite point set 𝑃 of at least 𝑔(𝑘) points has a subset 𝑄 for which the interior of the convex hull of the set 𝑄 contains exactly 𝑘 points in the set 𝑃. Moreover, they also showed the results that 𝑔(1)=1 and 𝑔(2)=4. In 1974, Bonnice [3] showed that 𝑔(𝑘)≥3𝑘−1 for all integer 𝑘≥3. In 2008, Wei and Ding [7] showed that 𝑔(𝑘)≥3𝑘 for all integer 𝑘≥3. Moreover, in 2009, they [8] also showed that 𝑔(3)=9. In 2011, Sroysang [9] showed that 𝑔(𝑘)≥4𝑘 for all integer 𝑘≥4. Moreover, in 2012, he [10] also showed that 𝑔(𝑘)≥𝑘² for all integer 𝑘≥4.

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

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

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

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

2. Preliminaries

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

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

We denote notations as follows: 𝐼(𝑃)∶= the set of interior points of the set 𝑃, 𝑖(𝑃)∶= the number of elements in the set 𝐼(𝑃), 𝐶𝐻(𝑃)∶= the convex hull of the set 𝑃, 𝑖𝑛𝑡𝐶𝐻(𝑃)∶= the interior of the set 𝐶𝐻(𝑃), 𝑉(𝑃)∶= the set of vertices of the set 𝐶𝐻(𝑃), 𝑣(𝑃)∶= the number of elements in the set 𝑉(𝑃).

For 𝑄⊆𝑃, 𝐼^∗(𝑄)∶=𝐼(𝑃)∩𝑖𝑛𝑡𝐶𝐻(𝑄), 𝑖^∗(𝑄)∶= the number of elements in the set 𝐼^∗(𝑄).

For 𝑥,𝑦,𝑧∈𝑃, Δ𝑥𝑦𝑧∶= the triangle with vertices 𝑥, 𝑦, and 𝑧.

An edge of the set 𝑃 is an edge in 𝐶𝐻(𝑃). A subset 𝑄 of the set 𝑃 is called a 𝑘-int subset if 𝑖^∗(𝑄)=𝑘.

Note that there is 𝑄⊆𝑃 such that 𝑖(𝑄)≠𝑖^∗(𝑄).

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

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

For any positive integer 𝑘≥3, ℎ(𝑘)=min𝑠∶𝑖(𝑃)≥𝑠⟹∃𝑄⊆𝑃s.t.𝑖^∗(𝑄)=𝑘or𝑘+3 (1)

A finite planar point set 𝑃 is called a deficient point set of type 𝑃(𝑚,𝑠,𝑘,𝑛) and denoted by 𝑃=𝑃(𝑚,𝑠,𝑘,𝑛) if 𝑣(𝑃)=𝑚,𝑖(𝑃)=𝑠, and 𝑖^∗(𝑄)∉{𝑘,𝑛} for all 𝑄⊆𝑃.

An edge 𝑥𝑦 of the set 𝑃(3,𝑠,3,3) is of type 𝑘 if there exists a subset 𝑄 of the set 𝑃 with 𝑖^∗(𝑄)=𝑘 such that the edge 𝑥𝑦 is an edge of the set 𝑄.

Proposition 2 (see [8]). Every edge of a deficient point set of type 𝑃(3,7,3,3) 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 𝑃 of at least 8 interior points has a subset 𝑄 for which the interior of the convex hull of the set 𝑄 contains exactly 3 or 6 points in the set 𝑃.

Lemma 3. ℎ(3)≤9.

Proof. Let 𝑃 be a finite planar point set such that 𝑖(𝑃)≥9.
By Proposition 1, there is a subset 𝑄 of the set 𝑃 such that 𝑖^∗(𝑄)=3. Then 𝑖^∗(𝑄)∈{3,6}. Hence, ℎ(3)≤9.

Lemma 4. ℎ(3)≥8.

Proof. This suffices to show the existence of a deficient point set of type 𝑃(3,7,3,6). We construct a deficient point set 𝑃 of type 𝑃(3,7,3,6) as shown in Figure 1. Hence, ℎ(3)≥8.

Figure 1 

A deficient point set of type 𝑃(3,7,3,6).

