5. Let $T$ be a triangle with vertices at points with integer coordinates such that each side of $T$ contains exactly $m$ points with integer coordinates. If the area of $T$ is less than 2020, determine the largest possible value of $m$.


Solution. Let us note that if the points with integer coordinates $(a, b)$ and $(c, d)$ lie on the same line, then the points

$$
(a+k(c-a), b+k(d-b)), k \in \mathbb{Z}
$$

also lie on the same line, and the distance between two adjacent points among them is equal. Let the desired triangle (with the maximum value of $m$) be denoted by $A B C$.

From the condition of the problem and the above discussion, it follows that the distances between adjacent points on the sides of triangle $A B C$ must be equal. Otherwise, from all the distances between adjacent points with integer coordinates on one of the sides, we choose the minimum, and due to the above discussion, we get that the number of points with integer coordinates is greater than $m$.

In each of the points with integer coordinates, we draw lines parallel to the sides of triangle $A B C$. Due to similarity, each of the lines drawn through the points on a side of $A B C$ must pass through some of the integer points on the other two sides. As a result of this, we get a network formed by congruent triangles. Let the area of one of these triangles be $P$. Then

$$
P_{A B C}=(1+3+\ldots+(2 m-3)) P
$$

and we have

$$
(m-1)^{2} P<2020
$$

According to this, the problem reduces to finding a triangle with integer coordinates with the minimum area.

Lemma. The area of a triangle with vertices at points with integer coordinates is greater than or equal to $\frac{1}{2}$.

Proof. The lemma is a direct consequence of Pick's theorem, which states: Let a polygon be given whose vertices have integer coordinates. If $i$ is the number of points with integer coordinates that are inside the polygon, and $b$ is the number of points with integer coordinates that are vertices of the polygon or lie on its sides, then its area is $P=i+\frac{b}{2}-1$. Therefore, $P=\frac{k}{2}$ for some $k \in \mathbb{N}$.

From inequality (2), we get that

$$
(m-1)^{2}<4040 \text { or } m \leq 64
$$

The value $m=64$ is achieved only when $P=\frac{1}{2}$. A triangle with area $P=\frac{1}{2}$ is a triangle with vertices at the points $(0,0),(1,0)$, and $(0,1)$. Due to the network given at the beginning, one of the possible triangles $A B C$ will have vertices at the points with coordinates $(0,0),(63,0)$, and $(0,63)$.