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Fixed a sign error, added missing "=0" to the necessary conditions.
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Matthew Bolan
Matthew Bolan
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It is possible to modify Uranix's lovely answer to show that the only $n \ge 3$ for which there is a regular $n$-gon with a rational point on each side are $n=3,4,8$ (for which simple constructions have already been demonstrated here).

Observing that one can obtain a regular $n$-gon by extending $n$ sides of a regular $nm$-gon, it suffices to prove impossibility in the case in which $n$ is $16,6,9$, or an odd prime $p > 3$. Fix such an $n$ and suppose for sake of contradiction such an $n$-gon exists. Let $q_i$ be a rational point on the $i$th side (oriented counterclockwise), $p$ the center of the polygon, and $T$ rotation by $-\frac{2 \pi}{n}$. It will also be convenient to define $c(k) = \cos\left(\frac{2\pi k}{n}\right)$ and $s(k) = \sin\left(\frac{2\pi k}{n}\right)$.

Now exactly as in Uranix's answer the points $$ T^i(q_i-p) $$ lie in a common line. Thus for each $k$, the sum $$ a_k = \sum_{i=1}^n (T^i(q_i-p)) \vee (T^{i+k}(q_{i+k}-p)) $$ is equal to $0$ (where indices are taken modulo $n$). Now using $x \vee T^ky = c(k) x \vee y - s(k) x \cdot y$ we have \begin{align*} a_k &= \sum_{i=1}^n (q_i-p) \vee (T^{k}(q_{i+k}-p))\\ &= \sum_{i=1}^n c(k) (q_{i}-p) \vee (q_{i+k}-p) - s(k) (q_{i}-p) \cdot (q_{i+k}-p)\\ &= c(k)\sum_{i=1}^n q_{i} \vee q_{i+k} - s(k) \sum_{i=1}^n q_{i}\cdot q_{i+k}-s(k)\sum_{i=1}^n 2 p\cdot (q_i - p ).\\ \end{align*}\begin{align*} a_k &= \sum_{i=1}^n (q_i-p) \vee (T^{k}(q_{i+k}-p))\\ &= \sum_{i=1}^n [c(k) (q_{i}-p) \vee (q_{i+k}-p) - s(k) (q_{i}-p) \cdot (q_{i+k}-p)]\\ &= c(k)\sum_{i=1}^n q_{i} \vee q_{i+k} - s(k) \sum_{i=1}^n q_{i}\cdot q_{i+k}+2s(k)\sum_{i=1}^n p\cdot (q_i - p ).\\ \end{align*} We wish to be rid of those terms depending on $p$. Denote $V_k = \sum_{i=1}^n q_{i} \vee q_{i+k}$ and $D_k = \sum_{i=1}^n q_{i} \cdot q_{i+k}$, and observe that \begin{align*} 0 &= 2s(m)a_k - 2s(k)a_m \\&= 2s(m)c(k)V_k - 2s(k)c(m)V_m-2s(m)s(k)(D_m-D_k) \\&= s(m+k)(V_k-V_m)+s(m-k)(V_k+V_m)-(c(m-k)-c(m+k))(D_m-D_k) \end{align*} This is finally our necessary condition for solutions to exist. Most of the argument works with $m=2,k=1$ where we have \begin{equation} \label{eqn:main} s(3)(V_1-V_2)+s(1)(V_1+V_2)-(c(1)-c(3))(D_2-D_1)\tag{$*$} \end{equation}\begin{equation} \label{eqn:main} s(3)(V_1-V_2)+s(1)(V_1+V_2)-(c(1)-c(3))(D_2-D_1)=0\tag{$*$} \end{equation} Note the crucial facts that $D_k,V_k$ are rational and that $V_1$ and $V_2$ are the volumes of the polytopes $q_1q_2q_3...$ and $q_1q_3q_5...$ and thus positive when $n > 4$.

We now derive a contradiction for every listed $n$.

