Going through the official solution of problem C8 from the IMO 2013 shortlist. Here's the whole pdf, in case someone needs it: https://www.imo-official.org/problems/IMO2013SL.pdf
Could someone please explain the very last paragraph on this page, the one that starts with words "Next, any interval blackened by B…"? What I struggle with is exactly the first sentence in this paragraph, namely — why do all "such intervals" have lengths not exceeding $1/2^m$?
Here's the scenario… Before round 1 begins, we have $x_1=0$. Now A plays $m=2$, and B responds with blackening the interval $I_1^1=\left[\frac{1}{4},\frac{1}{2}\right]$. At the start of round 2 we still have $x_2=0$. Now A plays $m=3$, and B blackens $I_1^2=\left[\frac{1}{8},\frac{1}{4}\right]$. Now, at the start of round $r=3$ we still have $x_r=x_3=0$. If A were to play $m=3$ again, B would blacken $I_0^3=\left[0,\frac{1}{8}\right]$. This would make the whole interval $\left[0,\frac{1}{2}\right]$ blackened, so that $x_{r+1}=x_4=1/2$.
Now, let's apply the logic from the sentence that I struggle with. According to it, all intervals that were blackened before the 3rd round have different lengths not exceeding $1/2^3=1/8$, since in the 3rd round A uses $m=3$. Which is obviously wrong, since in the very first round B blackens the interval of length $1/4$.
How come?
Best Answer
I think you are right, but the induction claim still holds because the respective part of the solution can be corrected to
"...all such intervals have different lengths not exceeding $\alpha$, so the total amount of ink used for them is less than $2\alpha$. Thus, the amount of ink used for the segment $[0, x_{r+1}]$ does not exceed the sum of $2\alpha$, $3x_r$ (used for $[0, x_r]$), and $1/2^m$ used for the segment $I^r_0$. In total it gives at most $3x_r+2\alpha+1/2^{m}<3(x_r+\alpha)=3x_{r+1}$".