Balkan Mathematical Olympiad

All solutions encountered in the competition (including the Official one) make use of the following

LEMMA. A triangle can be covered by a strip of breadth 1 if and only if (at least) one of its altitudes is at most 1 long.

Proof. The other implication being trivial, consider a triangle $ABC$ covered by a strip of breadth 1. The perpendicular lines through $A,B,C$ on the lines that define the strip can determine two situations one of them meets a side of the triangle $ABC$ at an interior point (Figure 1) one side of the triangle $ABC$ contains one of them perpendiculars (Figure 2)

Figure 1 Figure 2

In both cases notice one of the altitudes (in the figures – $AE$) is at most 1 long. $\square$

Now either choose points $A,B\in S$ such that the distance $AB$ be maximal, or choose points $A,B,C\in S$ such that the area of triangle $ABC$ be maximal.

In the former case, for any other point $C$, the distance from $C$ to $AB$ is at most 1 (when $ABC$ is a non-degenerate triangle, then this is the least altitude of the triangle), hence the set of points can be covered by a strip of breadth 2, determined by two parallel lines to $AB$, situated at distance 1 on each side of it.

In the latter case, it is easy to show all points of the set are included in the triangle $A^{\prime }{B}^{\prime }{C}^{\prime }$, determined by parallels through $A,B,C$ to the opposite sides (in effect, an old chestnut). Since one of the altitudes of $ABC$ is at most 1 long, one of the altitudes of $A^{\prime }{B}^{\prime }{C}^{\prime }$ will be at most 2 long, and we close the argument as above.