Lemma 5. Let 𝑃 be a finite planar point set. Assume that 𝑣(𝑃)=3 and 𝑖(𝑃)=8. Then the set 𝑃 has a 3-int or 6-int subset.

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

Figure 2 

8-I-monster.

Figure 3 

8-II-monster.

Figure 4 

8-III-monster.

Lemma 6. Let 𝑃 be a finite planar point set. Assume that 𝑣(𝑃)=4 and 𝑖(𝑃)=8. Then the set 𝑃 has a 3-int or 6-int subset.

Proof. Let 𝑉(𝑃)={𝑥,𝑦,𝑧,and𝑤} be such that vertices 𝑥,𝑦,𝑧,𝑤 are put into counterclockwise positions, respectively (see Figure 5).
Suppose that each subset of the planar point set 𝑃 is not a 6-int subset. Then the sets Δ𝑥𝑦𝑧, Δ𝑦𝑧𝑤, Δ𝑧𝑤𝑥, and Δ𝑤𝑥𝑦 are not 6-int subsets.
If Δ𝑥𝑦𝑧 is a 2-int subset, then the set Δ𝑧𝑤𝑥 is a 6-int subset. Then the set Δ𝑥𝑦𝑧 is not a 2-int subset. Similarly, the sets Δ𝑦𝑧𝑤, Δ𝑧𝑤𝑥, and Δ𝑤𝑥𝑦 are not 2-int subsets.
Let 𝑇={Δ𝑥𝑦𝑧,Δ𝑦𝑧𝑤,Δ𝑧𝑤𝑥,Δ𝑤𝑥𝑦}.
To show that the set 𝑃 has a 3-int subset, we divide into seven cases.
Case  1. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is a 3-int subset.
In this case, the set 𝑃 has a 3-int subset.
Case  2. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is a 5-int subset.
Without loss of generality, we assume that 𝐴=Δ𝑥𝑦𝑧. Then the set Δ𝑧𝑤𝑥 is a 3-int subset. Thus, the set 𝑃 has a 3-int subset.
Case  3. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is a 7-int subset.
Without loss of generality, we assume that 𝐴=Δ𝑥𝑦𝑧. Then the set Δ𝑧𝑤𝑥 is a 1-int subset. If the set 𝐴 has a 3-int subset, then the set 𝑃 has a 3-int subset. Assume that the set 𝐴 is a deficient point set of type 𝑃(3,7,3,3). By Proposition 2, there is a subset 𝐵 of the set 𝐶𝐻(Δ𝑥𝑦𝑧) with 𝑖^∗(𝐵)=2 such that the edge 𝑥𝑧 is an edge of the set 𝐵. Let 𝑄=𝐵∪Δ𝑧𝑤𝑥. Then 𝑖^∗(𝑄)=3. Thus, the set 𝑃 has a 3-int subset.
Case  4. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is a 1-int subset.
Without loss of generality, we assume that 𝐴=Δ𝑥𝑦𝑧. Then the set Δ𝑧𝑤𝑥 is a 7-int subset. Similar to Case 3, the set 𝑃 has a 3-int subset.
Case  5. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is an 8-int subset.
By Lemma 5, the set 𝑃 has a 3-int subset.
Case  6. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is a 0-int subset.
Without loss of generality, we assume that 𝐴=Δ𝑥𝑦𝑧. Then the set Δ𝑧𝑤𝑥 is an 8-int subset. By Lemma 5, the set 𝑃 has a 3-int subset.
Case  7. The sets Δ𝑥𝑦𝑧, Δ𝑦𝑧𝑤, Δ𝑧𝑤𝑥, and Δ𝑤𝑥𝑦 are 4-int subsets.
If one of them has a 3-int subset, then the set 𝑃 has a 3-int subset. Assume that they are deficient point sets without a 3-int subset. If the edge 𝑥𝑧 of the set Δ𝑥𝑦𝑧 is of type 2, then we obtain that 𝑖^∗(𝑃⧵{𝑦})=6. It follows that, the edge 𝑥𝑧 of the set Δ𝑥𝑦𝑧 is of type 0 or type 1. If the edge 𝑥𝑧 of the set Δ𝑥𝑦𝑧 is of type 0, then the edge 𝑥𝑦 of the set Δ𝑥𝑦𝑧 is of type 3, so the set 𝑃 has a 3-int subset. Next, we will assume that the edge 𝑥𝑧 of the set Δ𝑥𝑦𝑧 is of type 1. Similarly, it suffices to assume that the edge 𝑥𝑧 of the set Δ𝑧𝑤𝑥 is only of type 1, the edge 𝑦𝑤 of the set Δ𝑦𝑧𝑤 is only of type 1, and the edge 𝑦𝑤 of the set Δ𝑤𝑥𝑦 is only of type 1. Hence, we obtain only one possible configuration as shown in Figure 6.
However, there is a subset 𝑄 of 𝑃 such that 𝑖^∗(𝑄)=3, as shown in Figure 7. Thus, the set 𝑃 has a 3-int subset.
Therefore, the set 𝑃 has a 3-int or 6-int subset. This proof is completed.

