Solution 1. (Galois) The conclusion follows in two steps: fix $n \ge 3$ an integer. Claim: $n$ is frameable $\implies \cos \frac{2\pi}{n} \in \mathbb{Q}$. Proof. Suppose the $n$-gon is $A_0A_1A_2\dots A_{n-1}$ with side length $a>0$ and we have a family of parallel lines with distance between consecutive lines as $1$. Let $\ell$ be the line which passes through $A_0$ and $j, k, m$ be the orthogonal distances from $A_1, A_2, A_{n-1}$ to $\ell$. Let $\alpha=\measuredangle (\ell, A_0A_1)$ and $t=a^2$. Observe that $$\cos \left(\left(\frac{2\pi}{n}+\alpha\right)-\alpha\right)=\cos \left(\frac{2\pi}{n}+\alpha\right)\cos \alpha-\sin \left(\frac{2\pi}{n}+\alpha\right)\sin \alpha=\frac{\sqrt{(t-k)(t-j)}}{t}-\frac{jk}{t}$$which simplifies to $t\cdot \left(\cos^2 \frac{2\pi}{n}-1\right)=-\left(j^2+k^2-2jk\cos \frac{2\pi}{n}\right)$. Similarly, working with $\measuredangle (A_0A_1, A_0A_{n-1})$ one gets $$t\cdot \left(\cos^2 \frac{2\pi}{n}-1\right)=-\left(m^2+j^2+2mj\cos \frac{2\pi}{n}\right).$$Equating, we get the desired result. $\blacksquare$ Claim: $\cos \frac{2\pi}{n} \in \mathbb{Q} \implies n \in \{3, 4, 6\}$. Proof. Let $z=e^{\frac{2\pi i}{n}}$ and $x=\cos \frac{2\pi}{n}$. Note that $2x=z+z^{-1}$ so $[\mathbb{Q}(z):\mathbb{Q}(x)]=2$ and $[\mathbb{Q}(x):\mathbb{Q}]=1$ since $x \in \mathbb{Q}$. We know from algebraic NT that the minimal polynomial of $z$ is the $n$th cyclotomic polynomial, with degree $\phi(n)$. So $$\phi(n)=[\mathbb{Q}(z):\mathbb{Q}]=[\mathbb{Q}(z):\mathbb{Q}(x)] \cdot [\mathbb{Q}(x):\mathbb{Q}]=2$$forces $\phi(n)=2$ so $n \in \{3, 4, 6\}$. $\blacksquare$ Constructions for these are easy to do. $\blacksquare$ Sketch of Solution 2. (Ingenious descent, official solution) Again, constructions are easy to do so we won't bother. We prove that $n \geqslant 7$ is not frameable. Suppose the $n$-gon is $A_1A_2\dots A_n$ with side length $a>0$. Pick a point $O$ away from these, on some line. Construct points $B_1, \dots, B_n$ such that $OA_iA_{i+1}B_i$ is a parallelogram, for all $i$, indices modulo $n$. Note that $B_1B_2\dots B_n$ is a regular $n$-gon, with vertices lying on these lines. Note that $\frac{B_1B_2}{A_1A_2}=\frac{B_1B_2}{OB_2}=\left(2\sin \frac{\pi}{n}\right)=k<1$ as $n>6$. Iterate this procedure $m$ times to get a regular $n$-gon with side length $ak^m$ whose vertices lie on these lines. Since the distance between any two lines exceeds some fixed $\varepsilon>0$, by choosing $m$ sufficiently large, we get $ak^m<\varepsilon$, a contradiction! For $n=5$, reflect $B_i$ in $O$ to get $B_i'$. This lets us construct a frameable decagon; contradiction! $\blacksquare$ Remark. Easily my favourite problem on the test During test-solve, both me and NVT found the Galois solution, but missed the official one; we feel it is really tricky to come up with. Hope kids liked it as much as I did

Is it possible to get a bijection between frame-ability and tessellate-ability?

Hope I'm not missing something. This can also be done by Niven's theorem, which makes the problem both easy and 'elementary'. Constructions for $n = 3, 4, 6$ are easy to obtain so we'll omit them. For $n \ge 5$, label the vertices $1, ..., n$ modulo $n$. Then note that chords $V_2V_n$ and $V_3V_{n-1}$ are parallel and distinct. Their projections on the axis perpendicular to the drawn axes are commensurable (i.e. rational multiples of each other). So their lengths are themselves commensurable. This means $sin(2pi/n)$ and $sin(4pi/n)$ are commensurable. Then $cos(2pi/n)$ is rational. Then Niven's theorem forces $n = 6$, and we're done.

