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Collinear Interior Lattice Points in Triangles Satisfying $B(T)\in\{4,5\}$

This paper classifies the integers $k$ for which all lattice triangles with 4 or 5 boundary points and $k$ interior points have collinear interior points, proving that only $k \in \{1, 2, 5\}$ satisfy this condition for 4 boundary points while no such integers exist for 5 boundary points.

Original authors: Jonathan Sakunkoo, Annabella Sakunkoo, Dana Paquin

Published 2026-08-04

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CC BY 4.0

Original authors: Jonathan Sakunkoo, Annabella Sakunkoo, Dana Paquin

Original paper licensed under CC BY 4.0 (http://creativecommons.org/licenses/by/4.0/). ✨ This is an AI-generated explanation of the paper below. It is not written or endorsed by the authors. For technical accuracy, refer to the original paper. Read full disclaimer

Imagine a world where you can only draw shapes using dots arranged in a perfect grid, like the squares on a piece of graph paper. In this "lattice" world, geometry isn't about smooth curves or infinite lines; it's about counting. A famous rule called Pick's Theorem tells us that if you know how many dots sit on the edge of a shape and how many sit inside it, you can calculate the shape's exact area. It's a bit like knowing the number of bricks on a wall's perimeter and the number of bricks hidden inside to figure out the wall's total size.

But mathematicians are curious creatures. They don't just want to count; they want to know how those dots are arranged. Specifically, they are fascinated by "collinearity"—a fancy word for dots that all line up in a single, straight row. If you have a handful of dots inside a triangle, do they naturally fall into a straight line, or can they scatter in a messy, zigzag pattern? This question sits at the intersection of geometry (shapes), number theory (counting and patterns), and combinatorics (arrangements). It's a puzzle about whether the rules of the grid force points to behave in a rigid, orderly way, or if they have the freedom to dance around.

This paper, written by Jonathan Sakunkoo, Annabella Sakunkoo, and Dana Paquin, dives deep into this puzzle for triangles. They ask a very specific question: If you fix the number of dots on the triangle's edge, does that force all the dots inside to line up in a straight row? They focus on two specific scenarios: triangles with exactly 4 dots on the edge and triangles with exactly 5 dots on the edge.

Here is what they discovered. First, they looked at triangles with 4 boundary dots. They found that for most numbers of interior dots, the dots can scatter and refuse to line up. However, there are three special numbers where the dots must line up: if there are 1, 2, or 5 dots inside the triangle. If you have 1 dot, it's a line by itself. If you have 2, any two dots make a line. But the number 5 is the surprise star; the authors proved mathematically that if a triangle has 4 dots on its edge and 5 inside, those 5 interior dots are forced to form a perfect straight line. They also proved that for any number of interior dots greater than 5 (like 6, 7, 8, and so on), you can always find a triangle where the dots are messy and not collinear.

Then, they turned their attention to triangles with 5 boundary dots. Here, the story takes a sharp turn. The authors proved that no number of interior dots forces them to line up. Whether you have 3 dots inside, 6 dots, or 100 dots, there is always a way to arrange the triangle so that the interior dots are scattered and not in a straight line. Even more interestingly, they showed that if you try to have a triangle with 5 edge dots and a number of interior dots that isn't a multiple of 3 (like 4 or 5), such a triangle simply cannot exist on the grid at all. But for the ones that do exist, the "straight line" rule never holds.

The authors used a clever method of simplifying every possible triangle into a "canonical" (standard) version, like flattening a crumpled piece of paper into a neat rectangle, to make the math easier. They also used a special counting tool related to prime numbers to show exactly when the dots are forced to align. Their work is a complete mathematical proof, meaning they didn't just guess or simulate; they showed it must be true for all cases.

The big picture here is a structural contrast. With 4 dots on the edge, the grid is rigid enough to force a straight line in specific, rare cases (like when there are 5 interior dots). But with 5 dots on the edge, the grid becomes flexible enough to break that pattern entirely. The authors suggest that this sharp change between 4 and 5 might be the beginning of a larger, mysterious pattern in how grid points behave, leaving the door open for future explorers to investigate what happens with 6, 7, or even more boundary dots.

Technical Summary: Collinear Interior Lattice Points in Triangles Satisfying $B(T) \in \{4, 5\}$

Problem Statement
This paper investigates the geometric rigidity of interior lattice points within lattice triangles based on the number of boundary lattice points, denoted as $B(T)$ . An integer $k$ is defined as $B_n$ -collinear if at least one lattice triangle exists with exactly $n$ boundary points and $k$ interior points, and every such triangle possesses the property that all its $k$ interior points are collinear.