Figure 5 

Vertices 𝑥, 𝑦, and 𝑧, 𝑤 are put into counterclockwise positions.

Figure 6 

The edges 𝑥𝑧 and 𝑦𝑤 are only of type 1.

Figure 7 

A 3-int subset of the set in Figure 6.

Lemma 7. Let 𝑃 be a finite planar point set. Assume that 𝑣(𝑃)≥5 and 𝑖(𝑃)=8. Then the set 𝑃 has a 3-int or 6-int subset.

Proof. Let 𝑣(𝑃)=𝑚 and 𝑉(𝑃)={𝑣₁,𝑣₂,…,𝑣_𝑚} be such that vertices 𝑣₁,𝑣₂,…,𝑣_𝑚 are put into counterclockwise positions, respectively (see in Figure 8).
Suppose that each subset of the set 𝑃 is not a 6-int subset. Then the set Δ𝑣₁𝑣_𝑗𝑣_𝑗+1 is not a 6-int subset for all 𝑗.
Let 𝑇={Δ𝑣₁𝑣_𝑗𝑣_𝑗+1∣𝑗=2,3,…,𝑚}.
To show that the set 𝑃 has a 3-int subset, we divide into six cases.
Case  1. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is a 3-int subset.
In this case, the set 𝑃 has a 3-int subset.
Case  2. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is a 7-int subset.
Then the set Δ𝑣₁𝑣_𝑡𝑣_𝑡+1 is a 1-int subset for some 𝑡. Without loss of generality, we assume 𝐴=Δ𝑣₁𝑣_𝑗𝑣_𝑗+1 for some 𝑗>𝑡. If the set 𝐴 has a 3-int subset, then the set 𝑃 has a 3-int subset. Assume that the set 𝐴 is a deficient point set of type 𝑃(3,7,3,3). By Proposition 2, there is a subset 𝐵 of the set 𝐶𝐻(𝐴) with 𝑖^∗(𝐵)=2 such that the edge 𝑣₁𝑣_𝑗 is an edge of the set 𝐵. Let 𝑄=𝐵∪Δ𝑣₁𝑣_𝑡𝑣_𝑡+1. Then 𝑖^∗(𝑄)=3. Thus, the set 𝑃 has a 3-int subset.
Case  3. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is a 5-int subset.
We divide into three subcases.
Subcase  3.1. There is an element 𝐵 in the set 𝑇 such that the set 𝐵 is a 3-int subset.
In this subcase, the set 𝑃 has a 3-int subset.
Subcase  3.2. There exist elements 𝐵, 𝐶, and 𝐷 in the set 𝑇 such that the sets 𝐵, 𝐶, and 𝐷 are 1-int subsets.
It follows that 𝐴=Δ𝑣₁𝑣_𝑗𝑣_𝑗+1, 𝐵=Δ𝑣₁𝑣_𝑡𝑣_𝑡+1, 𝐶=Δ𝑣₁𝑣_𝑠𝑣_𝑠+1, and 𝐷=Δ𝑣₁𝑣_𝑟𝑣_𝑟+1, where 𝑗,𝑡,𝑠,𝑟∈{2,3,…,𝑚}. Without loss of generality, we assume that |𝑡−𝑗|=min{|𝑡−𝑗|,|𝑠−𝑗|,|𝑟−𝑗|}. Then the set 𝐴∪𝐵 is a 6-int subset. This is impossible.
Subcase  3.3. There exist elements 𝐵, 𝐶 in the set 𝑇 such that the set 𝐵 is a 1-int subset and the set 𝐶 is a 2-int subset.
It follows that 𝐴=Δ𝑣₁𝑣_𝑗𝑣_𝑗+1, 𝐵=Δ𝑣₁𝑣_𝑡𝑣_𝑡+1 and 𝐶=Δ𝑣₁𝑣_𝑟𝑣_𝑟+1 where 𝑗,𝑡,𝑟∈{2,3,…,𝑚}. If 𝑟 is not between 𝑡 and 𝑗, then the set 𝐴∪𝐵 is a 6-int subset which is impossible. Thus, 𝑟 is between 𝑡 and 𝑗. We choose 𝑄=𝐵∪𝐶. Then 𝑖^∗(𝑄)=3. Hence, the set 𝑃 has a 3-int subset.
Case  4. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is a 4-int subset.
We divide into five subcases.
Subcase  4.1. There is an element 𝐵 in the set 𝑇 such that the set 𝐵 is a 3-int subset.
In this subcase, the set 𝑃 has a 3-int subset.
Subcase  4.2. There exist elements 𝐵, 𝐶, 𝐷, and 𝐸 in the set 𝑇 such that the sets 𝐵, 𝐶, 𝐷, and 𝐸 are 1-int subsets.
