Home > Geometry, Problem Solving > Regular polygons and lattice points.

Regular polygons and lattice points.


Se say that a point in the plane is a lattice point if both its coordinates are integers. Prove that for n\neq 4 we can’t find a regular polygon with n egdes, its vertices being lattice points.

Solution:
Looking at Pick’s Formula, we see that two times the area of a polygon with vertices in lattice points is an integer. But if we take an equilateral triangle with side of length a with a^2 \in \mathbb{Z} (distance formula) then its area is \frac{a^2 \sqrt{3}}{4} which is not a rational number. Contradiction. As suggested in the comments, we need to take care of the case n=6 separately, but this is a simple consequence of n=3.
Suppose now n \geq 5,\ n\neq 6 and denote our polygon P_1...P_n. Then the angles of such a polygon are greater than 90^\circ. Therefore, if we construct Q_i such that \overrightarrow{P_iQ_i}=\overrightarrow{P_{i+1}P_{i+2}}. Then Q_1...Q_n is a regular polygon with vertices in lattice points, which is strictly inside the initial polygon. If P_1...P_n exists, then we can repeat this construction as many times we want, but this is a contradiction, since there are only a finite number of lattice points inside P_1..P_n.

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  1. Mathefreck
    November 15, 2013 at 11:04 am

    I don’t think this works for n = 6, because then Q_1 = Q_2 = … = Q_6. But then P_1P_2Q_1 would be an equilateral triangle, which provides the contradiction.

    • November 15, 2013 at 11:36 am

      Yes, you are right. The case n=6 should be treated separately, but as you point out, it is a consequence of the case n=3.

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