Man I thought that descent could work, just like in the proof to show that the only lattice $n$ gon is a square. But couldn't work out the details in time

sa2001 wrote: Hope I'm not missing something. This can also be done by Niven's theorem, which makes the problem both easy and 'elementary'. Constructions for $n = 3, 4, 6$ are easy to obtain so we'll omit them. For $n \ge 5$, label the vertices $1, ..., n$ modulo $n$. Then note that chords $V_2V_n$ and $V_3V_{n-1}$ are parallel and distinct. Their projections on the axis perpendicular to the drawn axes are commensurable (i.e. rational multiples of each other). So their lengths are themselves commensurable. This means $sin(2pi/n)$ and $sin(4pi/n)$ are commensurable. Then $cos(2pi/n)$ is rational. Then Niven's theorem forces $n = 6$, and we're done. I was thinking along those lines in the contest, but I didn't know Niven's theorem so thought it would useless to try to prove when $\sin$ would be rational (also I was caught off guard by the fact that we take $\sin 37^\circ$ to be $\frac{3}{5}$ (that's only an approximation))

anantmudgal09 wrote: Solution 1. (Galois) The conclusion follows in two steps: fix $n \ge 3$ an integer. Claim: $n$ is frameable $\implies \cos \frac{2\pi}{n} \in \mathbb{Q}$. Proof. Suppose the $n$-gon is $A_0A_1A_2\dots A_{n-1}$ with side length $a>0$ and we have a family of parallel lines with distance between consecutive lines as $1$. Let $\ell$ be the line which passes through $A_0$ and $j, k, m$ be the orthogonal distances from $A_1, A_2, A_{n-1}$ to $\ell$. Let $\alpha=\measuredangle (\ell, A_0A_1)$ and $t=a^2$. Observe that $$\cos \left(\left(\frac{2\pi}{n}+\alpha\right)-\alpha\right)=\cos \left(\cos \frac{2\pi}{n}+\alpha\right)\cos \alpha-\sin \left(\cos \frac{2\pi}{n}+\alpha\right)\sin \alpha=\frac{\sqrt{(t-k)(t-j)}}{t}-\frac{jk}{t}$$which simplifies to $t\cdot \left(\cos^2 \frac{2\pi}{n}-1\right)=-\left(j^2+k^2-2jk\cos \frac{2\pi}{n}\right)$. Similarly, working with $\measuredangle (A_0A_1, A_0A_{n-1})$ one gets $$t\cdot \left(\cos^2 \frac{2\pi}{n}-1\right)=-\left(m^2+j^2+2mj\cos \frac{2\pi}{n}\right).$$Equating, we get the desired result. $\blacksquare$ Claim: $\cos \frac{2\pi}{n} \in \mathbb{Q} \implies n \in \{3, 4, 6\}$. Proof. Let $z=e^{\frac{2\pi i}{n}}$ and $x=\cos \frac{2\pi}{n}$. Note that $2x=z+z^{-1}$ so $[\mathbb{Q}(z):\mathbb{Q}(x)]=2$ and $[\mathbb{Q}(x):\mathbb{Q}]=1$ since $x \in \mathbb{Q}$. We know from algebraic NT that the minimal polynomial of $z$ is the $n$th cyclotomic polynomial, with degree $\phi(n)$. So $$\phi(n)=[\mathbb{Q}(z):\mathbb{Q}]=[\mathbb{Q}(z):\mathbb{Q}(x)] \cdot [\mathbb{Q}(x):\mathbb{Q}]=2$$forces $\phi(n)=2$ so $n \in \{3, 4, 6\}$. $\blacksquare$ Constructions for these are easy to do. $\blacksquare$ Sketch of Solution 2. (Ingenius descent, official solution) Again, constructions are easy to do so we won't bother. We prove that $n \geqslant 7$ is not frameable. Suppose the $n$-gon is $A_1A_2\dots A_n$ with side length $a>0$. Pick a point $O$ away from these, on some line. Construct points $B_1, \dots, B_n$ such that $OA_iA_{i+1}B_i$ is a parallelogram, for all $i$, indices modulo $n$. Note that $B_1B_2\dots B_n$ is a regular $n$-gon, with vertices lying on these lines. Note that $\frac{B_1B_2}{A_1A_2}=\frac{B_1B_2}{OB_2}=\left(2\sin \frac{\pi}{n}\right)=k<1$ as $n>6$. Iterate this procedure $m$ times to get a regular $n$-gon with side length $ak^m$ whose vertices lie on these lines. Since the distance between any two lines exceeds some fixed $\varepsilon>0$, by choosing $m$ sufficiently large, we get $ak^m<\varepsilon$, a contradiction! For $n=5$, reflect $B_i$ in $O$ to get $B_i'$. This lets us construct a frameable decagon; contradiction! $\blacksquare$ Remark. Easily my favourite problem on the test During test-solve, both me and NVT found the Galois solution, but missed the official one; we feel it is really tricky to come up with. Hope kids liked it as much as I did I did by official method