It follows that 𝐴=Δ𝑣₁𝑣_𝑗𝑣_𝑗+1, 𝐵=Δ𝑣₁𝑣_𝑡𝑣_𝑡+1, 𝐶=Δ𝑣₁𝑣_𝑠𝑣_𝑠+1, 𝐷=Δ𝑣₁𝑣_𝑘𝑣_𝑘+1, and 𝐸=Δ𝑣₁𝑣_𝑟𝑣_𝑟+1, where 𝑗, 𝑡, 𝑠, 𝑟, 𝑘∈{2,3,…,𝑚}. Without loss of generality, we can assume that 𝑡<𝑠<𝑘<𝑟. If 𝑗<𝑠, then the set 𝐶∪𝐷∪𝐸 is a 3-int subset. If 𝑘<𝑗, then the set 𝐵∪𝐶∪𝐷 is a 3-int subset. Thus, the set 𝑃 has a 3-int subset if 𝑗<𝑠 or 𝑘<𝑗. Next, we will show that the statement “𝑠<𝑗<𝑘” is impossible. We suppose that 𝑠<𝑗<𝑘. Then the set 𝐴∪𝐶∪𝐷 is a 6-int subset which is a contradiction.
Subcase  4.3. There exist elements 𝐵 and 𝐶 in the set 𝑇 such that the sets 𝐵 and 𝐶 are 2-int subsets.
It follows that 𝐴=Δ𝑣₁𝑣_𝑗𝑣_𝑗+1, 𝐵=Δ𝑣₁𝑣_𝑡𝑣_𝑡+1and 𝐶=Δ𝑣₁𝑣_𝑟𝑣_𝑟+1 where 𝑗,𝑡,𝑟∈{2,3,…,𝑚}. Without loss of generality, we assume |𝑡−𝑗|<|𝑟−𝑗|. Then the set 𝐴∪𝐵 is a 6-int subset. This is impossible.
Subcase  4.4. There exist elements 𝐵, 𝐶, and 𝐷 in the set 𝑇 such that the sets 𝐵 and 𝐶 are 1-int subsets and the set 𝐷 is a 2-int subset.
It follows that 𝐴=Δ𝑣₁𝑣_𝑗𝑣_𝑗+1, 𝐵=Δ𝑣₁𝑣_𝑡𝑣_𝑡+1, 𝐶=Δ𝑣₁𝑣_𝑠𝑣_𝑠+1, and 𝐷=Δ𝑣₁𝑣_𝑟𝑣_𝑟+1, where 𝑗,𝑡,𝑠,𝑟∈  {2,3,…,𝑚}. Without loss of generality, we can assume that 𝑗<𝑟. Let 𝑘=max{𝑡,𝑠} and 𝑙=min{𝑡,𝑠}. If 𝑘<𝑟, then the set 𝐴∪𝐵∪𝐶 is a 6-int subset. If 𝑟<𝑙, then the set 𝐴∪𝐷 is a 6-int subset. Thus, we obtain that 𝑙<𝑟<𝑘. Then the set 𝐷∪Δ𝑣₁𝑣_𝑘𝑣_𝑘+1 is a 3-int subset. Hence, the set 𝑃 has a 3-int subset.
Subcase  4.5. There is an element 𝐵 in the set 𝑇⧵{𝐴} such that the set 𝐵 is a 4-int subset.
Without loss of generality, we assume that 𝐴=Δ𝑣₁𝑣_𝑗𝑣_𝑗+1 and 𝐵=Δ𝑣₁𝑣_𝑡𝑣_𝑡+1, where 𝑗<𝑡. Let 𝐶={𝑣₁,𝑣_𝑗,𝑣_𝑡,𝑣_𝑡+1} and 𝐷={𝑣₁,𝑣_𝑗,𝑣_𝑗+1,𝑣_𝑡+1}. Then the sets 𝐶 and 𝐷 are not 6-int subsets. If the set 𝐶 is an 8-int subset or the set 𝐵 is an 8-int subset then, by Lemma 6, the set 𝑃 has a 3-int subset. If the set 𝐶 is a 7-int subset, then the set Δ𝑣₁𝑣_𝑗𝑣_𝑡 is a 3-int subset, so the set 𝑃 has a 3-int subset. If the set 𝐷 is a 7-int subset, then the set Δ𝑣₁𝑣_𝑗+1𝑣_𝑡+1 is a 3-int subset, so the set 𝑃 has a 3-int subset. If the set 𝐶 is a 5-int subset, then the set Δ𝑣_𝑗𝑣_𝑗+1𝑣_𝑡 is a 3-int subset, so the set 𝑃 has a 3-int subset. If the set 𝐷 is a 5-int subset, then the set Δ𝑣_𝑗+1𝑣_𝑡𝑣_𝑡+1 is a 3-int subset, so the set 𝑃 has a 3-int subset. Next, we assume that the sets 𝐶 and 𝐷 are 4-int subsets. Then Δ𝑣₁𝑣_𝑗𝑣_𝑡+1 is a 0-int subset. Then the set {𝑣_𝑗,𝑣_𝑗+1,𝑣_𝑡,𝑣_𝑡+1} is an 8-int subset. By Lemma 6, the set 𝑃 has a 3-int subset.
Case  5. There is an element 𝐴 in the set 𝑇 such that the set 𝐴 is an 8-int subset.
By Lemma 5, the set 𝑃 has a 3-int subset.
Case  6. We have 𝑖^∗(𝐴)≤2 for all 𝐴∈𝑇.
If there exist elements 𝐴, 𝐵, 𝐶, and 𝐷 in the set 𝑇 such that the sets 𝐴, 𝐵, 𝐶, and 𝐷 are 2-int subsets where the sets 𝐴, 𝐵, 𝐶, and 𝐷 put into anticlockwise positions, then the set 𝐴∪𝐵∪𝐶 is a 6-int subset. Thus, we obtain that there is an element in the set 𝑇 such that it is a 1-int subset. It is easy to see that 𝑃 has a 3-int subset.
Therefore, the set 𝑃 has a 3-int or 6-int subset. This proof is completed.

