Let regular hexagon is divided into $6n^2$ regular triangles. Let $2n$ coins are put in different triangles such, that no any two coins lie on the same layer (layer is area between two consecutive parallel lines). Let also triangles are painted like on the chess board. Prove that exactly $n$ coins lie on black triangles.
Problem
Source: 2019 Saudi Arabia IMO TST I p3
Tags: combinatorics, Coloring
ckliao914
17.02.2022 12:19
Assume the height of every little triangle is $1$. Define the a,b,c-coordinate of a triangle as the minimum distance from a point on its boundary to the top side, the left-bottom side, the right-bottom side of the hexagon, respectively. The sum of the three coordinates of a black triangle is $3n-1$, while the sum of a white triangle is $3n-2$. The $2n$ triangles chosen has a-coordinate from $0$ to $2n-1$, so the sum of them is $2n^2-n$(b,c-coordinates, too). The sum of all the coordinates of the chosen triangle up(3 coordinates and $2n$ triangles) is $6n^2-3n$. If there are $x$ black triangles and $y$ white triangles chosen, then the sum of all coordinates can be computed again as $(3n-1)x+(3n-2)y$ with $x+y=2n$. Hence $6n^2-3n=3n(x+y)-x-2y, x+2y=3n$ We have $x=y=n$.