ubermensch wrote: Is it possible to get a bijection between frame-ability and tessellate-ability? I thought of that, since I recently read about the crystallographic restriction theorem (basically what you're suggesting) I proved that the pencil of parallel lines must be one of the medians of the triangles whose vertices are part of the polygon. In the contest I could only prove till this but now with some heavy trig bash I think I finished it.

Hexagrammum16 wrote: ubermensch wrote: Is it possible to get a bijection between frame-ability and tessellate-ability? I thought of that, since I recently read about the crystallographic restriction theorem (basically what you're suggesting) I proved that the pencil of parallel lines must be one of the medians of the triangles whose vertices are part of the polygon. In the contest I could only prove till this but now with some heavy trig bash I think I finished it. Huh... I remember getting a very similar statement, but I couldn't finish it during the exam either... it just reminded me of tesselatablity because of the similar config and of course the values of $n$ which work (I tried to prove that given frame-ability and translating it repeatedly from original polygon along the direction of the perpendicular bisector of each side such that set of parallel lines is the same must imply a way to get a tessellation, but my argument was vague at best)- perhaps you could post your completed solution?

I solved only part (a) I hope the marks distribution is generous enough to give atleast 3 marks for part (a) For n=3 Consider three consecutive lines. Rotate the one in the one by $30^{\circ}$ clockwise and counterclockwise. Now, we construct the triangle in the obvious way by intersecting the lines. Assume WLOG that the lines are $x=0,1,-1,2,-2\dots$ in the coordinate plane. Then, the points $(2,1), (-1,2), (-2,-1), (1,-2)$ form the vertices of a square with no two vertices on the same line. For $n=6$, I used the complex plane, and again, WLOG, let the lines be $Re(z)=0,1,-1,2,-1\dots$. Counter clockwise rotation by $60^{\circ}$ about the origin corresponds to a multiplication by $e=\frac{1}{2}+i\frac{\sqrt{3}}{2}$. We choose the points $a=4+\sqrt{3}$, $b=ae$, $c=ae^2$. Then, $a,b,c,-a,-b,-c$ form the vertices of a regular hexagon with vertices on different lines. Hence, we are done.

Just for fun, here's a cool way to construct for $n=4.$ Let the square be $ABCD.$ Consider an arbitrary line $\ell_a$ through $A.$ Draw $\ell_b,\ell_c,\ell_d$ through $B,C,D$ such that they are all parallel to $\ell_a.$ Let $x$ be the distance between $\ell_a,\ell_b,$ which by symmetry is the same as the distance between $\ell_d,\ell_c.$ Let $y$ be the distance between $\ell_b,\ell_d.$ Notice that when $\ell_a \equiv AB, x=0$ while $y>0.$ Further when $\ell_b \equiv BC, y=0$ while $x>0.$ By continuity there would exist a position of $\ell_a$ when $x=y.$ This is the desired configuration.