Firstly if $n$ is an odd prime and greater than $3$, we have that $\{s(1),s(3),c(1),c(3)\}$ are $\mathbb Q$-linearly independent since $\{e^{\pm\frac{2\pi i}{n}},e^{\pm 3\frac{2\pi i}{n}}\}$ are $\mathbb Q(i)$-linearly independent. Thus for \eqref{eqn:main} to hold we require $$(V_1-V_2)=(V_1+V_2)=(D_2-D_1)=0$$ so in particular $V_1=V_2=0$, contradicting positivity of the $V_k$.

If $n=6$, then $s(3)$ vanishes and \eqref{eqn:main} becomes $$ 0=\frac{\sqrt{3}}{2}(V_1+V_2)-\frac{3}{2}(D_2-D_1), $$ so again $V_1+V_2=0$, contradicting the positivity of the $V_k$.

If $n=9$ we use the alternate condition $$s(5)(V_2-V_3)+s(1)(V_2+V_3)-(c(1)-c(5))(D_3-D_2)$$$$s(5)(V_2-V_3)+s(1)(V_2+V_3)-(c(1)-c(5))(D_3-D_2)=0$$ for which the same argument as in the odd prime case applies to show $s(1), s(5), c(1)-c(5)$ are linearly independent, giving $(V_2-V_3)=(V_2+V_3)=(D_3-D_2)=0$ and thus $V_2=0$, again a contradiction.

Finally if $n=16$, then $c(1)$ and $c(3)$ are $\mathbb Q$-linearly independent, but $s(1) = c(3)$ and $s(3)=c(1)$, so \eqref{eqn:main} becomes $$ V_1-V_2 = D_2-D_1 = -V_1-V_2 $$ and thus $2V_1=0$ in this case, contradicting the positivity of the $V_k$.

This answer of course relies heavily on the stipulation that the rational points lie on each side. If this is removed there are solutions for at least the pentagon (as shown by Uranix) and the hexagon as well. I am curious if any other $n$ admit such solutions.

It is possible to modify Uranix's lovely answer to show that the only $n \ge 3$ for which there is a regular $n$-gon with a rational point on each side are $n=3,4,8$ (for which simple constructions have already been demonstrated here).

Observing that one can obtain a regular $n$-gon by extending $n$ sides of a regular $nm$-gon, it suffices to prove impossibility in the case in which $n$ is $16,6,9$, or an odd prime $p > 3$. Fix such an $n$ and suppose for sake of contradiction such an $n$-gon exists. Let $q_i$ be a rational point on the $i$th side (oriented counterclockwise), $p$ the center of the polygon, and $T$ rotation by $-\frac{2 \pi}{n}$. It will also be convenient to define $c(k) = \cos\left(\frac{2\pi k}{n}\right)$ and $s(k) = \sin\left(\frac{2\pi k}{n}\right)$.

Now exactly as in Uranix's answer the points $$ T^i(q_i-p) $$ lie in a common line. Thus for each $k$, the sum $$ a_k = \sum_{i=1}^n (T^i(q_i-p)) \vee (T^{i+k}(q_{i+k}-p)) $$ is equal to $0$ (where indices are taken modulo $n$). Now using $x \vee T^ky = c(k) x \vee y - s(k) x \cdot y$ we have \begin{align*} a_k &= \sum_{i=1}^n (q_i-p) \vee (T^{k}(q_{i+k}-p))\\ &= \sum_{i=1}^n c(k) (q_{i}-p) \vee (q_{i+k}-p) - s(k) (q_{i}-p) \cdot (q_{i+k}-p)\\ &= c(k)\sum_{i=1}^n q_{i} \vee q_{i+k} - s(k) \sum_{i=1}^n q_{i}\cdot q_{i+k}-s(k)\sum_{i=1}^n 2 p\cdot (q_i - p ).\\ \end{align*} We wish to be rid of those terms depending on $p$. Denote $V_k = \sum_{i=1}^n q_{i} \vee q_{i+k}$ and $D_k = \sum_{i=1}^n q_{i} \cdot q_{i+k}$, and observe that \begin{align*} 0 &= 2s(m)a_k - 2s(k)a_m \\&= 2s(m)c(k)V_k - 2s(k)c(m)V_m-2s(m)s(k)(D_m-D_k) \\&= s(m+k)(V_k-V_m)+s(m-k)(V_k+V_m)-(c(m-k)-c(m+k))(D_m-D_k) \end{align*} This is finally our necessary condition for solutions to exist. Most of the argument works with $m=2,k=1$ where we have \begin{equation} \label{eqn:main} s(3)(V_1-V_2)+s(1)(V_1+V_2)-(c(1)-c(3))(D_2-D_1)\tag{$*$} \end{equation} Note the crucial facts that $D_k,V_k$ are rational and that $V_1$ and $V_2$ are the volumes of the polytopes $q_1q_2q_3...$ and $q_1q_3q_5...$ and thus positive when $n > 4$.

