Taiwan IMO Selection Camp, 2022

All odd $n \ge 3$.

First we show that for every even $n \ge 4$ there exists an $n$-polygon violating the required statement. Consider a regular $k$-gon $A_0A_1...A_{k-1}$ with side length 1. Let $B_1, B_2, ..., B_{n/2-1}$ be the points symmetric to $A_1, A_2, ..., A_{n/2-1}$ with respect to the line $A_0A_{n/2}$. Then $\mathcal{P} = A_0A_1A_2...A_{n/2-1}A_{n/2}B_{n/2-1}B_{n/2-2}...B_2B_1$ is a convex $n$-gon whose sides all have length 1. If $k$ is big enough, $\mathcal{P}$ is contained in a strip of width $1/2$, which clearly does not contain any equilateral triangle of side length 1.

Assume now that $n = 2k + 1$. As the case $k = 1$ is trivially true, we assume $k \ge 2$ henceforth. Consider a convex $(2k+1)$-gon $\mathcal{P}$ whose sides all have length 1. Let $d$ be its longest diagonal. The endpoints of $d$ split the perimeter of $\mathcal{P}$ into two polylines, one of which has length at least $k+1$. Hence we can label the vertices of $\mathcal{P}$ so that $\mathcal{P} = A_0A_1...A_{2k}$ and $d = A_0A_\ell$ with $\ell \ge k+1$. We will show that, in fact, the polygon $A_0A_1...A_\ell$ contains an equilateral triangle of side length 1.

Suppose that $\angle A_\ell A_0 A_1 \ge 60^\circ$. Since $d$ is the longest diagonal, we have $A_1 A_\ell \le A_0 A_\ell$, so $\angle A_0 A_1 A_\ell \ge \angle A_\ell A_0 A_1 \ge 60^\circ$. It follows that there exists a point $X$ inside the triangle $A_0A_1A_\ell$ such that the triangle $A_0A_1X$ is equilateral, and this triangle is contained in $\mathcal{P}$. Similar arguments apply if $\angle A_{\ell-1} A_\ell A_0 \ge 60^\circ$.

From now on, assume $\angle A_\ell A_0 A_1 < 60^\circ$ and $\angle A_{\ell-1} A_\ell A_0 < 60^\circ$.

Consider an isocles trapezoid $A_0YZA_\ell$ such that $A_0A_\ell \parallel YZ$, $A_0Y = ZA_\ell = 1$, and $\angle A_\ell A_0Y = \angle ZA_\ell A_0 = 60^\circ$. Suppose that $A_0A_1 \dots A_\ell$ is contained in $A_0YZA_\ell$. Note that the perimeter of $A_0A_1 \dots A_\ell$ equals $\ell + A_0A_\ell$ and the perimeter of $A_0YZA_\ell$ equals $2A_0A_\ell + 1$.

Recall a well-known fact stating that if a convex polygon $\mathcal{P}_1$ is contained in a convex polygon $\mathcal{P}_2$, then the perimeter of $\mathcal{P}_1$ is at most the perimeter of $\mathcal{P}_2$. Hence we obtain
$$\ell + A_0 A_\ell \le 2A_0 A_\ell + 1, \quad \text{ i.e. } \ell - 1 \le A_0 A_\ell.$$
On the other hand, the triangle inequality yields
$$A_0 A_\ell < A_\ell A_{\ell+1} + A_{\ell+1} A_{\ell+2} + \dots + A_{2k} A_0 = 2k + 1 - \ell \le \ell - 1.$$
which gives a contradiction.

Therefore, there exists a vertex $A_m$ of $A_0A_1 \dots A_\ell$ which lies outside $A_0YZA_\ell$. Since
$$\angle A_\ell A_0 A_1 < 60^\circ = \angle A_\ell A_0 Y \quad \text{and} \quad \angle A_{\ell-1} A_\ell A_0 < 60^\circ = \angle Z A_\ell A_0, \qquad (1)$$
the distance between $A_m$ and $A_0A_\ell$ is at least $\sqrt{3}/2$.

Let $P$ be the projection of $A_m$ to $A_0A_\ell$. Then $PA_m \ge \sqrt{3}/2$, and by (1) we have $A_0P > 1/2$ and $PA_\ell > 1/2$. Choose points $Q \in A_0P$, $R \in PA_\ell$, and $S \in PA_m$ such that $PQ = PR = 1/2$ and $PS = \sqrt{3}/2$. Then $QRS$ is an equilateral triangle of side length 1 contained in $A_0A_1 \dots A_\ell$.