SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO

Here is a winning strategy for Olya. Denote by $L_n$ the total length of the white segments after the $n$-th round ($L_0$ is considered to be equal to 1). Statement. Olya can play in such a way that for $k=0,1,\dots ,1012$ after the $(2k)$-th round the total length of the white segments is greater than $\frac{1}{2}$ ($L_{2k}>\frac{1}{2}$), and for some $\delta >0$ at least one of the open intervals $]0,\delta [$ or $]1-\delta ,1[$ is completely white. Let us prove this statement by induction. The basis for $k=0$ is obvious. Assume that after the $(2k)$-th round the situation is as described in the statement. Consider Tolya's move in the $(2k+1)$-th round. If he chooses $l=0$ or $l=1$, then Olya can choose the same $l$ on her turn in the round $2k+2$, and the segment $[0,1]$ will not change its color, with the exception, perhaps, of two points in the case $l=0$.

Now suppose that Tolya chose $0<l<1$. Since at least one of the outermost open intervals is white, Olya can choose $J$ in such a way that after the $(2k+1)$-th round both outermost open intervals $([0,\delta ]$ or $]1-\delta ,1[$ for some $\delta >0$) were painted white. If $L_{2k+1}>\frac{1}{2}$, then Olya simply chooses $l=0$. Let's consider the case when $L_{2k+1}\le \frac{1}{2}$. Since the outermost open intervals of the interval $[0,1]$ are white, there exists $ϵ>0$ such that the intervals $]0,ϵ[$ and $]1-ϵ,1[$ are white. Then if Olya chooses the number $l\in ]1-ϵ,1[$, for example $l=1-\frac{ϵ}{2}$, then whatever segment $J$ Tolya chooses, he will completely cover all black intervals and leave one or two white intervals with a total length of $ϵ$. That is, $L_{2k+2}=(1-L_{2k+1})+ϵ>\frac{1}{2}$. It is also clear that Tolya will not be able to repaint both outermost intervals black at once.