We now derive a contradiction for every listed $n$.

Firstly if $n$ is an odd prime and greater than $3$, we have that $\{s(1),s(3),c(1),c(3)\}$ are $\mathbb Q$-linearly independent since $\{e^{\pm\frac{2\pi i}{n}},e^{\pm 3\frac{2\pi i}{n}}\}$ are $\mathbb Q(i)$-linearly independent. Thus for \eqref{eqn:main} to hold we require $$(V_1-V_2)=(V_1+V_2)=(D_2-D_1)=0$$ so in particular $V_1=V_2=0$, contradicting positivity of the $V_k$.

If $n=6$, then $s(3)$ vanishes and \eqref{eqn:main} becomes $$ 0=\frac{\sqrt{3}}{2}(V_1+V_2)-\frac{3}{2}(D_2-D_1), $$ so again $V_1+V_2=0$, contradicting the positivity of the $V_k$.

If $n=9$ we use the alternate condition $$s(5)(V_2-V_3)+s(1)(V_2+V_3)-(c(1)-c(5))(D_3-D_2)$$ for which the same argument as in the odd prime case applies to show $s(1), s(5), c(1)-c(5)$ are linearly independent, giving $(V_2-V_3)=(V_2+V_3)=(D_3-D_2)=0$ and thus $V_2=0$, again a contradiction.

Finally if $n=16$, then $c(1)$ and $c(3)$ are $\mathbb Q$-linearly independent, but $s(1) = c(3)$ and $s(3)=c(1)$, so \eqref{eqn:main} becomes $$ V_1-V_2 = D_2-D_1 = -V_1-V_2 $$ and thus $2V_1=0$ in this case, contradicting the positivity of the $V_k$.

This answer of course relies heavily on the stipulation that the rational points lie on each side. If this is removed there are solutions for at least the pentagon (as shown by Uranix) and the hexagon as well. I am curious if any other $n$ admit such solutions.

It is possible to modify Uranix's lovely answer to show that the only $n \ge 3$ for which there is a regular $n$-gon with a rational point on each side are $n=3,4,8$ (for which simple constructions have already been demonstrated here).

Observing that one can obtain a regular $n$-gon by extending $n$ sides of a regular $nm$-gon, it suffices to prove impossibility in the case in which $n$ is $16,6,9$, or an odd prime $p > 3$. Fix such an $n$ and suppose for sake of contradiction such an $n$-gon exists. Let $q_i$ be a rational point on the $i$th side (oriented counterclockwise), $p$ the center of the polygon, and $T$ rotation by $-\frac{2 \pi}{n}$. It will also be convenient to define $c(k) = \cos\left(\frac{2\pi k}{n}\right)$ and $s(k) = \sin\left(\frac{2\pi k}{n}\right)$.