Figure 8 

Vertices 𝑣1,𝑣2,…,𝑣𝑚 are put into counterclockwise positions.

Theorem 8. One has ℎ(3)=8.

Proof. By Lemmas 3 and 4, it follows that 8≤ℎ(3)≤9. By Lemmas 5, 6, and 7, we obtain that ℎ(3)≤8. Hence,   ℎ(3)=8.

4. Conclusion and Discussion

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

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

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

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

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

For any positive integer 𝑘≥4, we let ℎ^∗(𝑘) be the smallest integer such that every planar point set 𝑃 with no three collinear points and with at least ℎ^∗(𝑘) interior points has a subset 𝑄 for which the interior of the convex hull of the set 𝑄 contains exactly 3 or 𝑘 points of the set 𝑃.

For any positive integer 𝑘≥3, ℎ^∗(𝑘)=min𝑠∶𝑖(𝑃)≥𝑠⟹∃𝑄⊆𝑃s.t.𝑖^∗(𝑄)=3or𝑘. (2)

Thus, we have the following formulas: ℎ^∗(𝑘)ℎ=3if𝑘=4;_∗(𝑘)ℎ=7if𝑘=5,7;_∗(𝑘)ℎ=8if𝑘=6,8;_∗(𝑘)=9if𝑘≥9. (3)

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.