No, she can´t. Cinderella, on her turn, empties the two neighboring buckets with maximum sum of water. Assume the stepmother wins. She wins iff at some point three or more buckets have more than one liter of water each. That happens iff in her turn there are three buckets that have two liters of water between the three of them. Obviously this can`t happen in the first turn. Since Cinderella always empties the two neighboring buckets with maximum sum of water, Cinderella always empties at least $2/5$ of the total amount of water. Therefore, before Cinderella emptied her buckets, there had to be at least $2:3/5=10/3$ liters of water. WLOG assume this was the first time there was at least $10/3$ liters. Then on the previous turn, the stepmother added $1$ liter, so before the stepmother played there were $7/3$ liters. But the same calculation gives that the turn before there were at least $7/3:3/5=35/9$ liters. But $35/9>10/3$ , contradiction.

I wonder what is the minimum amount of water the stepmother needs to pour to ensure a bucket overflows?

jestrada wrote: No, she can´t. Cinderella, on her turn, empties the two neighboring buckets with maximum sum of water. Assume the stepmother wins. She wins iff at some point three or more buckets have more than one liter of water each. That happens iff in her turn there are three buckets that have two liters of water between the three of them. Obviously this can`t happen in the first turn. Since Cinderella always empties the two neighboring buckets with maximum sum of water, Cinderella always empties at least $2/5$ of the total amount of water. Therefore, before Cinderella emptied her buckets, there had to be at least $2:3/5=10/3$ liters of water. WLOG assume this was the first time there was at least $10/3$ liters. Then on the previous turn, the stepmother added $1$ liter, so before the stepmother played there were $7/3$ liters. But the same calculation gives that the turn before there were at least $7/3:3/5=35/9$ liters. But $35/9>10/3$ , contradiction. You claim she wins only if she can get 3 buckets with > 1 liter. But what about two buckets with > 1 liter, but they aren't consecutive? Also she can get 2-epsilon for any epsilon, but she cannot actually get 2.

Let the buckets be $B_1,B_2,...,B_5$. Suppose after some turn, 2 of the buckets, say $B_4,B_5$ are empty, and $B_1+B_3\leq 1$ and $B_2\leq 1$. Clearly after stepmother's turn, no buckets will overflow. Now after she pours the water in, $B_1+B_3+B_4+B_5\leq 2$ so by pigeonhole, WLOG let $B_1+B_4\leq 1$. Clearly $B_5\leq 1$, so Cinderella will empty $B_2,B_3$ and the condition still remains. Thus Cinderella can maintain this condition and the buckets can never overflow.

jestrada wrote: I wonder what is the minimum amount of water the stepmother needs to pour to ensure a bucket overflows? The answer to the question is 2. To prove this, I will show that Cinderella can have one bucket with $2-\epsilon$ liter in it. (in this case, $\epsilon = \frac{1}{2^{n}}$ for any positive integer $n$). Let Cinderella start with the move that leads to $(1/2,0,1/2,0,0)$. Than, he will have at least one bucket with $1/2$ liter in it. In his next move, Cinderella will get the position $(3/4,0,3/4,0,0)$ (worst case). Going on in this manner Cinderella can get $(\frac{1}{2^n}, 0 , \frac{1}{2^n}, 0 , 0 )$. And after Stepmothers move, Cinderella will get (in one of the buckets) $2-\frac{1}{2^n}$ liters.

I've done a typo Instead of $(\frac{1}{2^n} , 0 , \frac{1}{2^n} , 0 , 0 )$ I should have written $(1-\frac{1}{2^n} , 0 , 1-\frac{1}{2^n} , 0 , 0 )$.

If the stepmaother gets two non-consecutive buckets with more than 1lit of water she wins. So lets prove she can't have equal or more than2lit of water in the buckets at anytime.I used induction to prove this: Lets say in a turn we have S=2-K<2lit water totally.{The amount of water in each bucket is A_i} Lets prove that we will have S'<2lit water in the next round.{The amount of water added to each bucket in this turn is B_i} (A_1+B_1)+...+(A_5+B_5)=3-K Now lets say C_i=A_i+B_i Then: (C_1+C_2)+...+(C_5+C_1)=2*(3-K)=6-2K By pigeonhole priciple there exist two consecutive buckets with at least 1.2-(K/5)lit of water together. Cinderella empties these buckets and becuase 1.2-(K/5)>1-K now altogether there are S'<2lit of water. AND WE ARE DONE

Reading through this thread (and also official solutions http://www.imo-official.org/problems/IMO2009SL.pdf) it's difficult to see how people come about these invariants. Can anyone give some motivation behind them?

MatjazL wrote: Can anyone give some motivation behind them? The idea is to work backwards. First we know that after every turn, Cinderella needs to make sure the buckets have $a_1,\ldots,a_5\le 1$ or else the stepmother can make some $a_i>1$ overflow. Suppose that the stepmother then adds $b_1,\ldots,b_5$ to the buckets, where $b_1+\cdots+b_5=1$. Going one step further, this means that for any two nonadjacent buckets with $a_i,a_{i+2}$ (indices taken modulo 5) after Cinderella's turn, the stepmother wins if we have both $a'_i=a_i+b_i>1$ and $a'_{i+2}=a_{i+2}+b_{i+2}>1$, because Cinderella can then only empty at most one of the buckets $i,i+2$. So to prevent the stepmother from setting $(b_i,b_{i+2})=(1+\epsilon-a_i,1+\epsilon-a_{i+2})$, Cinderella needs to make sure $2+2\epsilon-a_i-a_{i+2}>1=b_1+\cdots+b_5$ for all $\epsilon>0$, or equivalently, $a_i+a_{i+2}\le 1$. Now you can easily find oneplusone's solution. (In fact, I think these ideas can easily show that the bucket size can't be reduced if Cinderella wants to win.)

