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Combinatorics Difficulty 7.3 National olympiad, round 2 Prove it Ukraine

Given a foundation that is in a form of a rectangle 2m×2n2m \times 2n that is divided into smaller 1×11 \times 1 squares. There is a gap of length 11 between any two adjacent squares. The foundation is covered by several layers of bricks of size 2×12 \times 1. Every layer consists of 2mn2mn bricks and each brick fully covers exactly one gap of length 11. Such cover is called *strong*, if every gap is covered by a brick at least in one of the layers. What is the minimum amount of layers that make a strong cover?
(Bogdan Rublyov)

Solution

Without loss of generality let 2mn2 \le m \le n. Consider a square AA of size 1×11 \times 1, that does not touch the sides of a big rectangle. All 4 sides of it have to be covered with different bricks, since half of each brick covers the square AA itself. Therefore, covering with less than 4 layers is not possible. It suffices to show that covering with 4 layers is possible (Fig. 19). First two layers are shown on the picture, the other two layers are given by dividing rectangle on 2×22 \times 2 squares. Then two different options of dividing such squares into bricks give two layers as in Fig. 19.

Clearly, in case of m=n=1m=n=1 it suffices to have 2 layers.

If m=1<nm=1<n covering with 3 layers exists. Example is shown in Fig. 20. Covering with less layers is not possible, since for those squares whose vertices does not coincide with a vertices of a big rectangle, 3 sides have to be covered, that is possible only with at least three layers.

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