AoPS - Problem

Source: 2022 Junior Macedonian Mathematical Olympiad P4

Tags: combinatorics

liliput

08.06.2022 00:30

An equilateral triangle $T$ with side length $2022$ is divided into equilateral unit triangles with lines parallel to its sides to obtain a triangular grid. The grid is covered with figures shown on the image below, which consist of $4$ equilateral unit triangles and can be rotated by any angle $k \cdot 60^{\circ}$ for $k \in \left \{1,2,3,4,5 \right \}$. The covering satisfies the following conditions: $1)$ It is possible not to use figures of some type and it is possible to use several figures of the same type. The unit triangles in the figures correspond to the unit triangles in the grid. $2)$ Every unit triangle in the grid is covered, no two figures overlap and every figure is fully contained in $T$. Determine the smallest possible number of figures of type $1$ that can be used in such a covering. Proposed by Ilija Jovcheski

Attachments:

As a sketch: $T$ contains $2022$ more upward-pointing unit triangles than downward-pointing unit triangles. As any rotation of figures of types 2,3,4 have the same number of upward- and downward-pointing triangles, this excess must be resolved by figures of type 1. The rotations of figure 1 have an excess of $\pm 2$ upward-pointing unit triangles, which means there are at least $2022/2 = 1011$ figures of type 1 in any covering. Any triangle of side length $2n$ may be decomposed into a "top" triangle of side length $2(n-1)$ and an isosceles trapezoid composed of the bottom two rows of the original triangle. This trapezoid may be covered with a figure of type 1, unrotated, on the far left, and $2n-2$ figures of type 2 rotated $60^\circ$ clockwise. Inductively, this gives a covering of the triangle of side length $2n$ using $n$ figures of type 1. Put together, this shows that the smallest number of figures of type 1 that can be used in any legitimate covering of $T$ is $1011$.