International Mathematical Olympiad Shortlisted Problems

No. Such a strategy for player $A$ does not exist.

We will present a strategy for player $B$ that guarantees that the interval $[0,1]$ is completely blackened, once the paint pot has become empty.

At the beginning of round $r$, let $x_r$ denote the largest real number for which the interval between $0$ and $x_r$ has already been blackened; for completeness we define $x_1=0$. Let $m$ be the integer picked by player $A$ in this round; we define an integer $y_r$ by $$\frac{y_r}{2^m}⩽x_r<\frac{y_r+1}{2^m}$$ Note that $I_0^r=[y_r/2^m,(y_r+1)/2^m]$ is the leftmost interval that may be painted in round $r$ and that still contains some uncolored point.

Player $B$ now looks at the next interval $I_1^r=[(y_r+1)/2^m,(y_r+2)/2^m]$. If $I_1^r$ still contains an uncolored point, then player $B$ blackens the interval $I_1^r$; otherwise he blackens the interval $I_0^r$. We make the convention that, at the beginning of the game, the interval $[1,2]$ is already blackened; thus, if $y_r+1=2^m$, then $B$ blackens $I_0^r$.

Our aim is to estimate the amount of ink used after each round. Firstly, we will prove by induction that, if before $r$th round the segment $[0,1]$ is not completely colored, then, before this move, (i) the amount of ink used for the segment $[0,x_r]$ is at most $3x_r$; and (ii) for every $m$, $B$ has blackened at most one interval of length $1/2^m$ to the right of $x_r$.

Obviously, these conditions are satisfied for $r=0$. Now assume that they were satisfied before the $r$th move, and consider the situation after this move; let $m$ be the number $A$ has picked at this move.

If $B$ has blackened the interval $I_1^r$ at this move, then $x_{r+1}=x_r$, and (i) holds by the induction hypothesis. Next, had $B$ blackened before the $r$th move any interval of length $1/2^m$ to the right of $x_r$, this interval would necessarily coincide with $I_1^r$. By our strategy, this cannot happen. So, condition (ii) also remains valid.

Assume now that $B$ has blackened the interval $I_0^r$ at the $r$th move, but the interval $[0,1]$ still contains uncolored parts (which means that $I_1^r$ is contained in $[0,1]$). Then condition (ii) clearly remains true, and we need to check (i) only. In our case, the intervals $I_0^r$ and $I_1^r$ are completely colored after the $r$th move, so $x_{r+1}$ either reaches the right endpoint of $I_1$ or moves even further to the right. So, $x_{r+1}=x_r+\alpha$ for some $\alpha >1/2^m$.

Next, any interval blackened by $B$ before the $r$th move which intersects $(x_r,x_{r+1})$ should be contained in $[x_r,x_{r+1}]$; by (ii), all such intervals have different lengths not exceeding $1/2^m$, so the total amount of ink used for them is less than $2/2^m$. Thus, the amount of ink used for the segment $[0,x_{r+1}]$ does not exceed the sum of $2/2^m$, $3x_r$ (used for $[0,x_r]$), and $1/2^m$ used for the segment $I_0^r$. In total it gives at most $3(x_r+1/2^m)<3(x_r+\alpha )=3x_{r+1}$. Thus condition (i) is also verified in this case. The claim is proved.

Finally, we can perform the desired estimation. Consider any situation in the game, say after the $(r-1)$st move; assume that the segment $[0,1]$ is not completely black. By (ii), in the segment $[x_r,1]$ player $B$ has colored several segments of different lengths; all these lengths are negative powers of $2$ not exceeding $1-x_r$; thus the total amount of ink used for this interval is at most $2(1-x_r)$. Using (i), we obtain that the total amount of ink used is at most $3x_r+2(1-x_r)<3$. Thus the pot is not empty, and therefore $A$ never wins.