Belarusian Mathematical Olympiad

b) it is not true.

a) Let $ABCDDEF$ be a hexagon with all obtuse inner angles and parallel opposite sides ($AB\parallel DE$, $BC\parallel EF$, $CD\parallel FA$, see Fig. 1). Consider the greatest side of this hexagon (one of such sides if there are more than one). Let it be the side $AB$. Since the hexagon $ABCDDEF$ is convex, it completely lies in one of the half-planes with respect to the line $AB$; denote this half-plane by $\Pi$.



Fig. 1

Suppose that none of the lines intersecting and perpendicular to the side $AB$ intersect the side $DE$. Construct the perpendiculars $h_A$ and $h_B$ in the half-plane $\Pi$ (see Fig. 2). By our assumption, there are no points of the side $DE$ in the half-strip $P_0$. In particular, either point $D$ lies to the right of the ray $h_A$ or point $E$ lies to the left of the ray $h_B$. Without loss of generality, suppose that $D$ lies to the right of $h_A$ (see Fig. 2). Then the vertex $C$ lies neither to the right of the ray $h_A$ nor to the left of the ray $h_B$. Indeed, otherwise we obtain $BC>BA$ (see Fig. 3) and $CD>BA$ (see Fig. 4), respectively. But any of these inequalities contradicts to the choice of the side $AB$ as the greatest side of the hexagon. So the vertex $C$ lies in the half-strip $P_0$, but in this case the angle $ABC$ is not obtuse, contrary to the problem condition. Therefore, there exists a straight line such that it is perpendicular to the sides $AB$ and $DE$ and intersects both of them.

b) Construct the convex hexagon $ABCDDEF$ satisfying the problem condition (all inner angles are obtuse and opposite sides are parallel) such that there exists exactly one pair of the opposite sides intersected by the perpendicular straight line.



Fig. 5 Fig. 6

Consider the isosceles triangle $KLM$ with the base $LM$ and the acute angle $LKM$ (see Fig. 5). Let $KH$ be the altitude of the triangle $KLM$. Let points $P$ and $Q$ be the inner points of the segments $MH$ and $LH$, respectively, such that they are symmetric with respect to $H$. Let $E$ be the foot of perpendicular from $P$ into $KL$, and let $D$ be the some point $A$ on the segment $LQ$ and some point $B$ on the segment $MP$. Construct the line $l(A)$ passing through $A$ parallel to $KM$, and construct the line $l(B)$ passing through $B$ parallel to $KL$. Let $F$ be the intersection point of $l_A$ and $KL$, and let $G$ be the intersection point of $l_B$ and $KM$ (see Fig. 5). Then (see Fig. 6) it is easy to see that the hexagon $ABCDDEF$ is required.