Consider the equilateral triangular lattice in the complex plane defined by the Eisenstein integers; let the ordered pair $(x,y)$ denote the complex number $x+y\omega$ for $\omega=e^{2\pi i/3}$. We define an $\omega$-chessboard polygon to be a (non self-intersecting) polygon whose sides are situated along lines of the form $x=a$ or $y=b$, where $a$ and $b$ are integers. These lines divide the interior into unit triangles, which are shaded alternately black and white so that adjacent triangles have different colors. To tile an $\omega$-chessboard polygon by lozenges is to exactly cover the polygon by non-overlapping rhombuses consisting of two bordering triangles. Finally, a tasteful tiling is one such that for every unit hexagon tiled by three lozenges, each lozenge has a black triangle on its left (defined by clockwise orientation) and a white triangle on its right (so the lozenges are BW, BW, BW in clockwise order). a) Prove that if an $\omega$-chessboard polygon can be tiled by lozenges, then it can be done so tastefully. b) Prove that such a tasteful tiling is unique. Victor Wang.
Problem
Source: ELMO Shortlist 2012, C8
Tags: AMC, USAMO, polynomial, combinatorics
13.10.2016 09:25
For (a): Notice that we can "rotate"three lozenges and every time we do this,the white triangle which the black one connected to will change to another anti-clockwisely. So that after finite rotation,the triangle on the border will no longer be changed,so by induction we can get a tasteful tiling.
13.10.2016 09:51
For b: Construct a graph using every triangle as vertices,and two will connect if they are of the same lozenges.supose there are two different tasteful tiling,get them together as a new gragh(deleting the double side.obviously the new gragh consists of several non-intersecting circles.choose one circle without other triangles in it,there will be some point in it (the point of the triangles initially),and they form a tree when connected by the side of the triangles.concider one of the leafs of the tree,it can be seen one of the gragh contains a "bad "hexagon.contradiction! SORRY FOR MY POOR ENGLISH.
19.01.2017 11:22
P-H-David-Clarence wrote: choose one circle (maybe you mean cycle) without other triangles in it. Really, can you always do this?