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 $\vec{a}=\overrightarrow{AB}$, $\vec{b}=\overrightarrow{BC}$, ..., $\vec{f}=\overrightarrow{FA}$. Let $M$ and $N$ be the midpoints of the sides $AB$ and $DE$ respectively. We have $$\overrightarrow{MN}=\frac{1}{2}\vec{a}+\vec{b}+\vec{c}+\frac{1}{2}\vec{d}\text{ and }\overrightarrow{MN}=-\frac{1}{2}\vec{a}-\vec{f}-\vec{e}-\frac{1}{2}\vec{d}.$$ Thus we obtain $$\overrightarrow{MN}=\frac{1}{2}(\vec{b}+\vec{c}-\vec{e}-\vec{f}).\stackrel{◯}{1}$$ From the given property, we have $$\overrightarrow{MN}=\frac{\sqrt{3}}{2}(|\mathbf{a}|+|\mathbf{d}|)\ge \frac{\sqrt{3}}{2}|\mathbf{a}-\mathbf{d}|.\stackrel{◯}{2}$$ 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 $$|\mathbf{y}-\mathbf{z}|\ge \sqrt{3}|\mathbf{x}|.\stackrel{◯}{3}$$ Similarly, we see that $$|\mathbf{z}-\mathbf{x}|\ge \sqrt{3}|\mathbf{y}|,\stackrel{◯}{4}$$ $$|\mathbf{x}-\mathbf{y}|\ge \sqrt{3}|\mathbf{z}|.\stackrel{◯}{5}$$ Note that $$\stackrel{◯}{3}\iff |\mathbf{y}{|}^2-2\mathbf{y}\cdot \mathbf{z}+|\mathbf{z}{|}^2\ge 3|\mathbf{x}{|}^2,$$ $$\stackrel{◯}{4}\iff |\mathbf{z}{|}^2-2\mathbf{z}\cdot \mathbf{x}+|\mathbf{x}{|}^2\ge 3|\mathbf{y}{|}^2,$$ $$\stackrel{◯}{5}\iff |\mathbf{x}{|}^2-2\mathbf{x}\cdot \mathbf{y}+|\mathbf{y}{|}^2\ge 3|\mathbf{z}{|}^2.$$ 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 $\overrightarrow{PQ}=\mathbf{x}$, $\overrightarrow{QR}=\mathbf{y}$, $\overrightarrow{RP}=\mathbf{z}$. 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 $\angle ABC=\angle BCD=\cdots =\angle FAB=120^{\circ }$.