Very nice explanation, thanks .

My solution appears to be similar to the second official one, although there is some casework in mine. I will present it here, with some motivation spread throughout. First, let's check the weaker case where Cinderella can remove any two of the buckets. Now for the stepmother to win, at some point there must be three buckets, each with total greater than $1$. The apparent weakness of this problem suggests a purely greedy approach: Cinderella takes the two buckets with the largest amount of water. Thus the total amount of water after Cinderella goes and the Stepmother goes again is at most $\tfrac{3}{5}n+1$, where $n$ is the amount of water after the Stepmother's previous turn. But if $n\le 3$, then $\tfrac{3}{5}n+1\le \tfrac{14}{5}<3$, so it's easy. But this is not optimal; since $\tfrac{14}{5}<3$ we can find a better bound. We just solve $n=\tfrac{3}{5}n+1$ or $n=\tfrac{5}{2}$. So Cinderella can always keep the total amount of water at most $2.5$ (in this weaker case). Store this away. This ends up having no bearing on the actual problem, but I wanted to explain why we're motivated to keep pushing with our approach. So in the actual problem, the Stepmother has to fill two non-adjacent buckets such that each has at least a liter of water in it (otherwise Cinderella will remove any with volume larger than $1$.) Then before the Stepmother's turn, the sum of the volume of water in those two buckets was at least $1$. So all Cinderella has to do is ensure that the five sums $B_1+B_3,B_1+B_4,B_2+B_4,B_2+B_5,B_3+B_5$ are each at most one after any of her turns. (It is at this point where perhaps the more intelligent, reserved approach leads to oneplusone's excellent solution. On the other hand, rushing forward like an angry bull accomplishes the same thing.) Let's suppose Cinderella embarks upon this strategy. Note that if the Stepmother ever makes all of these sums more than $1$, Cinderella will lose, since she can only remove $4$ of the problems. So, along with our inspiration of our earlier work, we devise the additional requirement for Cinderella: $B_1+B_2+B_3+B_4+B_5\le 1.5$ (after her turn) because then, after the Stepmother's turn the sum is at most $2.5$ and hence at least one of the non-adjacent sums is at most $1$. So we have created a strategy for Cinderella which we are hoping will work: (Let $S=\{ B_1+B_3,B_1+B_4,B_2+B_4,B_2+B_5,B_3+B_5\}$ for convenience.) For every item $x\in S$, Cinderella ensures that $x\le 1$. Also, she ensures that $B_1+B_2+B_3+B_4+B_5\le 1.5$. It is obvious that if she follows this strategy she will never lose. It suffices to show that there is always a pair of adjacent buckets, such that removing them satisfies our conditions. We just do this by casework, on which elements in $S$ are greater than one after the Stepmother's turn. Assume that Cinderella has successfully followed her strategy on every previous turn (obviously it is possible on the first turn) and we prove she can successfully follow it now. I have included Asymptote diagrams for convenience. If 4/5 elements in $S$ are greater than $1$:

WLOG $B_2+B_5\le 1$. Now if $B_1> \tfrac{1}{2}$ then $2.5\ge B_1+(B_2+B_4)+(B_3+B_5)> 2.5$, contradiction. So $B_1\le \tfrac{1}{2}$ and $B_1+(B_2+B_5)\le 1.5$, so Cinderella can remove $B_3, B_4$ and is done. If 3/5 elements in $S$ are greater than $1$:

By Pigeonhole one vertex must be used twice, WLOG it is $B_1$. Now there are two cases, depending on if the last connection does or does not touch one of $B_3,B_4$. If it does not, then $B_2+B_5\le 1$. Now if $B_3+B_4>1$ then $2\ge (B_2+B_4) +(B_3+B_5) > 2$, contradiction. So $B_3 +B_4\le 1$. Now if $B_2+B_3+B_4, B_5+B_3+B_4>1.5$, then $3\ge (B_2+B_4)+(B_3+B_5)+(B_3+B_4)>3$, contradiction. So one of the pairs $B_1,B_5$ and $B_1,B_2$ will work for Cinderella. Now if one of $B_3,B_4$ is touched, WLOG it is $B_4$. Now if $B_5>\tfrac{1}{2}$ then $2.5\ge B_5+(B_1+B_3)+(B_2+B_4)>2.5$, contradiction. So $B_5\le \tfrac{1}{2}$. Now if $B_3+B_4+B_5, B_1+B_2+B_5>1.4$ then $3< B_5+(B_1+B_2+B_3+B_4+B_5)\le 3$, contradiction. So there is a pair that works for Cinderella. If 2/5 elements in $S$ are greater than $1$:

Note that this is symmetrical to the 3/5 case, so we have found both possibilities. In the first, if $B_1\le \tfrac{1}{2}$, then $B_3,B_4>\tfrac{1}{2}$, then $B_2,B_5\le \tfrac{1}{2}$, so removing $B_3,B_4$ wins for Cinderella. Otherwise $B_1>\tfrac{1}{2}$. Now if $B_1+B_2+B_5,B_3+B_4+B_5,B_2+B_3+B_4>1.5$ then $4.5< 2(B_1+B_2+B_3+B_4+B_5)-B_1\le 4.5$, contradiction. So there is a pair that wins for Cinderella. In the second, if $B_1> \tfrac{1}{2}$ then $2.5\ge B_1+(B_2+B_4)+(B_3+B_5)>2.5$, contradiction, so $B_1\le \tfrac{1}{2}$. Then $B_1+(B_2+B_5)\le 1.5$, so removing $B_3,B_4$ wins for Cinderella. If 1/5 elements in $S$ are greater than $1$:

