Problem 834

Let the total amount of cheese be 50 (pounds). Players $A$ (first) and $B$ are involved. It is clear that in the division, both would benefit from a greedy algorithm (each time the largest piece is chosen). In this case, $B$ will get the second and fourth largest pieces by weight.

Let's denote their total weight by $R$.

We will show how $A$ can guarantee himself 30 pounds.

First, $A$ divides the cheese into pieces weighing 30 and 20.

Let the weight of the smaller piece cut by $B$ later be $u$. If $u \leq 5$, then $A$ can obtain the set $L=\{30-u, 20-u, u, u\}$, where $0 < u \leq 5$. If $5 < u \leq 10$, then $A$ can obtain the set $M=\{30-u, 20-u, u, u\}$, where $10 > u > 5$.

In both cases, the sum $R$ of the pieces in the 2nd and 4th positions is 20. After the cut, $R$ can only increase if at least one of these pieces is replaced by a larger piece, moving to a position with a higher number. However, the piece $20-d$ cannot move "to the right": $a$ cannot be cut into two parts larger than $20-d$ (in both cases $20-d > a / 2$). Moreover, in case $L$, both parts $d$ cannot be moved to the 5th position. However, in case $M$, one of the two larger parts can be cut into two pieces larger than $d$. Let's consider two sub-cases.

1) The part $a=20$ is cut into pieces $s, 20-s$, where $d < s < 10$ (since $d < 10$ and $20-s > d$). Then $R < 20$.

Now we will show how $B$ can guarantee himself 20 pounds (i.e., ensure $R \geq 20$).

Let the pieces after the first cut be $x$ and $y \geq x$. We will call pieces weighing at least 20 "large".

If $x \leq 10$, player $B$ cuts $y$ in half, and if $x \geq 20$, he cuts a 20-pound piece from $y$. In both cases, two large pieces $a \geq b \geq 20$ are formed. If $A$ does not cut them, then $B$ will not cut them either, and will be able to take one of them. If $A$ cuts one of them, then $B$ will cut the other into proportional parts. These four parts form two pairs $a_{1} \geq b_{1}$ and $a_{2} \geq b_{2}$, where $b_{1} + b_{2} = b$. The smaller pieces from the pairs are guaranteed to $B$, so $R \geq b \geq 20$.

If $10 \leq x \leq 20$, $B$ gets the set $\{20, x, y-20\}$, where all pieces are at least 10.

If $A$ then cuts the 20-pound piece, $B$ (cutting 10 from one of the "old pieces") gets the set $\{10+a \geq 10+b \geq 10 \geq 10-b \geq 10-a\}$, in which $R=20$.

Otherwise, before $B$, the set is $\{20, a, b, c\}$, where $a \geq b \geq c$. By cutting $b$ from 20, he transitions to the set $\{a, b, b, 20-b, c\}$. Since $a+b+c=30$, then $b+c \leq 20, a+b \geq 20$, so $c \leq 20-b \leq a$ and $R=b+(20-b)=20$.

## Answer

The first is 0.6, the second is -0.4 of the total piece.