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Geometry Difficulty 8.5 Shortlist Prove it Balkan Mathematical Olympiad

By a strip of breadth $b$ we mean a closed part of the plane consisting of all points that lie between two parallel lines at distance $b$ from each other. Let $S$ be a finite set of $n$ ( $n \ge 4$ ) points in the plane, such that any three points from $S$ can be covered by a strip of breadth 1. Prove that $S$ can be covered by a strip of breadth 2.

Solution

Firstly we shall prove the following statement.

Lemma. If a triangle can be covered by a strip of breadth $b$ , then at least one altitude of the triangle is at most $b$ long.

Proof. At least one of the perpendicular lines through the vertices of the triangle to the border lines of the strip meets the opposite side of the triangle. Therefore the segment between that vertex and the meeting point with the opposite side is of length at most $b$ . The altitude corresponding to that vertex is thus also of length at most $b$ . The Lemma is proved. $\Box$

As a corollary, the least breadth of a strip that can cover a triangle is equal to the length of its shortest altitude.

Choose now points $A$ and $B$ from $S$ at maximal distance from each other. For any other point $C$ from $S$ the side $AB$ will be the longest of the triangle $ABC$ . Therefore the altitude from $C$ on $AB$ will be the shortest. According to Lemma, it is at most $1$ long, since the triangle $ABC$ can be covered by a strip of breadth $1$ , by hypothesis.

Hence $S$ will be covered by a strip of breadth $2$ with borders parallel to $AB$ , at distance $1$ on both sides of $AB$ . $\Box$

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