Building upon prior work by Li and Paquin, which established that for $B(T)=3$ , the only $B_3$ -collinear integers are $\{1, 2, 4, 7\}$ , this study aims to completely classify the $B_4$ -collinear and $B_5$ -collinear integers. The central question is whether the combinatorial constraints imposed by the boundary count force the interior points into a collinear configuration.

Methodology
The authors employ a combination of canonical lattice representations, Pick's Theorem, and arithmetic properties of a generalized totient function.

Canonical Form: The authors adapt the canonical framework established by Li and Paquin. Any lattice triangle with $n$ boundary points can be transformed into a "canonical" triangle with vertices at $(0,0)$ , $(1,0)$ , and $(a,b)$ without altering the number of interior points, boundary points, or the collinearity of interior points.
- For $B(T)=4$ , the height $b$ is fixed by Pick's Theorem as $b = 2k + 2$ . A choice of $a$ is "valid" if $\{ \gcd(a, b), \gcd(a-1, b) \} = \{1, 2\}$ .
- For $B(T)=5$ , the height is $b = 2k + 3$ . A choice of $a$ is "valid" if $\{ \gcd(a, b), \gcd(a-1, b) \} = \{1, 3\}$ .
Non-Collinearity Criteria: The paper establishes a criterion to prove that interior points are not collinear. By analyzing the intersection of the canonical triangle with the line $x=1$ , the authors demonstrate that if there exists a valid parameter $a$ within a specific range (specifically $3 \le a \le k$ ), the triangle will contain at least two interior points on the line $x=1$ and at least one interior point off this line. Since two points define a unique line, the presence of a third point off that line proves non-collinearity.
Arithmetic Tools (Alder's Function): To determine the existence of such valid parameters $a$ , the authors utilize Alder's generalized totient function, $g(m)$ , defined as the count of integers $0 \le r < m$ such that $\gcd(m, r) = \gcd(m, 2r+1) = 1$ .
- The number of valid $a$ values between $3$ and $k$ for the $B(T)=4$ case is shown to be exactly $g(k+1) - 1$ .
- The function $g$ is proven to be multiplicative, allowing for the analysis of $g(k+1)$ based on the prime factorization of $k+1$ .

Key Results

Theorem 1 (Case $B(T)=4$ ): The only $B_4$ -collinear integers are 1, 2, and 5.
- Proof Logic: For $k=1$ and $k=2$ , collinearity is trivial. For $k=3$ and $k=4$ , explicit counterexamples (non-collinear triangles) are provided. For $k=5$ , the authors prove that the only valid $a$ values are congruent to $2$ or $11 \pmod{12}$ . In both cases, the interior points lie on a single median line, ensuring collinearity. For all $k > 5$ , the authors show that $g(k+1) > 1$ , implying the existence of a valid $a$ in the range $[3, k]$ . By the non-collinearity criterion, this guarantees the existence of a triangle with non-collinear interior points.
Theorem 2 (Case $B(T)=5$ ): There are no $B_5$ -collinear integers.
- Proof Logic: First, the authors show that for a lattice triangle with 5 boundary points to exist, the number of interior points $k$ must be a multiple of 3 (derived from the condition that $b = 2k+3$ must be divisible by 3).
- For any $k$ that is a multiple of 3, the authors construct a specific canonical triangle $T_k$ with vertices $(0,0)$ , $(1,0)$ , and $(4, 2k+3)$ . They prove that for $k \ge 3$ , this triangle contains at least two interior points on the line $x=1$ and at least one interior point off that line. Consequently, no integer $k$ forces collinearity for $B(T)=5$ . Integers not divisible by 3 are excluded by definition as no such triangles exist.

Significance and Claims
The paper claims to provide a complete classification for the first three non-trivial boundary cases ( $n=3, 4, 5$ ). The results reveal a sharp structural contrast:

While $B(T)=3$ and $B(T)=4$ exhibit specific constraints where certain interior counts force collinearity, $B(T)=5$ disrupts this pattern entirely.
The authors argue that the interaction between boundary structure and interior-point geometry is more subtle than previously understood, as increasing the boundary count from 4 to 5 eliminates the possibility of forced collinearity for any $k$ .

The paper concludes by identifying the classification of $B_6$ -collinear integers and the general behavior as $n \to \infty$ as open questions for future research, noting that the methods developed here serve as a foundation for these inquiries. The authors explicitly state that their computational verification (checking $k$ up to $10^6$ for $B(T)=4$ ) supports the theoretical classification but is not part of the formal proof.

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