Obviously this is the only case. WLOG $B_2\ge B_5\implies B_2\ge \tfrac{1}{2}\implies B_4\le \tfrac{1}{2}$. So $B_4+(B_5+B_3)\le 1.5$, so removing $B_1,B_2$ works for Cinderella. Now in the 0/5 case, she removes the pair with maximal sum. Obviously this sum is at least $\tfrac{2}{5}$ of the total, so after her turn there is at most $\tfrac{3}{5}\cdot 2.5=1.5$ liters of water which works. And since the whole point of the total sum restriction was to exclude the 5/5 case, we are done. By the way, this casework looks very long but it's very very quick while you're doing it, especially since all of the arguments are similar.

Just notice that if we in some moment have $Bi,Bi+1,Bi+2$ we must have $Bi+Bi+2<=1$ and $Bi+1$.It is obvious that $(0,0,0,0,0)$ satisfayes this,so suppose that in some moment we have $Bi+Bi+2<=1$ and $Bi+1<=1$.Now,it is easy to see(it is the same as in oneplusone solution) that we can again obtain for some j $Bj+Bj+2<=1$ and $Bj+1$,so we are finished.

We will prove that Cinderella can always empty two neighboring buckets such that the following invariant holds: Invariant. For any diagonal of the pentagon, the sum of the amount of water in the two edge buckets is at most $1$ liter. This invariant is clearly satisfied at the beginning. Now, suppose at some moment the Stepmother distributes $1$ liter arbitrarily over the five buckets. Lemma 1. For any four buckets, there exist two connected by a diagonal with sum of water at most $1$ liter. Proof. Supposing the contrary, we find the sum of the water in the four buckets is greater than $2$ liters. However, before the Stepmother's move, the sum of water of two of the buckets connected by a diagonal is at most $1$ liter, and the Stepmother can add at most $1$ liter distributed among the four buckets, so the final sum of water should be at most $2$ liters, contradiction. Lemma 2. If two buckets have more than $1$ liter each, they must be adjacent. Proof. If the two buckets are not adjacent, then before the Stepmother's move, the sum of water in the buckets is at most $1$ liter. After the Stepmother's move, the sum of water in the buckets is at most $2$ liters, so some bucket has at most $1$ liter of water, contradiction. Now suppose there is a bucket $B$ with more than $1$ liter. Among the remaining buckets, there exist two, $C$ and $E$, with sum of water at most $1$ liter by Lemma 1. There is a third bucket $D$ such that $C, D, E$ are consecutive. Because $B$ and $D$ are not adjacent, $D$ has at most $1$ liter by Lemma 2. Thus, remove the buckets other than $C, D, E$, and notice $C, D, E$ satisfy the invariant. If there is no bucket with more than $1$ liter, then choose $B$ arbitrarily. Among the remaining buckets, there exist two, $C$ and $E$, with sum of water at most $1$ liter by Lemma 1. There is a third bucket $D$ such that $C, D, E$ are consecutive. By our assumption, $D$ has at most $1$ liter. Thus, remove the buckets other than $C, D, E$, and notice $C, D, E$ satisfy the invariant. Hence, the invariant always holds, so in particular, no bucket can have more than $1$ liter before the Stepmother's move. Then any bucket has at most $2$ liters after the Stepmother's move, so no bucket can overflow.

Please check my proof

The wicked stepmother can not enforce a bucket flow. Let $S$ be the set of the sums of amount of water in the five pairs of buckets that are not adjacent. Let $s$ be sum of the pair of non neighboring buckets that remains after Cinderella takes two neighboring buckets to the river. Cinderella's strategy: Cinderella can prevent overflow by using the following strategy: (1) If there are buckets with more than $1 L$ water, she will remove one of them and one bucket adjacent so that $s$ is smallest possible, (2) otherwise she chooses the sum in $S$ that is smallest and empties buckets so that $s$ is this sum. $A:$ If after some round, $s >1$, then in this round after stepmother's step,all sums were $>1$.

$B:$ If overflow, previously existed a time when all five numbers in $S$ are $>1$

$C:$ Impossible for all five numbers in $S$ to be $>1$

Now $B$ and $C$ implies that the stepmother can't enforce a bucket flow.

Could someone explain Solution 2 from the official solutions (I'd post a link to the official solutions, but as a newly registered, I can't - the link can be found in the post of MatjazL above though)? I struggle at the penultimate paragraph starting with words "Now suppose...". Assuming the first inequality $y_0+y_1+y_2\leq\frac{3}{2}$ holds and we empty buckets $B_3$ and $B_4$, how come condition 1' is satisfied for buckets $B_1$ and $B_3$? After emptying we have $y_1+y_3=y_1$ and it needs to be $\leq1$, but I fail to see why $y_1\leq1$. Thank you.

