IMO Problem Shortlist

No, the Stepmother cannot enforce a bucket overflow and Cinderella can keep playing forever. Throughout we denote the five buckets by $B_0,B_1,B_2,B_3$, and $B_4$, where $B_k$ is adjacent to bucket $B_{k-1}$ and $B_{k+1}$ ($k=0,1,2,3,4$) and all indices are taken modulo $5$. Cinderella enforces that the following three conditions are satisfied at the beginning of every round:

(1) Two adjacent buckets (say $B_1$ and $B_2$) are empty.

(2) The two buckets standing next to these adjacent buckets (here $B_0$ and $B_3$) have total contents at most $1$.

(3) The remaining bucket (here $B_4$) has contents at most $1$.

These conditions clearly hold at the beginning of the first round, when all buckets are empty.

Assume that Cinderella manages to maintain them until the beginning of the $r$-th round ($r\ge 1$). Denote by $x_k$ ($k=0,1,2,3,4$) the contents of bucket $B_k$ at the beginning of this round and by $y_k$ the corresponding contents after the Stepmother has distributed her liter of water in this round.

By the conditions, we can assume $x_1=x_2=0$, $x_0+x_3\le 1$ and $x_4\le 1$. Then, since the Stepmother adds one liter, we conclude $y_0+y_1+y_2+y_3\le 2$. This inequality implies $y_0+y_2\le 1$ or $y_1+y_3\le 1$. For reasons of symmetry, we only consider the second case.

Then Cinderella empties buckets $B_0$ and $B_4$.

At the beginning of the next round $B_0$ and $B_4$ are empty (condition (1) is fulfilled), due to $y_1+y_3\le 1$ condition (2) is fulfilled and finally since $x_2=0$ we also must have $y_2\le 1$ (condition (3) is fulfilled).

Therefore, Cinderella can indeed manage to maintain the three conditions (1)-(3) also at the beginning of the $(r+1)$-th round. By induction, she thus manages to maintain them at the beginning of every round. In particular she manages to keep the contents of every single bucket at most $1$ liter. Therefore, the buckets of $2$-liter capacity will never overflow.

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Alternative solution.

We prove that Cinderella can maintain the following two conditions and hence she can prevent the buckets from overflow:

(1') Every two non-adjacent buckets contain a total of at most $1$.

(2') The total contents of all five buckets is at most $\frac{3}{2}$.

We use the same notations as in the first solution. The two conditions again clearly hold at the beginning. Assume that Cinderella maintained these two conditions until the beginning of the $r$-th round. A pair of non-neighboring buckets $(B_i,B_{i+2}),i=0,1,2,3,4$ is called critical if $y_i+y_{i+2}>1$. By condition (2), after the Stepmother has distributed her water we have $y_0+y_1+y_2+y_3+y_4\le \frac{5}{2}$. Therefore,

$$(y_0+y_2)+(y_1+y_3)+(y_2+y_4)+(y_3+y_0)+(y_4+y_1)=2(y_0+y_1+y_2+y_3+y_4)\le 5$$

and hence there is a pair of non-neighboring buckets which is not critical, say $(B_0,B_2)$. Now, if both of the pairs $(B_3,B_0)$ and $(B_2,B_4)$ are critical, we must have $y_1<\frac{1}{2}$ and Cinderella can empty the buckets $B_3$ and $B_4$. This clearly leaves no critical pair of buckets and the total contents of all the buckets is then $y_1+(y_0+y_2)\le \frac{3}{2}$. Therefore, conditions (1') and (2') are fulfilled.

Now suppose that without loss of generality the pair $(B_3,B_0)$ is not critical. If in this case $y_0\le \frac{1}{2}$, then one of the inequalities $y_0+y_1+y_2\le \frac{3}{2}$ and $y_0+y_3+y_4\le \frac{3}{2}$ must hold. But then Cinderella can empty $B_3$ and $B_4$ or $B_1$ and $B_2$, respectively and clearly fulfill the conditions.

Finally consider the case $y_0>\frac{1}{2}$. By $y_0+y_1+y_2+y_3+y_4\le \frac{5}{2}$, at least one of the pairs $(B_1,B_3)$ and $(B_2,B_4)$ is not critical. Without loss of generality let this be the pair $(B_1,B_3)$. Since the pair $(B_3,B_0)$ is not critical and $y_0>\frac{1}{2}$, we must have $y_3\le \frac{1}{2}$. But then, as before, Cinderella can maintain the two conditions at the beginning of the next round by either emptying $B_1$ and $B_2$ or $B_4$ and $B_0$.