Problem 6. (8 points) Two friends are sharing a piece of cheese, and of course, each wants to take as much as possible. At first, they take turns, with each choosing one of the existing pieces and cutting it into two. The cutting ends when the pieces total five. Then the friends take turns again, each picking and taking one piece. The one who cuts first also picks first and gets one more piece. What is the optimal strategy for each, and how much cheese does it guarantee them, regardless of the other's moves?
Solution: Let there be a total of 50 kg of cheese. Players I and II play, and let $R_{1}$ and $R_{2}$ be the results of I and II. Everywhere below, we will assume that $x \leq y$ and $a \leq$ $b \leq c$. It is clear that after cutting, each player takes the largest existing piece. Then, when cutting pieces

$$
N(a, b)=\{10,10-a, 10-b, 10+a, 10+b\}
$$

the players will get $R_{1}=30$ and $R_{2}=20$. Furthermore, when cutting pieces $\left\{a, b_{1} \geq b_{2}, c_{1} \geq c_{2}\right\}$, we will have $R_{1} \geq a+b_{2}+c_{2}$.

We will show how I can achieve $R_{1} \geq 30$. With his first move, he brings the game to $\{30,20\}$. Depending on the second player's move, with his second move I can bring the game to one of the two positions:

$$
K(a)=\{20,10+a, 10,10-a\}, \quad \text { where } \quad a<10
$$

or

$$
L(a)=\{30,10+a, 10,10-a\}, \quad \text { where } \quad a<10
$$

If $R_{1}-R_{2}>20$, then II cuts to $\{a, b, b\}$, where $a<10$, and by repeating the move for equal pieces, achieves $R_{1}-R_{2} \leq a<10$. In the opposite case, II cuts to $\{p, q, 20\}$, where $p$ and $q \geq 10$. If in response I cuts the 20 kg piece, then II brings the game to $N(a, b)$. Otherwise, before II is the position $\{a, b, c, 20\}$, where $a+b+c=30$. He brings the game to $\{a, b, c, b, 20-b\}$. It is clear that $a+b \leq 20$ and $b+c \geq 20$, so $a \leq 20-b \leq c$ and thus gets $R_{2}=b+(20-b)=20$.