66th Belarusian Mathematical Olympiad

Answer: 1.

Show that among any two intersecting main diagonals of the hexagon at most one diagonal can be good. Suppose, contrary to our claim, that there are two good intersecting main diagonals. Without loss of generality, we assume that $AD$ and $BE$ are good diagonals of the hexagon $ABCDEF$ (see Fig. 1). Then $ABCD$ and $BCDE$ are circumscribed quadrilaterals, therefore the sums of their opposite sides are equal. Hence, $AB+CD=BC+AD$, $BC+DE=BE+CD$. Summing both the equalities we obtain $$AB+DE=AD+BE.(1)$$ Let $M$ be the intersection point of $AD$ and $BE$. Then by the triangle inequality we have $$\begin{gathered} AD+BE=(AM+MD)+(BM+ME)=(AM+BM)+(DM+EM)> \\ >AB+DE, \end{gathered}$$ contrary to (1).

Fig. 1 Fig. 2

If the diagonal of the hexagon divides it into two quadrilaterals, then this diagonal connects two opposite vertices of the hexagon ("main" diagonal). But any two such diagonals intersect each other, hence the hexagon has at most one good diagonal.

There are many ways to construct the hexagon with one good diagonal. For example, we draw the circle and circumscribe the trapezoid $ABCD$, $AD\parallel BC$, $AD>BC$ (see Fig. 2). Let $E$ and $F$ be symmetric to $C$ and $B$, respectively, with respect to the line $AD$. It is evident that $AD$ is a good diagonal of the hexagon $ABCDEF$.