@above: I also think that the official solution is confusing, here is my solution: Label the buckets 1 through 5 in clockwise order Condition 2: Cinderella can keep the total water below 5/2 This is easy to see since if you sum the 5 pairs of adjacent buckets, you get twice the total water. Thus cinderella can empty at least 40% of the total water by emptying the largest pair. The witch can only add 1 liter per turn so the water is less than 5/2 Condition 1: Cinderella can keep every non-adjacent pair sums to less than 1 Note that if every non-adjacent pair summed to less than 1 the previous turn, this turn, after the witch goes, every non-adjacent pair sums to less than 2.(witch can only add 1 liter) Since the total sum is less than 5/2, there must exist a non-adjacent pair that sums to less than 1.(The sum of all the pairs is less than 5 since each bucket counted twice) Suppose $B_1+B_3<1$ wlog, if $B_2$ has less than 1 liter, then we are done(empty $B_4,B_5$) if $B_2>=1$ then $B_1+B_3+B_4+B_5<3/2$ Thus either $B_1+B_4$ or $B_3+B_5$ is less than 1. Suppose $B_1+B_4<1$ We know that $B_5<1$ (otherwise $B_2+B_5>2$) and so we can just empty $B_2,B_3$. It is easy to see with these 2 conditions the witch cannot win. Edit: How is this a C5? Meanwhile I can't even do an A2

We claim Cinderella can win. Suppose the position after her move is $a,b,c,0,0$. We call this position strong if $b,a+c<1$. Clearly, if a position is strong, stepmother can't immediately make it overflow on her turn. In fact, Cinderella will be able to do her move such that position remains strong, so the process will continue forever, as desired. Suppose stepmother distributes the water by $x_1,\ldots,x_5$, so the new position is \[a+x_1,b+x_2,c+x_3,x_4,x_5.\]We claim Cinderella can make the position strong by removing either $a+x_1,b+x_2$, or $b+x_2,c+x_3$. If this was not the case, we would have $c+x_3+x_5\ge 1$, and $a+x_1+x_4\ge 1$. This means we would have $a+c+x_1+x_3+x_4+x_5\ge 2$, but this is clearly false as $x_1+x_3+x_4+x_5\le 1$ and $a+c<1$. Thus, the new position can be made to be strong, so the game continues indefinitely.

We claim Cinderella can always keep the buckets from overflowing. Suppose that Cinderella just emptied two consecutive buckets, so the remaining are now $(0,0,a,b,c)$ upto cyclic shifting. We claim that if $a+c<1$ and $b<1$, then Cinderella can keep the buckets from overflowing. Clearly, if $b>1$, Stepmother adds all the water to it and wins. Suppose $a+c>1$. Then we claim Stepmother can win. Indeed, she can simply add water to $a$ and $c$ to make them each slightly greater than 1; call the new config $(0,0,a',b,c')$. Now, if Cinderella empties $(a',b)$ or $(b,c')$, one of $a',c'>1$ will remain. Then Stepmother can add 1 to this bucket, making it overflow. We also must show that if $a+c<1,b<1$, then any subsequent position can also satisfy this. Say the new position is $(x,y,a',b',c')$ for some $a',b',c',x,y$. Suppose we remove $a',b'$; the resulting position is $(0,0,c',x,y)$, and if we remove $b',c'$, the resulting position is $(0,0,x,y,a')$. Either $x+a'<1$ or $c'+y<1$, since $x+y+a'+c'<2$ because we added at most 1. Hence, at least one of the two is valid. The game continues infinitely.

We claim Cinderella can always prevent an overflow. Suppose after her turn that the buckets have values $(x_1,x_2,x_3,0,0)$. We claim that we can actually always force $x_1+x_3<1$, $x_2<1$, which of course directly implies the result. Suppose that the current configuration already satisfies the criterion. Now, the Stepmother adds $(d_1,d_2,d_3,d_4,d_5)$ to the buckets. If $d_4$ or $d_5$ is $1$, then $d_1=d_2=d_3=0$, so we can just empty buckets $(4,5)$ and get back where we started. On the other hand, if $d_4,d_5<1$, I claim we can either empty buckets $(1,2)$ or $(2,3)$ to preserve the condition. If not, then we must have $d_5+d_3+x_3\ge 1$, $d_4+d_1+x_1\ge 1\implies (d_1+d_3+d_4+d_5)+(x_1+x_3)\ge 2$. However, we know $d_1+d_3+d_4+d_5=1-d_2\le 1$, $x_1+x_3<1$, so this is a contradiction, and Cinderella can actually always enforce the condition, as desired.

edit: why is (almost) every solution the same Cinderella always wins. Consider all possible bucket states after Cinderella has moved. Clearly two of them must have no water; say that the buckets contain $0, b, a, c, 0$ liters of water in that order. We say that a state is \emph{winnable} if $a < 1$ and $b+c<1$. Clearly the Stepmother cannot win the move after a winnable state. We claim that if we are on a winnable state, then no matter what the Stepmother does, Cinderella can move such that we return to a winnable state. Say the Stepmother adds water so that the buckets now contain $w_1, w_2 + b, w_3 + a, w_4 + c, w_5$ liters of water. Now assume for the sake of contradiction that $w_2 + w_5 + b \ge 1$ and $w_1 + w_4 + c \ge 1$. This means that $w_1 + w_2 + w_2 + w_5 + b + c \ge 1$, but \[ w_1 + w_2 + w_2 + w_5 + b + c = 1 - w_3 + b + c < 2 - w_3 < 2,\]contradiction. Now we have $w_1, w_5 < 1$ (otherwise Cinderella can just empty the buckets with $w_1, w_5$), so either $w_1, w_2 + b, 0, 0, w_5$ or $w_1, 0, 0, w_4 + c, w_5$ is winnable, and we are done. $\blacksquare$

I claim that Cinderella can always prevent a bucket overflow. Let $a_1, a_2, a_3, a_4, a_5$ be the amount of water in each bucket. Note that when each round begins, two neighboring buckets, say $a_1$ and $a_2$, will be empty. Define a state as special if the following two properties hold: $a_4 < 1$ $a_3 + a_5 < 1$ Notice that if the state of the game is special, the Stepmother cannot win that round. The original game state is clearly a special state. We will show that given such a state, after one round Cinderella can always force a special state. On the Stepmother's move, let $w_i$ be the amount of water the Stepmother puts into $a_i$ for $1 \leq i \leq 5$. Now the amount of water in each bucket is $w_1, w_2, a_3+w_3, a_4+w_4,a_5+w_5$. Assume for the sake of contradiction that both $$a_3+w_3+w_1 \geq 1, a_5+w_5 + w_2 \geq 1.$$Summing the two inequalities gives $$(w_1+w_2+w_3+w_5) + (a_3+a_5) \geq 2,$$which is a clear contradiction since $w_1+w_2+w_3+w_5 \leq 1$ and $a_3+a_5 <1$ by hypothesis. Now Cinderella can remove the water from either buckets $a_3,a_4$ and $a_4,a_5$ to return the game to a winnable state.

I claim that Cinderella always wins. Define a $\textit{turn}$ as Stepmother move followed by a Cinderella move. After each turn, two buckets will be empty, say $\mathcal{B}_1$ and $\mathcal{B}_{2}$. I claim that among the remaining buckets $\mathcal{B}_3, \mathcal{B}_4, \mathcal{B}_5$, Cinderella can keep $\mathcal{B}_3 + \mathcal{B}_{4} < 1$ and $\mathcal{B}_5 < 1$. Then, in the stepmother's next move, she cannot force any bucket to overflow. To do so, we show that in a state of five buckets with water levels $(0, 0, x, y, z)$ satisfying $x + y < 1$ and $z < 1$ (suppose we call these states $\textbf{safe}$), that after a turn, Cinderella can force another safe state. Since the original state after one turn is clearly safe, we would be done through an inductive argument as Cinderella keeps forcing safe states. Now, from $(0, 0, x, y, z)$, say the Stepmother turns it to $(\epsilon_1, \epsilon_2, x + \epsilon_3, y + \epsilon_4, z + \epsilon_5)$ where $\epsilon_1 + \ldots + \epsilon_5 = 1$. At this point, I claim that Cinderella can either empty $(x + \epsilon_3, z + \epsilon_5)$ or $(y + \epsilon_4, z + \epsilon_5)$, one of which will yield a safe state. Suppose not, and that $(0, 0, \epsilon_1, \epsilon_2, y + \epsilon_4)$ and $(0, 0, \epsilon_1, \epsilon_2, x + \epsilon_3)$ are both not safe. FTSoC assume\[\epsilon_1 + (y + \epsilon_4) \geq 1 \text{ and } \epsilon_2 + (x + \epsilon_3) \geq 1.\]Summing yields $2 \leq x + y + (1 - \epsilon_5) < x + y + 1 < 2$, a contradiction. So one of $\epsilon_1 + (y + \epsilon_4), \epsilon_2 + (x + \epsilon_3)$ is $< 1$ and all $\epsilon_i$ are $< 1$ so Cinderella can force a safe state, as desired. $\blacksquare$

Suppose that, at some point in time, it is the stepmother's turn and the buckets (in order) contain $a_1,a_2,a_3,0,0$ liters of water. Suppose further that $a_1+a_3\le 1$ and $a_2\le 1.$ Now assume the stepmother adds $b_i$ liters to the $i$th bucket. Note that $$b_4\le b_1+b_2+b_3+b_4+b_5=1,$$$$b_5\le b_1+b_2+b_3+b_4+b_5=1,$$$$(a_3+b_3+b_5)+(a_1+b_1+b_4)\le (a_1+a_3)+(b_1+b_2+b_3+b_4+b_5)\le 2.$$Therefore, one of the following is true $$ a_3+b_3+b_5\le 1\text{ and } b_4\le 1 $$$$ a_1+b_1+b_4\le 1\text{ and } b_5\le 1 $$Thus, by emptying buckets $1$ and $2$ or buckets $2$ and $3,$ Cinderella can ensure that the condition we assumed originally is still satisfied. This implies that Cinderella can prevent any bucket from containing more than $1$ liter, so the stepmother can never make a bucket overflow.

No. For the stepmother to win, at some point a diagonal must have at least $1$ liter of water in total. Label the vertices by 1 through 5, and diagonal $i$ connects vertex $i$ with $i+2$. Let $a_i$ be the amount of water that diagonal $i$ currently contains, and $b_i$ be the amount of water that bucket $i$ currently contains. After Cinderella has emptied two buckets (assume WLOG they are on vertices 1 and 2), we have $a_1=b_3,a_2=b_4,a_3=b_3+b_5,a_4=b_4,a_5=b_5$. I will show that if currently, $a_3\le 1$ and $b_4\le 1$, then a symmetric condition can be held on her next turn, ensuring that $\max(a)$ does not reach $1$. Since $a_1+a_5=a_3\le 1$, after Stepmother's turn at least one of $a_1\le 1,a_5\le 1$ must hold. Suppose that $a_5\le 1$; then, clear buckets 3,4. Obviously bucket 2 can't have more than 1 liter of water either.

This took so freaking long it's really not even funny how dumb I am. I say that a lot but like this is just killing me. Here's a joke. Cindereeeeeeeeeeeeeeella the princess, *CAN HER FOOT FIT*, Cindereeeeeeeeeeeeeeella the princess, $\boxed{\text{Yes, she can!}}$ Let the buckets have $(b_1, b_2, b_3, b_4, b_5)$ amounts of water, where $b_i$ is adjacent to be $b_{i+1}$ and $b_6 = b_1$. The key claim is Claim: Cinderella can guarantee that after her move there are two adjacent empty buckets, two diagonally opposite buckets (not one of the above empty buckets) have volume sum of $< 1$, and the final bucket has volume $< 1$. Proof: Obviously this is true after Cinderella's first move. Now WLOG $a_4$, $a_5$ are empty, $a_1+a_3, a_2 < 1$. Then after the Stepmother's move, $a_1+a_2+a_3+a_4+a_5 < 2$. If $a_1+a_3<1$, then $a_2+a_4+a_5>1$. If $a_5>1$, then empty $a_4$ and $a_5$; elseways, empty $a_2$ and $a_4$. Otherwise, $a_2+a_4+a_5<1$ and empty $a_1$ and $a_3$. All of these operations result in the property being preserved. The above claim guarantees she cannot lose since $a_1+a_2+a_3+a_4+a_5$ is always less than $2$, and so the Stepmother can never overflow a bucket.

Call a bucket position promising if it is the stepmother's move, and there are two adjacent empty buckets, two buckets adjacent to those empty buckets who combined have at most one liter, which we will call the critical pair, and the final bucket alone, which we will call dangerous, with at most one liter. Note that the starting position is promising. After the stepmother has moved, the buckets other than the dangerous bucket will sum to at most two. Thus, Cinderella can find two pairs of non-adjacent buckets, one adjacent to the dangerous bucket and one not. These two pairs are disjoint, so one of them sums to at most one, and we will make this our new critical pair. The one adjacent to both the critical pair is less than one so it can become our new dangerous bucket, and Cinderella can dump the other two buckets. Thus, no matter what the stepmother does, Cinderella can keep a promising position promising.

apparently i have never done a 2009 C sl problem before this The answer is no. Represent a bucket state by a cyclic $5$-tuple, which is initially $(0,0,0,0,0)$. Call a state happy if it is a cyclic permutation of some $(0,a_1,a_2,a_3,0)$ where $a_1+a_3 \leq 1$ and $a_2 \leq 1$. I claim that if the bucket state is happy right before the Stepmother adds water, then Cinderella can guarantee that it is happy after the water is added and she empties the buckets. First, note that the Stepmother can never immediately overflow a bucket from a happy state. Therefore, if we go from $(0,a_1,a_2,a_3,0)$ to $(b_0,b_1,b_2,b_3,b_4)$, so $a_1+a_2+a_3+1=b_0+b_1+b_2+b_3+b_4$, then $b_0+b_1+b_3+b_4 \leq 2$, hence we either have $b_0+b_3 \leq 1$ or $b_1+b_4 \leq 1$; WLOG the former. Then clearly $b_4 \leq 1$, so Cinderella can empty the buckets $b_1,b_2$ (abusing notation) to once again obtain a happy configuration. Since the initial configuration is clearly happy, it follows that Cinderella can repeat this strategy indefinitely and the Stepmother cannot enforce a bucket overflow. $\blacksquare$ Remark: The capacity of $2$ can't be decreased: the Stepmother make a bucket reach $2-\varepsilon$ with the following strategy: she initially repeatedly distributes water in a way to make all of the buckets have an equal amount of water. Since this common amount of water clearly goes from $x$ to $\tfrac{3x+1}{5}$, it will eventually exceed $\tfrac{1}{2}-\varepsilon$. Then in the state $(0,\tfrac{1}{2}-\varepsilon,\tfrac{1}{2}-\varepsilon,\tfrac{1}{2}-\varepsilon,0)$, the Stepmother adds water to reach $(0,1-\varepsilon,\tfrac{1}{2}-\varepsilon,1-\varepsilon,0)$, and thus after Cinderella empties her choice of buckets there is some bucket with $1-\varepsilon$ liters which can be then made to contain $2-\varepsilon$ liters. Remark: One way to think about this problem is that if we are in a happy state at some point, we can help out the Stepmother by transforming it to $(0,a,1,1-a,0)$ for some appropriate choice of $a$, but despite this, the Stepmother still cannot win—the advantage of doing this is that the problem becomes easier to think about. A similar idea can be found in USA TST 2015/3. To be completely honest I don't know how I came up with this idea, but it was related to the fact that we can "basically" reach $(0,\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},0)$, which is better than the original state, so we might as well give it to the Stepmother (since it also doesn't look like she can win from here). A bit of playing around revealed that I could reach something like $(0,\tfrac{5}{8},\tfrac{5}{8},0,0)$, and it seemed like the Stepmother couldn't guarantee a win here either. Furthermore, it looked like the strategy for this case also worked for $(0,1,1,0,0)$, so there was some sort of idea of "bouncing" indefinitely between these two configurations. It doesn't work, since there are more possibilities, but after realizing that the $a_1+a_3 \leq 1$ condition was really what was important and also noticing that $(0,\tfrac{1}{2},1,\tfrac{1}{2},0)$ is more or less equivalent to $(0,\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},0)$ the definition of happy configurations was extremely motivated.

this actually took like 3 hours and i hate this problem even though it seems so easy; I'm literally livid. We claim that a configuration of safe buckets (a,b,c,0,0):a+c<1,b<1 (considered cyclically, and just make sure a,c non-adjacent) means the Stepmother never wins. In the next stage we get (d,e,f,g,h). Removing d and e, we get (0,0,f,g,h), while removing e and f we get (0,0,g,h,d); because d+f+g+h<2 (added at most 1 in this, and a+c was <1), either f+h<1 or d+g<1; either way, the condition is satisfied. Finally, note that g or h cannot be greater than 1, so the other condition is always kept. All that remains is that in the initial position after the Stepmother moves the position is safe.

