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 :wow: