Bulgarian Mathematical Olympiad

Consider the following two colorings:

(1) Point $(x,y)$ is in color $i$, $1-i-2$ if and only if $x\equiv i(\mathrm{mod}2)$.

(2) Point $(x,y)$ is in color $i$, $1-i-3$ if and only if $x\equiv i(\mathrm{mod}3)$.

It is clear that if $S$ exists then $2S$ is an integer. Coloring (1) implies that $S$ could be $1$, $2$, $3$ or greater and coloring (2) shows that $S$ could be $3/2$, $3$ or greater. Therefore if $S$ exists then $S\ge 3$.

We prove that for arbitrary coloring there exists a triangle of area $3$ having all its vertices in one and the same color. Note that for some $d\in \{1,2,3\}$ there exist points $A=(x,y)$ and $B=(x+d,y)$ of the same color. Indeed, it suffices to consider the points $(0,0),(1,0),(2,0),(3,0)$. If $m\equiv AB$ and $l$ is parallel to $m$ at distance $\frac{6}{d}$ from $m$ then the line $l\parallel Ox$ is either having points of only two colors or we have a triangle of area $3$.

We say that a color $c$ permits distance $a$ in the line $l$ if there exist two points on $l$ having color $c$ and at distance $a$ apart.

If there exists a distance $a\in \{1,2,3,6\}$ that is permitted by the two colors on $l$ then the line $p\parallel l$ at a distance $\frac{6}{a}$ from $l$ is having all its points in the same color. If such a distance does not exist then one of the colors $l$ permits all distances in the set $\{2,3,6\}$.

(Indeed, assume that color $c_1$ does not permit distance $1$, and there exists at least one point $P_0$ from $l$ having color $c_1$. Two neighbors of $P_0$ - points $P_{-1}$ and $P_1$ are colored in $c_2$, so $c_2$ permits distance $2$. Therefore $c_1$ does not permit distance $2$ and points $P_{-2}$ and $P_2$ are of color $c_2$. As above $c_2$ does not permit distance $3$. $c_1$ does not permit distance $3$ and $P_{-3}$ and $P_3$ are of color $c_2$. Thus $c_2$ permits all distances in $\{2,3,6\}$.)

In both cases there exists a line $p\parallel Ox$ and color $c$ from $p$ that permits all distances in the set $\{2,3,6\}$.

Consider the lines $u_1,u_2,u_3$, parallel to $p$ and such that $u_i$, $i=1,2,3$ is at distance $i$ above $p$. It follows from the above that all such lines are having points only in two colors. Finally, consider all nine points of intersection of lines $u_1,u_2,u_3$ with lines $x=0$, $x=3$, $x=6$. It is easy to see that there exists a triangle of area $3$ having all its vertices colored in the same color.