Now exactly as in Uranix's answer the points $$ T^i(q_i-p) $$ lie in a common line. Thus for each $k$, the sum $$ a_k = \sum_{i=1}^n (T^i(q_i-p)) \vee (T^{i+k}(q_{i+k}-p)) $$ is equal to $0$ (where indices are taken modulo $n$). Now using $x \vee T^ky = c(k) x \vee y - s(k) x \cdot y$ we have \begin{align*} a_k &= \sum_{i=1}^n (q_i-p) \vee (T^{k}(q_{i+k}-p))\\ &= \sum_{i=1}^n [c(k) (q_{i}-p) \vee (q_{i+k}-p) - s(k) (q_{i}-p) \cdot (q_{i+k}-p)]\\ &= c(k)\sum_{i=1}^n q_{i} \vee q_{i+k} - s(k) \sum_{i=1}^n q_{i}\cdot q_{i+k}+2s(k)\sum_{i=1}^n p\cdot (q_i - p ).\\ \end{align*} We wish to be rid of those terms depending on $p$. Denote $V_k = \sum_{i=1}^n q_{i} \vee q_{i+k}$ and $D_k = \sum_{i=1}^n q_{i} \cdot q_{i+k}$, and observe that \begin{align*} 0 &= 2s(m)a_k - 2s(k)a_m \\&= 2s(m)c(k)V_k - 2s(k)c(m)V_m-2s(m)s(k)(D_m-D_k) \\&= s(m+k)(V_k-V_m)+s(m-k)(V_k+V_m)-(c(m-k)-c(m+k))(D_m-D_k) \end{align*} This is finally our necessary condition for solutions to exist. Most of the argument works with $m=2,k=1$ where we have \begin{equation} \label{eqn:main} s(3)(V_1-V_2)+s(1)(V_1+V_2)-(c(1)-c(3))(D_2-D_1)=0\tag{$*$} \end{equation} Note the crucial facts that $D_k,V_k$ are rational and that $V_1$ and $V_2$ are the volumes of the polytopes $q_1q_2q_3...$ and $q_1q_3q_5...$ and thus positive when $n > 4$.

We now derive a contradiction for every listed $n$.

Firstly if $n$ is an odd prime and greater than $3$, we have that $\{s(1),s(3),c(1),c(3)\}$ are $\mathbb Q$-linearly independent since $\{e^{\pm\frac{2\pi i}{n}},e^{\pm 3\frac{2\pi i}{n}}\}$ are $\mathbb Q(i)$-linearly independent. Thus for \eqref{eqn:main} to hold we require $$(V_1-V_2)=(V_1+V_2)=(D_2-D_1)=0$$ so in particular $V_1=V_2=0$, contradicting positivity of the $V_k$.

If $n=6$, then $s(3)$ vanishes and \eqref{eqn:main} becomes $$ 0=\frac{\sqrt{3}}{2}(V_1+V_2)-\frac{3}{2}(D_2-D_1), $$ so again $V_1+V_2=0$, contradicting the positivity of the $V_k$.

If $n=9$ we use the alternate condition $$s(5)(V_2-V_3)+s(1)(V_2+V_3)-(c(1)-c(5))(D_3-D_2)=0$$ for which the same argument as in the odd prime case applies to show $s(1), s(5), c(1)-c(5)$ are linearly independent, giving $(V_2-V_3)=(V_2+V_3)=(D_3-D_2)=0$ and thus $V_2=0$, again a contradiction.

Finally if $n=16$, then $c(1)$ and $c(3)$ are $\mathbb Q$-linearly independent, but $s(1) = c(3)$ and $s(3)=c(1)$, so \eqref{eqn:main} becomes $$ V_1-V_2 = D_2-D_1 = -V_1-V_2 $$ and thus $2V_1=0$ in this case, contradicting the positivity of the $V_k$.

This answer of course relies heavily on the stipulation that the rational points lie on each side. If this is removed there are solutions for at least the pentagon (as shown by Uranix) and the hexagon as well. I am curious if any other $n$ admit such solutions.

Source Link
Matthew Bolan
Matthew Bolan
  • 531
  • 1
  • 4
  • 13

It is possible to modify Uranix's lovely answer to show that the only $n \ge 3$ for which there is a regular $n$-gon with a rational point on each side are $n=3,4,8$ (for which simple constructions have already been demonstrated here).

Observing that one can obtain a regular $n$-gon by extending $n$ sides of a regular $nm$-gon, it suffices to prove impossibility in the case in which $n$ is $16,6,9$, or an odd prime $p > 3$. Fix such an $n$ and suppose for sake of contradiction such an $n$-gon exists. Let $q_i$ be a rational point on the $i$th side (oriented counterclockwise), $p$ the center of the polygon, and $T$ rotation by $-\frac{2 \pi}{n}$. It will also be convenient to define $c(k) = \cos\left(\frac{2\pi k}{n}\right)$ and $s(k) = \sin\left(\frac{2\pi k}{n}\right)$.