We claim that the answer is yes. Rename the Wicked Stepmother to IAmTheHazard, or IATH for short. Cinderella adopts the following strategy. She first finds the maximum bucket, and the adjacent bucket next to the maximum bucket, then empties that pair. Claim: Draw a pentagram between the $5$ buckets. Cinderella wins if she can always guarantee that the sum of the liters in two buckets adjacent on the pentagram is less than $1$ after her turn. Proof. Work backwards. Cinderella can only lose if after IATH's turn, there are $2$ non-adjacent buckets both with at least $1$ liter each. Then, before IATH's turn, the result follows. $\blacksquare$ We prove the following things inductively on the number of turns. Claim: After her move, Cinderella can guarantee that all sums on the pentagram are less than $1$. Furthermore, she can guarantee that the sum is at less than $2$. Proof. Inductively, it follows that after IATH's turn there are less than $3$ liters of water. Label the buckets $b_1, b_2, b_3, b_4, b_5$. First suppose that WLOG $b_3 \ge 1$. It then follows that $(b_1 + b_5) + (b_2 + b_4) < 2$. Cinderella can choose either $b_2$ or $b_4$ such that the remaining diagonal has sum less than $1$. Now, suppose that $b_i < 1$. This implies that the remaining buckets are less than $1$ as well. Then, Cinderella first finds the diagonal of the pentagram with minimal sum. Since the sum of all $5$ diagonals is at most $2 \cdot \frac52 = 5$, some minimal pair has sum less than $1$. WLOG up to rotation suppose that $b_2 + b_4 < 1$ is the minimal pair. Then Cinderella empties $b_1$ and $b_5$. Then, $b_2 + b_3 + b_4 < 1 + 1 = 2$ as desired. $\blacksquare$

Procees is so fun. $\boxed{\text{The Stepmother will never be able to enforce a bucket overflow.}}$ Heuristics First we try and bound how ``close" the Stepmother may come to achieving her goal. Consider the following sequence of moves with $\overset{S}{\mapsto}$ denoting the Stepmother's moves, and $\overset{C}{\mapsto}$ denoting Cinderella's moves: \begin{align*} (0, 0, 0, 0, 0) \overset{S}\mapsto \left(0, \frac{1}{2}, 0, \frac{1}{2}, 0 \right) \overset{C}\mapsto \left(0, \frac{1}{2}, 0, 0, 0 \right) \overset{S}\mapsto \left(0, \frac{3}{4}, 0, \frac{3}{4}, 0 \right) \overset{C}\mapsto \left(0, \frac{3}{4}, 0, 0, 0 \right) \overset{S}\mapsto \left(0, \frac{7}{8}, 0, \frac{7}{8}, 0 \right) \end{align*}and so on. It is clear that the Stepmother will tend to some bucket having $1$ liter of water after Cinderella's move, at which moment she may place her liter of water into that bucket, forcing it to overflow. Thus the bound of $2$ is sharp, and in fact this suggests that the problem is engineered so that the Stepmother will never achieve her goal. Cinderella's Strategy Now working backwards we see that the Stepmother wishes to end at $(a, b, c, d, 2)$ or some permutation. However then, During Cinderella's turn she was faced with some $(a, b, c, d, e)$ where bucket $e$ and at least one non-adjacent bucket, say $c$ both had at least $1$ liter of water. Then she removed $c$, and one adjacent bucket to $c$, say $d$. Thus the Stepmother played on a board of the form $(a, b, 0, 0, e)$. Then it suffices to show that Cinderella can always prevent such a state from occurring. To prevent the Stepmother from reaching such a state, Cinderella has two possible strategies namely, Remove the bucket with greatest quantity of water, and the adjacent buckets with the most water. Remove the two adjacent buckets with greatest sum of water. It is easy to see the former strategy is unhelpful, as the Stepmother can continuously ``raise" three buckets together, and win. Thus we proceed with the second strategy. Proof of Victory Note that latter strategy guarantees that Cinderella removes at least $\frac{2}{5}$ of the water on the board. Then after Cinderella's $n$-th turn there is at most $w_n$ liters of water on the board, we have the recursion $w_{n+1} = \frac{3}{5}w_n + \frac{3}{5}$. Note that $w_1 = \frac{2}{5}$. Now we in fact claim that $w_n < 2$ for all $n$. Indeed note that for $w_n$ to be greater than $2$ we require, \begin{align*} w_n = \frac{3}{5}w_{n-1} + \frac{3}{5} > 2 \implies w_{n-1} > \frac{7}{3} > 2 \end{align*}which upon downwards induction implies $w_1 = \frac{2}{5} > 2$ which is clearly false. Then note that this implies at most $1$ bucket has more than $1$ liter of water. Now assume that after the Stepmothers $n$-th move, we have a board of the form $(a, b, c, d, e)$ with $e > 1$ and $a+b+c+d < 1$, which follows from $a+b+c+d+e < 2$. Then it suffices to show we can never have, \begin{align*} \max(a + b, b + c, c + d) > \max(a + e, d + e) \end{align*}However this is obvious so we are done. Namely, after any of Cinderella's moves, there is never a bucket left with at least $1$ liter of water. $\square$

