Exercise 6.3.11* (Regular polygons with rational vertices)

Question

Prove that no $n$ points with rational coordinates $(x,y)$ can be chosen in the Euclidean plane to form the vertices of a regular polygon with $n$ sides, except in the case $n = 4$.

\bigskip
Hint. If $n=3$, the area of such a triangle can be shown to be rational by the use of one standard elementary formula, but irrational by another. For values of $n$ other than $3,4$, or $6$, a similar contradiction can be obtained by applying the law of cosines to a triangle formed by two adjacent vertices and the center of the polygon.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

\begin{proof} 
For $n = 3$, let $c 
e 0$ denote the side of an equilateral triangle, and $(a_i,b_i)$ the coordinates of the vertices $A_i,\ (i = 1,2,3)$. Assume for the sake of contradiction that these coordinates are rational.

The height of this triangle is $ h = c \sqrt{3}/2$, and the area is 
$$\mathscr{A} = \frac{1}{2} c h = \frac{c^2}{4} \sqrt{3}.$$
(Alternatively, this can be obtained by Heron's formula.)

Moreover $c^2 = (a_2 - a_1)^2 + (b_2 -b_1)^2 \in \Q^*$, and $\sqrt{3} \in \R \setminus \Q$, thus $\mathscr{A} \in \R \setminus \Q$.
The area is also half of the area of the parallelogram with sides $A_1A_2,A_1A_3$. This gives
\begin{align*}
\mathscr{A} &= \frac{1}{2} \left| \det(\overrightarrow{A_1A_2}, \overrightarrow{A_1A_3}ight |\
&=\frac{1}{2} \begin{vmatrix} a_2 - a_1 & a_3 -a_1\ b_2 - b_1 & b_3 - b_1 \end{vmatrix}.\
\end{align*}
Since the coordinates $a_i,b_i$ are rational, this shows that $\mathscr{A} \in \Q$. This is a contradiction, so no equilateral triangle has rational vertices.

\bigskip
Assume now that $n \geq 4$, and that the vertices $A_i$ of the regular polygon $(A_1,\ldots, A_n)$ have rational coordinates $(a_i,b_i)$.  Let $O$ denote the center of the regular polygon. Since $O$ is the isobarycenter of $A_1,\ldots, A_n$, its coordinates $(x_0,y_0) = ((a_1+\cdots+a_n)/n, (b_1+\cdots + b_n)/n)$ are rational. Consider the triangle $A_1 A_2 O$, with area $\mathscr{A}$.

Let $c$ be the length of $A_1A_2$ and $r$ be the length of $OA_1$, so that $c^2$ and $r^2$ are rational. The law of cosines applied to this triangle gives
\begin{align*}
c^2 = 2r^2\left(1 -\cos \left(\frac{2\pi}{n}ight) ight).
\end{align*}
Since $r >0$, this shows that $\cos \left(\frac{2\pi}{n}ight)  \in \Q$, where $n\geq 3$. By Theorem 6.16, this implies $n = 3, 4$ or $6$.

The case $n = 3$ has been treated. Of course, for $n = 4$, there exists squares with rational vertices, for instance $(0,0), (0,1), (1,1),(1,0)$.

It remains the case $n = 6$. If a regular hexagon has rational vertices, then $A_1A_2O$ is an equilateral triangle with rational vertices: this is impossible by the first part, so there don't exist a regular hexagon with rational vertices.

\bigskip
In conclusion,  no $n$ points with rational coordinates can be chosen in the Euclidean plane to form the vertices of a regular polygon with $n$ sides, except in the case $n = 4$.
\end{proof}

Exercise 6.3.11* (Regular polygons with rational vertices)

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Exercise 6.3.11* (Regular polygons with rational vertices)

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