Easily my favourite problem in INMO 2020. Objectively, (b) is the hardest problem in the paper. Let $n$ be a framable number, and let $\mathcal{P}$ be an $n$-gon framed in the lines given. Set up a complex coordinate system with the lines being parallel to the $y$-axis and adjacent lines being one unit apart. Suppose $0$ is the centre of $\mathcal{P}$ (we can do this because the origin will also have its real part rational, and then we can scale and translate), and let $z = a+ib$ be one of its vertices. Then, all its vertices are $z,z\omega, z\omega^2, \cdots, z\omega^{n-1}$, where $\omega = e^{\frac{2i\pi}{n}}$. We want the real part of $z\omega^k$ to be an integer for all $0 \le k \le n-1$. This implies $$a \cos \frac{2k\pi}{n} - b \sin \frac{2k\pi}{n} \in \mathbb{Z}$$Put $k=0$ to get $a \in \mathbb{Z}$. Put $k=1$ and $k=n-1$, and add to get $$a \cos \frac{2\pi}{n}, b \sin \frac{2\pi}{n} \in \mathbb{Q}$$If $a \neq 0$, then we get $\cos \frac{2\pi}{n} \in \mathbb{Q}$. Otherwise $a=0$, so $b \neq 0$ and $b \sin \frac{2i\pi}{n} \in \mathbb{Q}$. Put $k=1,2$ and divide (this is allowed because $\sin \frac{2\pi}{n} \neq 0$) to get $\cos \frac{2\pi}{n} \in \mathbb{Q}$ anyway. Thus, if $n$ is framable, then $\cos \frac{2\pi}{n} \in \mathbb{Q}$. Let us now solve the parts given. (a) This is easy. (b) and (c) Suppose $n \ge 3$ and $\cos \frac{2\pi}{n}$ is rational. Let $2 \cos \frac{2\pi}{n} = \frac{p}{q}$ for coprime integers $p,q$, with $q > 0$. Then, $\alpha = \cos \frac{2\pi}{n} + i \sin \frac{2\pi}{n}$ has minimal polynomial $qx^2-px+q \in \mathbb{Z}[x]$. Also, $\alpha$ is a root of $x^{n-1}+x^{n-2} + \cdots + 1 \in \mathbb{Z}[x]$ too. This means $qx^2-px+q$ is a factor of $x^{n-1}+x^{n-2} + \cdots + 1$ in $\mathbb{Z}[x]$. This means $q= 1$, so $2 \cos \frac{2\pi}{n}$ is an integer. This forces $\cos \frac{2\pi}{n} = \pm 1, \frac{\pm 1}{2}, 0$, which implies $n=3,4,6$. Of course, (c) can be solved separately too; $\cos \frac{2\pi}{5} = \frac{\sqrt{5}-1}{4}$ is irrational.

Can anyone please send a diagram for n=6? Thanks @below

Hexagrammum16 wrote: Can anyone please send a diagram for n=6? I'll tell you complex coordinates of the $6$ points. $\sqrt{3}-5i, -\sqrt{3}+5i, 3\sqrt{3}-i, -3\sqrt{3}+i, \sqrt{3}+4i, -\sqrt{3}-4i$. Here, my lines are parallel to the $x$-axis.

Can we directly mention Niven's Theorem in INMO?

Here's the proof for Niven's theorem I'd read when I first found out about the theorem: First prove by induction that $2cos(nx)$ is an $n$-degree monic integer polynomial in $2cos(x)$ (use $cos((n+2)x) + cos(nx) = 2cos((n+1)x)cosx$). Now, if $q$ is rational, then $2cos(nq*pi)$ is $0$ for some natural number $n$. Thus, $2cos(q*pi)$ is a root of a monic integer polynomial. By rational root theorem, $2cos(q*pi)$ must be an integer if it is rational, which gives us the desired values $cos(q*pi) = -1, -1/2, 0, 1/2, 1$. Fun fact: We can use the identity $cos(2x) = 2cos^2(x) - 1$ to determine all rational $q$ such that $cos^2(q*pi)$ is rational.

Coordinate Solution- Let the polygon be $A_1A_2,\cdots,A_n$ . Set up a coordinate system with $A_1\leftrightarrow(0,0)$ and $A_2\leftrightarrow(1,0)$. We get that $A_3\leftrightarrow(1+$cos$\frac{2\pi}n, $sin$\frac{2\pi}n)$ and $A_4\leftrightarrow(2$cos$^2\frac{2\pi}n+$cos$\frac{2\pi}n, $sin$\frac{2\pi}n(1+2$cos$\frac{2\pi}n)) $ Now let the $n$ be frameable. Let the slope of the parallel lines be $m$. We note that all the $x-intecepts$ of these lines must be rational . Hence for $A_3$ we get $1+$cos$\frac{2\pi}n + m $sin$\frac{2\pi}n = r$ And for $A_4$ we get $2$cos$^2\frac{2\pi}n+ $cos$\frac{2\pi}n+m$sin$\frac{2\pi}n(1+2$cos$\frac{2\pi}n) =s $ Where $r,s\in\mathbb{Q}$ Eliminating $m$sin$\frac{2\pi}n$ we get that- $(r-1-$cos$\frac{2\pi}n)(1+2$cos$\frac{2\pi}n)=s-2$cos$^2\frac{2\pi}n- $cos$\frac{2\pi}n$ From this we have $2(r-1)$cos$\frac{2\pi}n=s-r+1$ But as each point lies on only one line the $x-intercept$ of $A_3$ cannot be $1$ (as $n\geq 3$) and hence $r\neq 1$ Hence we get that cos$\frac{2\pi}n$ is rational. After this the proof is same as in #2 solution 1

