Triangle with same black and white areas Suppose we have an infinite chessboard with the usual black/white coloring. A triangle $T$ with area $a$ is given with vertices at corners of some cells. Prove that there exists another triangle $T'$ similar to $T$ with area $2a$, also with vertices at corners of some cells, such that the white area and the black area in $T'$ are both $a$.
One idea is to double all of the coordinates of $T$ (change $(x,y)$ to $(2x,2y)$ for all $x,y$), but this results in a triangle with four times the area, not twice. To get twice we have to multiply by $\sqrt{2}$, but this does not result in integer coordinates.
 A: Update (Full solution)
You may assume that one vertex of the given lattice triangle $T$ is $(0,0)$. Apply the linear transformation $A$ with matrix
$$[A]=\left[\matrix{1&-1\cr 1&1\cr}\right]$$ to $T$. This amounts to a counterclockwise rotation by  $45^\circ$ and a linear scaling by  $\sqrt{2}$. The resulting triangle $T'$ is again a lattice triangle, is similar to $T$, and has area  $2a$. In addition the vertices ${\bf z}_i=(x_i,y_i)$ of $T'$ obey the parity condition $$x_i+y_i=0\qquad({\rm mod}\ 2)\ .\tag{1}$$
Call a point ${\bf z}=(x,y)\in{\mathbb Z}^2$ even if it satisfies $(1)$.
Claim: If the vertices ${\bf 0}$, ${\bf a}$, ${\bf b}$ of a lattice triangle $T$  are all even then its b/w-ratio  is ${1\over2}$. 
Proof. The midpoint ${\bf m}:={1\over2}({\bf a}+{\bf b})$ is either a lattice point or the center of a lattice square. Let ${\bf z}\mapsto R\,{\bf z}$ denote the reflection of the  plane in ${\bf m}$. Then ${\bf z}$ and $R\,{\bf z}$ have the same color. It follows that  the parallelogram $P$ with vertices ${\bf 0}$, ${\bf a}$, ${\bf a}+{\bf b}$, ${\bf b}$ obtained by reflecting $T$ in ${\bf m}$ has the same b/w-ratio as $T$. 
It is easily seen that the translations ${\bf z}\mapsto {\bf z}+{\bf a}$ and  ${\bf z}\mapsto {\bf z}+{\bf b}$ are color preserving as well. It follows that the plane can be completely tiled with color-congruent translational copies of $P$. From this we deduce that the b/w-ratio of $P$, and hence of $T$, must be ${1\over2}$.