this feels very dumb hopefully not a fakesolve also my writeup skills decided to leave my body so this is garbage Cinderella wins (yay! in hindsight of course she does, it's Cinderella) I actually had the following on my paper from working backwards: WS: All a+b>1 below C: Two nonadjacent a,b with a,b <= 1 and a+b>1 WS: Two nonadjacent >1 C: One value >1 which is essentially necessary and sufficient if the stepmother wins. So basically if Cinderella can always have any diagonal with sum at most $1$ she will be fine. There are basically just two cases to Cinderella's strategy: Remove all the $1$s if any exist. Also opportunistically choose the neighbor if we can (to describe later). This depends on having no two nonadjacent $1$s. Otherwise, take nonadjacent $a+b\le 1$. They share a neighbor $u$ and there are two other vertices $v,w$. Then Cinderella writes $v,w:=0$. This depends on having the sum be at most $\frac{5}{2}$. The second step depends on the total sum always being at most $\frac{5}{2}$. This isn't too hard to show: if the sum before a turn is $s$, then the first step will not increase the sum. Furthermore the second step will leave $(a+b)+u\le 2$ so we are fine. The first step is also not too bad. We'll just show that the wicked stepmother can't have two nonadjacent values more than $1$ after she moves. This is not hard, thankfully: consider the configuration before the wicked stepmother's last move (so before the previous turn). If the wicked stepmother did not create a value more than $1$, then Cinderella performs the second step. The resulting configuration is fine. If the wicked stepmother did create a value more than $1$, assume there was only one of these (for minimality). Then Cinderella removed it and also we can assume the permutation is $(1,a,d,c,b)$ or something. Then we would require $a+c>1$ and $b+d>1$ to fail which is impossible since the sum is then more than $3$. bALLS there's something wrong here, see #29 for the idea. okay i think with better writeup skill this would have been easier

Rename the Stepmother to Glinda and Cinderella to Elphaba. In this scenario, Glinda is attempting to kill Elphaba by spilling one of the buckets of water(water melts evil witches). We will show that Elphaba is capable of stopping Glinda from spilling the buckets and thus surviving. First we notice that if it is Glinda's turn and two non-adjacent buckets have more than one liter of water combined, then Glinda wins. Call a configuration of buckets Good if the buckets in order are $(b_1, b_2, b_3, b_4, b_5) = (0, 0, a, b, c)$ with $a + c < 1$ by the previous condition and $a, b, c < 1$, and call a configuration Wicked if it succeeds a Good turn. If $b > 1$ then Glinda can win directly. Our goal is to show that Glinda is not capable of winning on a Good turn, and Elphaba will always be able turn a Wicked config into a Good one. Say that only the condition $a + c < 1$ is violated on a Wicked turn. Then Elphaba can choose either $(b_3, b_4)$ or $(b_4, b_5)$, depending on whether $b_1 + b_3 > 1$ or $b_2 + b_5 > 1$ respectively(note that both can't occur due to $b_3 + b_5 < 1$). So WLOG the new scenario is $(x, y, 0, 0, c)$ and clearly $x < 1$ and $c + y < 1$ so it is Good. If $a, b, c < 1$ is violated at the end of the Good turn then nothing changes, so we are done(Glinda will never be able to kill Elphaba).

Call any segment with the endpoints on any two nonadjacent buckets a diagonal. Note that the Stepmother must have had a diagonal in which both buckets have $>1$ liters, or else Cinderella can just empty both of the buckets with $>1$ liters. Now, to get to this diagonal of both $>1$ liters, note that before the Stepmother added to these two buckets, their sum must have been $>1.$ Now, note that because this filling comes after an emptying, before the emptying all diagonals must have summed to $>1.$ Otherwise, Cinderella would have emptied in a way that made none of the diagonals sum to $>1.$ Note that after this emptying, there is only one diagonal left that sums to $>1.$ It is not hard to show that after each emptying, the remaining diagonal sums to $<1,$ so we're done$.\blacksquare$

Strange that c5 is easier than c1... The answer is no, the Stepmother cannot enforce a bucket overflow. Notice that the amount of water that is in each bucket is $(a,b,c,0,0)$ at the beginning of each round (where we rotated the buckets so that the $0$s are at the end). The main claim is that Cinderella can always keep $$a+c<1$$and $$b<1$$at the end of every round. Proving this will solve the problem, as on the next move the Stepmother cannot immediately win. To prove this, we use induction. The base case is obvious. Assume that this holds. After the Stepmother makes a move, the configuration looks like $$(a+x_1,b+x_2,c+x_3,x_4,x_5)$$for some $x_1,x_2,x_3,x_4,x_5$ that sum to $1$. If the Stepmother decides that one of $x_1,x_2,\dots,x_5$ equals $1$, then Cinderella can empty that bucket and an adjacent one (it doesn't matter which). Otherwise, Cinderella will remove either $(a+x_1,b+x_2)$ or $(b+x_2,c+x_3)$. This will leave the configuration $$(0,0,c+x_3,x_4,x_5)$$or $$(a+x_1,0,0,x_4,x_5)$$Since $x_4<0$ and $x_5<0$, it remains to show that one of the following holds: $$c+x_3+x_5<1$$$$a+x_1+x_4<1.$$However, their sum is less than $2$, which finishes the inductive step.