Answer $n\in\{3,4,6\} $ Solution At first, it can be noticed that $3$, $4$ and $6$ are frameable. As we can put the vertices of a triangle, a square and a regular hexagon into the given lines. $n=3$ is frameable $n=4$ is frameable. Let $E $ be midpoint of $A_1(C), A_2(D)$ $n=6$ is frameable We need to prove that other numbers are not frameable. Consider $A_1A_2\cdots A_n $ $(n\ge 5) $ is a regular polygon with $n$ sides. Let $\ell_i (i=1,2,\cdots,n)$ be the drawn line passes through $A_i $ and $d_{(i,j)}(1\le i <j\le n)$ be the number of drawn lines between $A_i $ and $A_j $. Note, $\frac {d (1,2)+1}{d (3,n)+1}=\frac {A_1A_2}{A_3A_n}\implies\frac {A_1A_2}{A_3A_n}\in\mathbb Q$ Let $H_1$, $H_2$ be the points in $\overline {A_1A_2} $ and $\overline {A_nA_3} $ respectively, such that $A_1H_1\perp A_nA_3$ and $A_2H_2\perp A_nA_3$. $\frac {A_nA_3}{A_1A_2}=1+2\cdot\frac {A_nH_1}{A_nH_1}=1+2\sin\measuredangle A_nA_1H_1$ $\frac {A_nA_3}{A_1A_2}\in\mathbb Q\implies \sin\measuredangle A_nA_1H_1$ must be rational. Now, $\angle A_nA_1H_1=\frac{\pi}{2}-\angle A_1A_nA_3=\frac {\pi}{2}-\frac {2}{n}\cdot \pi=\pi\cdot\frac {n-4}{2n} $ It is easy to see that $0 <\sin\left (\pi\cdot\frac {n-4}{2n}\right)<1(n\ge 5)$ Applying$\text { Niven's theorem}\implies\sin(\pi)\frac {n-4}{2n}=\tfrac 12\implies\frac {n-4}{2n}=\tfrac 16\implies n=6$ If $n\ge 5$, and $n\neq 6$, then $n $ is not frameable. So, $3,4,6$ are all frameable.

Making things easier How about I consider two lines.. A straight perpendicular line which is equal to a units and other slanting lines also equal to k units In other parallels I mean By this 3 is framable by taking two slant lines and distance between them on the parallel line as k too Same for 4 as a square of side n.a can be formed where n is a natural number For 6- 4 slant lines are used which act as transversal and distance between the vertices of parallel line be k Greater than 6 not possible as those figures will have both slant and perpendiculars which are not equal in length as we know perpendicular is shortest line between two parallel lines For 5 I don't know how to explain tbh

