A squiggle is composed of six equilateral triangles with side length $1$ as shown in the figure below. Determine all possible integers $n$ such that an equilateral triangle with side length $n$ can be fully covered with squiggles (rotations and reflections of squiggles are allowed, overlappings are not). [asy][asy] import graph; size(100); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; draw((0,0)--(0.5,1),linewidth(2pt)); draw((0.5,1)--(1,0),linewidth(2pt)); draw((0,0)--(3,0),linewidth(2pt)); draw((1.5,1)--(2,0),linewidth(2pt)); draw((2,0)--(2.5,1),linewidth(2pt)); draw((0.5,1)--(2.5,1),linewidth(2pt)); draw((1,0)--(2,2),linewidth(2pt)); draw((2,2)--(3,0),linewidth(2pt)); dot((0,0),ds); dot((1,0),ds); dot((0.5,1),ds); dot((2,0),ds); dot((1.5,1),ds); dot((3,0),ds); dot((2.5,1),ds); dot((2,2),ds); clip((-4.28,-10.96)--(-4.28,6.28)--(16.2,6.28)--(16.2,-10.96)--cycle);[/asy][/asy]
Problem
Source: Baltic Way 2007
Tags: geometry, geometric transformation, rotation, reflection, parallelogram, modular arithmetic, combinatorics proposed
02.12.2010 01:08
It is easy to see that there are only two ways of starting filling a triangle with squiggles at an angle. Depending on which one of the two bottom angles of a squiggle coincides with the angle of the triangle, there are either 2 squiggles that form a 2x3 parallelogram or 4 squiggles that form a 3x4 parallelogram. The angles next to those parallelograms are similar and can thus only be filled with parallelograms again. So squiggles cannot fill any triangle.
19.01.2012 20:03
we can do this for all $n=0(mod12)$
20.01.2012 00:50
Oh, I see that I was wrong for the 3x4 parallelogram. As we need $6\mid n^2$, we have $n\equiv0\pmod 6$ as necessary condition. Color the cells alternatingly in black and white. Then a squiggle contains either $4$ white and $2$ black cells or vice versa. The big triangle contains $\binom{n+1}2$ cells of one and $\binom n2$ cells of the other color. So these numbers must be both even, which means $n\equiv0\pmod 4$. This means that altogether we need $n\equiv0\pmod {12}$. Thus it is sufficient to find a construction for $n=12$. Do you have one? Seems to be fairly asymmetric.
22.01.2012 22:29
i remember that i construct an example for n=12 but i can't do this now i do it very fluky