A Beautiful Combi

[Nirjhor]

Suppose we have a collection of infinite stones. On the Cartesian plane, a stone is kept on $(0,0)$. In a move, we can remove a stone from $(m,n)$ (provided that it contains a stone) and keep two stones on $(m+1,n)$ and $(m,n+1)$ (provided that these two points don't have any stones). Prove that at any point in the whole process, there will always be a stone placed on some lattice point $(a,b)$, with $a+b\le 3$. (The point $(a,b)$ is not fixed)

Source: some TST.

[Phlembac Adib Hasan]

The idea:

If the statement is false, then there will be at least four stones on line $x=1$ at some point. And three of them have to be moved from the $(3,0)(0,0)(0,3)$ triangle. Notice how this moving process affects line $x=2$ (and the later vertical lines)

[Nirjhor]

My solution relies on a weighting invariance.

Assign the stone at $(x,y)$ the weight $\dfrac{1}{2^{x+y}}$. The reason for this choice is that under any valid move, the total weight remains constant, and that is $1$. So at some point in the process, if we exclude the points $(a,b)$ with $a+b\le 3$ then the total weight at that point is clearly less than \[\left(\sum_{i=0}^\infty \sum_{j=0}^\infty \dfrac{1}{2^{i+j}}\right)-\left(1+\dfrac{2}{2}+\dfrac{3}{4}+\dfrac{4}{8}\right)=\dfrac{3}{4}<1\] which is impossible. Here \[\sum_{i=0}^\infty \sum_{j=0}^\infty \dfrac{1}{2^{i+j}}=\left(\sum_{i=0}^\infty \dfrac{1}{2^i}\right)\left(\sum_{j=0}^\infty \dfrac{1}{2^j}\right)=2\times 2=4. \]