At first of all suppose $\Omega$ be the plane in complex plane consists all equidistant parallel lines . Suppose the parallel lines are parallel to the real axis. This is because we always get a transformation $T(ax+by-c)\to e +si$ where $s$ is a real constant . also suppose two consecutive lines are one unit apart to each other . Suppose $P_0,\cdots ,P_{n-1}$ are n ($n>2$)points of n sided regular polygon . Suppose $P_i =z .\zeta ^i$ for all $i \in \{0,1,2,\cdots ,n -1\}$. Where $z$ is a complex number with $z=x+yi$ Also denote the parallel lines as $l_i$ iff $P_i \in l_i$ . So we can see that $l_k \equiv \ Im ({P_k})=y \cos \frac{2\pi k}{n} +x \sin \frac{2\pi k}{n}$ . So, $l_0 \equiv y \cdots (1) $ . $l_1 \equiv y \cos \frac{2\pi }{n} +x \sin \frac{2\pi }{n} \cdots (2)$ $l_{n-1} \equiv y \cos \frac{2\pi }{n} -x \sin \frac{2\pi }{n} \cdots (3)$. From $(2),(3)$ get , $x\sin \frac{2\pi}{n} ,y \cos \frac{2\pi}{n} $ both are rational number . So, from $(2),(1)$ get $y$ is also a rational number . $\implies \cos \frac{2\pi}{n} $ is rational. Part (a) We get $\cos \frac{2\pi }{3} =\frac{-1}{2} \in \mathbb{Q}$ and similarly we get $ \cos \frac{2\pi}{n} $ is rational for $n=4,6$ . So $n=3,4,6$ are Frameable . Part (c) We get $\cos \frac{2\pi}{5} = \frac{\sqrt{5}-1}{2} \in (\mathbb{R}- \mathbb{Q})$. So ,$n=5$ is not Frameable . Part (b). Claim: we always get a polynomial $P_n$ ($n\ge 1$)with integer cofficient such that $P_n(2\cos t) = 2 \cos nt$. In general $\deg (P_n)=n$ and its a Monic polynomial. Proof. It can be done by simple induction . We can get some examples for $n=1,2$. $P_1(x)=x $ . $P_2(x)=x^2-2 $ and $P_2(2 \cos t) =2 \cos 2t$. Note that $2\cos (kx) (2\cos x) - 2\cos ((k-1) x)$. $\implies \boxed{ P_{k+1} (x)= x P_k(x) -P_{k-1}(x)}$. Here $\deg (P_{k+1} ) =k+1$. So our claim is proved $\blacksquare$. Now $P_n(2\cos \frac{2\pi}{n})=2\cos (2\pi)=2$. If $2\cos \frac{2\pi} {n} =\frac{p}{q}$ is rational then $P_n-2$ has a rational root then , $ | p| \mid 2$ and $|q| \mid 1$ so, for $|\frac{p}{q}|$ we have only two chice that is $|\cos \frac{2\pi}{n}| =1,\frac{1}{2}$. So ,for $n=2,0,3,6$ the above is possible. So, for any $n\ge 7$ are not Frameable .

Orient the diagram so that all parallel lines are horizontal. Suppose a regular $n$-gon is framable, and without loss let it have side length $1$. Let $A$, $B$, $C$, and $D$ be consecutive vertices of the polygon (where possibly $A\equiv D$). Then $ABCD$ is an isosceles trapezoid and, in particular, $AD\parallel BC$. But since all the parallel lines are equidistant, the vertical distance between $A$ and $D$ is a rational multiple of the vertical distance between $B$ and $C$; it follows that the length $AD$ is a rational multiple of the length $BC$, i.e. $AD = \lambda\in\mathbb Q$. But now a bit of right-triangle trig on trapezoid $ABCD$ yields that $\cos(\tfrac{2\pi}n) = \tfrac{\lambda-1}2$, which is rational. Hence $n\in\{3,4,6\}$, and we proceed as in the other solutions.

djmathman wrote: Orient the diagram so that all parallel lines are horizontal. Suppose a regular $n$-gon is framable, and without loss let it have side length $1$. Let $A$, $B$, $C$, and $D$ be consecutive vertices of the polygon (where possibly $A\equiv D$). Then $ABCD$ is an isosceles trapezoid and, in particular, $AD\parallel BC$. But since all the parallel lines are equidistant, the vertical distance between $A$ and $D$ is a rational multiple of the vertical distance between $B$ and $C$; it follows that the length $AD$ is a rational multiple of the length $BC$, i.e. $AD = \lambda\in\mathbb Q$. But now a bit of right-triangle trig on trapezoid $ABCD$ yields that $\cos(\tfrac{2\pi}n) = \tfrac{\lambda-1}2$, which is rational. Hence $n\in\{3,4,6\}$, and we proceed as in the other solutions. Nice but i could not understand the last paragraph

what i did was that for Part A i did a periodic tiling for 3,4,6

Consider this plane to be a ruled sheet of paper. Assume that a regular $n$-gon has been placed on the paper. Let $O$ be the center of this polygon. Rotate the plane through an angle $360^{\circ} / n$ centered at the point $O$. Then each vertex of the $n$-gon lies at the intersection point of two grid lines: the old and the new (obtained after the rotation) parallel lines. Now, this is just embedding in the unit lattice, but this is possible only for $n=3,4$ or $6$.

