International Mathematical Olympiad

Proof I We first prove the following lemma.
Lemma Consider a triangle $PQR$ with $\angle QPR \ge 60^\circ$. Let $L$ be the midpoint of $QR$. Then $PL \le \frac{\sqrt{3}}{2} QR$. The equality holds if and only if the triangle $PQR$ is equilateral.
Proof of the lemma
Let $S$ be the point such that the triangle $QRS$ is equilateral, where the points $P$ and $S$ lie in the same half-plane bounded by the line $QR$.
Then the point $P$ lies inside the circumcircle (including the circumference of the circle) of the triangle $QRS$, which lies inside the circle with center $L$ and radius $\frac{\sqrt{3}}{2} QR$.
This completes the proof of the lemma.
The main diagonals of a convex hexagon form a triangle though the


Thus we may choose two of these three diagonals that form an angle greater than or equal to $60^\circ$. Without loss of generality, we may assume that the diagonals $AD$ and $BE$ of the given hexagon $ABCDEF$ satisfy $\angle APB \ge 60^\circ$, where $P$ is the intersection of these two diagonals. Then, using the lemma, we obtain
$$MN = \frac{\sqrt{3}}{2}(AB + DE) \ge PM + PN \ge MN,$$
where $M$ and $N$ are the midpoints of $AB$ and $DE$ respectively. Thus it follows from the lemma that the triangles $ABP$ and $DEP$ are equilateral.
Therefore the diagonal $CF$ forms an angle greater than or equal to $60^\circ$ with one of the diagonals $AD$ and $BE$. Without loss of generality, we may assume that $\angle AQF \ge 60^\circ$, where $Q$ is the intersection of $AD$ and $CF$. Arguing in the same way as above, we infer that the triangles $AQF$ and $CQD$ are equilateral. This implies that $\angle BRC = 60^\circ$, where $R$ is the intersection of $BE$ and $CF$. Using the same argument as above for the third time, we obtain that the triangles $BCR$ and $EFR$ are equilateral.
This completes the proof.

Proof II Let $ABCDEF$ be the given hexagon and let $\vec{a} = \vec{AB}$, $\vec{b} = \vec{BC}$, ..., $\vec{f} = \vec{FA}$.

Let $M$ and $N$ be the midpoints of the sides $AB$ and $DE$ respectively. We have
$$\vec{MN} = \frac{1}{2}\vec{a} + \vec{b} + \vec{c} + \frac{1}{2}\vec{d} \text{ and } \vec{MN} = -\frac{1}{2}\vec{a} - \vec{f} - \vec{e} - \frac{1}{2}\vec{d}.$$
Thus we obtain
$$\vec{MN} = \frac{1}{2}(\vec{b} + \vec{c} - \vec{e} - \vec{f}). \qquad \textcircled{1}$$
From the given property, we have
$$\vec{MN} = \frac{\sqrt{3}}{2}(|\mathbf{a}| + |\mathbf{d}|) \ge \frac{\sqrt{3}}{2}|\mathbf{a} - \mathbf{d}|. \qquad \textcircled{2}$$
Set $\mathbf{x} = \mathbf{a} - \mathbf{d}$, $\mathbf{y} = \mathbf{c} - \mathbf{f}$, $\mathbf{z} = \mathbf{e} - \mathbf{b}$. From ① and ②, we obtain
$$|\mathbf{y} - \mathbf{z}| \ge \sqrt{3} |\mathbf{x}|. \qquad \textcircled{3}$$
Similarly, we see that
$$|\mathbf{z} - \mathbf{x}| \ge \sqrt{3} |\mathbf{y}|, \qquad \textcircled{4}$$
$$|\mathbf{x} - \mathbf{y}| \ge \sqrt{3} |\mathbf{z}|. \qquad \textcircled{5}$$
Note that
$$\textcircled{3} \Leftrightarrow |\mathbf{y}|^2 - 2\mathbf{y} \cdot \mathbf{z} + |\mathbf{z}|^2 \ge 3|\mathbf{x}|^2,$$
$$\textcircled{4} \Leftrightarrow |\mathbf{z}|^2 - 2\mathbf{z} \cdot \mathbf{x} + |\mathbf{x}|^2 \ge 3|\mathbf{y}|^2,$$
$$\textcircled{5} \Leftrightarrow |\mathbf{x}|^2 - 2\mathbf{x} \cdot \mathbf{y} + |\mathbf{y}|^2 \ge 3|\mathbf{z}|^2.$$
By adding up the last three inequalities, we obtain
$$- |\mathbf{x}|^2 - |\mathbf{y}|^2 - |\mathbf{z}|^2 - 2\mathbf{y} \cdot \mathbf{z} - 2\mathbf{z} \cdot \mathbf{x} - 2\mathbf{x} \cdot \mathbf{y} \ge 0,$$
or $-|\mathbf{x} + \mathbf{y} + \mathbf{z}|^2 \ge 0$. Thus $\mathbf{x} + \mathbf{y} + \mathbf{z} = 0$ and the equality holds in every inequality above. Hence we conclude that
$$\mathbf{x} + \mathbf{y} + \mathbf{z} = 0,$$
$$|\mathbf{y} - \mathbf{z}| = \sqrt{3} |\mathbf{x}|, \mathbf{a} \parallel \mathbf{d} \parallel \mathbf{x},$$
$$|\mathbf{z} - \mathbf{x}| = \sqrt{3} |\mathbf{y}|, \mathbf{c} \parallel \mathbf{f} \parallel \mathbf{y},$$
$$|\mathbf{x} - \mathbf{y}| = \sqrt{3} |\mathbf{z}|, \mathbf{e} \parallel \mathbf{b} \parallel \mathbf{z}.$$
Suppose that $PQR$ is the triangle such that $\vec{PQ} = \mathbf{x}$, $\vec{QR} = \mathbf{y}$, $\vec{RP} = \mathbf{z}$. We may assume $\angle QPR \ge 60^\circ$, without loss of generality. Let $L$ be the midpoint of $QR$. Then $PL = \frac{1}{2}|\mathbf{z} - \mathbf{x}| = \frac{\sqrt{3}}{2}|\mathbf{y}| = \frac{\sqrt{3}}{2} QR$. It follows from the lemma in Proof I that the triangle $PQR$ is equilateral. Thus we have $\angle ABC = \angle BCD = \dots = \angle FAB = 120^\circ$.