Exercise 1317Geometry › Analytic geometryOlympiadAdded 5 Nov 2026

Pick's theorem and lattice triangles

#Pick's theorem#lattice points#area#equilateral triangle

Statement

Pick's theorem. A polygon with vertices at lattice points, $I$ lattice points in its interior and $B$ lattice points on its boundary, has area $I+\frac B2-1$ .

Using it:

Compute the area and the number of interior lattice points of the triangle with vertices $(0,0)$ , $(5,2)$ , $(2,3)$ .
Prove that a lattice triangle having no lattice point other than its vertices has area $\frac12$ .
Prove that there is no equilateral triangle with all three vertices at lattice points.

Hint

For part 1, the number of lattice points on a segment from $(0,0)$ to $(p,q)$ (endpoints included) is $\gcd(p,q)+1$ . For part 3, compare the area of an equilateral triangle with the rational numbers produced by Pick's formula.

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Solution

1. The triangle. Its area is $\frac12|5\cdot3-2\cdot2|=\frac{11}2$ . The sides are the vectors $(5,2)$ , $(-3,1)$ , $(-2,-3)$ , with gcds $1,1,1$ , so $B=3$ . Pick: $\frac{11}2=I+\frac32-1$ , so $I=5$ . The interior lattice points are $(1,1),(2,1),(2,2),(3,2),(4,2)$ .

2. Triangles without other lattice points. $I=0$ , $B=3$ , so area $=\frac12$ .

3. No lattice equilateral triangle. Suppose the vertices are lattice points and the side has squared length $n=p^2+q^2$ . The area is $\frac{\sqrt3}4\,n$ , irrational. But by Pick, the area of a lattice polygon is a rational $I+\frac B2-1$ . Contradiction.

Remark. The square is the only regular polygon with lattice vertices in the plane.

Lesson. Pick's theorem forces rationality of areas of lattice polygons: any area containing $\sqrt3$ is impossible.

Pick's theorem and lattice triangles

Statement

Solution

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