The key is that the result is equivalent to $\cos \frac{2\pi}n$ being rational. To see this, consider any four consecutive vertices $A_1A_2A_3A_4$ and note that $A_1A_4=rA_2A_3$ for some rational $r$. By dropping the altitudes from $A_2, A_3$ this is equivalent to the conclusion. On the other hand, we must have $n \in \{3, 4, 6\}$ by Niven's theorem. It is easy to see that regular $n$-gons for these values of $n$ actually work.

Consider the problem in the complex plane. By arbitrarily rotating and scaling, we may assume that the lines are parallel to the real axis of the complex plane, and that the points are defined as \[ \text{cis } \theta, \text{cis} \left( \theta + \frac{2\pi}{n} \right), \dots, \text{cis} \left( \theta + (n-1) * \frac{2\pi}{n} \right) \]in particular, then we require \[ \frac{\sin \left( \theta + 2\cdot \frac{2\pi}{n} \right) - \sin \left( \theta - 2\cdot \frac{2\pi}{n} \right)}{\sin \left( \theta + \frac{2\pi}{n} \right) - \sin \left( \theta - \frac{2\pi}{n} \right)} \]to be rational; using sum to product, this is equivalent to \[ \frac{\cos \theta \sin\left( 2\cdot \frac{2\pi}{n} \right)}{\cos \theta \sin \left( \frac{2\pi}{n} \right)} = \frac{\sin \left( 2\cdot \frac{2\pi}{n} \right)}{\sin \left( \frac{2\pi}{n} \right)} \]but now, just use the fact $\sin \left( 2\cdot \frac{2\pi}{n} \right) = 2 \sin \left( \frac{2\pi}{n} \right) \cos \left( \frac{2\pi}{n} \right)$. Thus, we have \[ \cos \left( \frac{2\pi}{n} \right) \in \mathbb{Q} \]which, by Niven's theorem, implies that $n = 3, 4, 6$. Now the construction for $n = 3$ is just $1, w, w^2$ where $w^2 + w + 1 = 0$ with the lines parallel to the real axis through each point (they are obviously equally spaced apart). For $n = 4$, we consider $0, 2+i, -1+2i, 1+3i$ and the lines parallel to the real axis intersecting the imaginary axis at integer multiples of $i$. For $n = 6$, consider the angle $\theta < 60^\circ$ satisfying $2\sin \theta = \sin \left( 60^\circ - \theta \right)$; in particular, for this $\theta$, $\sin \theta = \frac{\sqrt{3}}{\sqrt{28}}$. Then consider the points \[ A_1 = 0, A_2 = \text{cis } \theta, A_3 = \text{cis } \theta + \text{cis } \left( 60 + \theta \right), \dots \]notice that by construction, the imaginary component of $A_2 - A_1$ is half that of $A_6 - A_1$. By symmetry, it is also the case that $A_5 - A_4$ has an imaginary component half of $A_6 - A_4$. I claim that the imaginary component of $A_2A_3$ is also a rational multiple of $A_1A_2$. Indeed, it is just \[ \sin \left( \theta + 60^\circ \right) = \sin \theta \cos 60^\circ + \cos \theta \sin 60^\circ = \frac{6\sqrt{3}}{2\sqrt{28}} \]so then just take the lines parallel to the real axis, intersecting the imaginary axis at multiples of $\frac{\sqrt{3}}{\sqrt{28}}$. $\blacksquare$

Here we go: By sufficient dilating and rotating, consider the vertices of the regular $n$-gon to be in unit circle, with the black lines parallel to the imaginary axis. Consider the points to be of the form: $\text{cis}(\theta + \frac{2 \pi k}{n})$. We must have: $\frac{\text{cis} (\theta + \tfrac{2\pi k}{n}) - \text{cis} (\theta - \tfrac{2 \pi k}{n})}{\text{cis} (\theta + \frac{2\pi}{n}) - \text{cis} (\theta - \frac{2 \pi}{n})} \in \mathbb Q.$ Expanding it and plugging $k=2$ gives us: $\cos(\tfrac{2 \pi}{n}) \in \mathbb Q$. Due to Niven's theorem, we must have $n=3, 4, 6$ and we are done.