Now exactly as in Uranix's answer the points $$ T^i(q_i-p) $$ lie in a common line. Thus for each $k$, the sum $$ a_k = \sum_{i=1}^n (T^i(q_i-p)) \vee (T^{i+k}(q_{i+k}-p)) $$ is equal to $0$ (where indices are taken modulo $n$). Now using $x \vee T^ky = c(k) x \vee y - s(k) x \cdot y$ we have \begin{align*} a_k &= \sum_{i=1}^n (q_i-p) \vee (T^{k}(q_{i+k}-p))\\ &= \sum_{i=1}^n c(k) (q_{i}-p) \vee (q_{i+k}-p) - s(k) (q_{i}-p) \cdot (q_{i+k}-p)\\ &= c(k)\sum_{i=1}^n q_{i} \vee q_{i+k} - s(k) \sum_{i=1}^n q_{i}\cdot q_{i+k}-s(k)\sum_{i=1}^n 2 p\cdot (q_i - p ).\\ \end{align*} We wish to be rid of those terms depending on $p$. Denote $V_k = \sum_{i=1}^n q_{i} \vee q_{i+k}$ and $D_k = \sum_{i=1}^n q_{i} \cdot q_{i+k}$, and observe that \begin{align*} 0 &= 2s(m)a_k - 2s(k)a_m \\&= 2s(m)c(k)V_k - 2s(k)c(m)V_m-2s(m)s(k)(D_m-D_k) \\&= s(m+k)(V_k-V_m)+s(m-k)(V_k+V_m)-(c(m-k)-c(m+k))(D_m-D_k) \end{align*} This is finally our necessary condition for solutions to exist. Most of the argument works with $m=2,k=1$ where we have \begin{equation} \label{eqn:main} s(3)(V_1-V_2)+s(1)(V_1+V_2)-(c(1)-c(3))(D_2-D_1)\tag{$*$} \end{equation} Note the crucial facts that $D_k,V_k$ are rational and that $V_1$ and $V_2$ are the volumes of the polytopes $q_1q_2q_3...$ and $q_1q_3q_5...$ and thus positive when $n > 4$.

We now derive a contradiction for every listed $n$.

Firstly if $n$ is an odd prime and greater than $3$, we have that $\{s(1),s(3),c(1),c(3)\}$ are $\mathbb Q$-linearly independent since $\{e^{\pm\frac{2\pi i}{n}},e^{\pm 3\frac{2\pi i}{n}}\}$ are $\mathbb Q(i)$-linearly independent. Thus for \eqref{eqn:main} to hold we require $$(V_1-V_2)=(V_1+V_2)=(D_2-D_1)=0$$ so in particular $V_1=V_2=0$, contradicting positivity of the $V_k$.

If $n=6$, then $s(3)$ vanishes and \eqref{eqn:main} becomes $$ 0=\frac{\sqrt{3}}{2}(V_1+V_2)-\frac{3}{2}(D_2-D_1), $$ so again $V_1+V_2=0$, contradicting the positivity of the $V_k$.

If $n=9$ we use the alternate condition $$s(5)(V_2-V_3)+s(1)(V_2+V_3)-(c(1)-c(5))(D_3-D_2)$$ for which the same argument as in the odd prime case applies to show $s(1), s(5), c(1)-c(5)$ are linearly independent, giving $(V_2-V_3)=(V_2+V_3)=(D_3-D_2)=0$ and thus $V_2=0$, again a contradiction.

Finally if $n=16$, then $c(1)$ and $c(3)$ are $\mathbb Q$-linearly independent, but $s(1) = c(3)$ and $s(3)=c(1)$, so \eqref{eqn:main} becomes $$ V_1-V_2 = D_2-D_1 = -V_1-V_2 $$ and thus $2V_1=0$ in this case, contradicting the positivity of the $V_k$.

This answer of course relies heavily on the stipulation that the rational points lie on each side. If this is removed there are solutions for at least the pentagon (as shown by Uranix) and the hexagon as well. I am curious if any other $n$ admit such solutions.