11.10. Given a natural number $n \geqslant 2$. Petya and Vasya play the following game. Petya chooses $2 n$ (not necessarily distinct) non-negative numbers $x_{1}, x_{2}, \ldots, x_{2 n}$, the sum of which is 1. Vasya arranges these numbers in a circle in some order of his choosing. After that, he calculates the products of pairs of adjacent numbers and writes on the board the largest of all $2 n$ obtained products. Petya wants the number on the board to be as large as possible, while Vasya wants it to be as small as possible. What number will be on the board with the correct play?

(A. Khryabrov)
Answer: $\frac{1}{8(n-1)}$.

Solution: If Petya chooses the numbers $0, \frac{1}{2}, \frac{1}{4(n-1)}, \frac{1}{4(n-1)}, \ldots, \frac{1}{4(n-1)}$, then no matter how Vasya arranges these numbers, the number $\frac{1}{2}$ will be paired with the number $\frac{1}{4(n-1)}$. Therefore, one of the products will be $\frac{1}{8(n-1)}$, and the others will not exceed it. Thus, the number $\frac{1}{8(n-1)}$ will appear on the board.

We will show how Vasya can obtain a number on the board that is not greater than $\frac{1}{8(n-1)}$ for any numbers. Let's enumerate the numbers in descending order: $x_{1} \geqslant x_{2} \geqslant \ldots \geqslant x_{2 n}$. Place the number $x_{1}$ at some point on the circle, and then place the numbers $x_{2}, x_{3}, \ldots, x_{n}$ clockwise from $x_{1}$ through empty spaces. Now place the number $x_{2 n}$ between $x_{1}$ and $x_{n}$; then place the numbers $x_{2 n-1}, x_{2 n-2}, \ldots, x_{n+1}$ clockwise from $x_{2 n}$ in the empty spaces. The products of the pairs of adjacent numbers will be: $x_{n} x_{2 n}$,

$$
x_{1} x_{2 n}, x_{2} x_{2 n-1}, x_{3} x_{2 n-2}, \ldots, x_{k} x_{2 n-k+1}, \ldots, x_{n} x_{n+1}
$$

and

$$
x_{1} x_{2 n-1}, x_{2} x_{2 n-2}, x_{3} x_{2 n-3}, \ldots, x_{k} x_{2 n-k}, \ldots, x_{n-1} x_{n+1}
$$

Since $x_{k} x_{2 n-k+1} \leqslant x_{k} x_{2 n-k}$, the largest product can only be in the second row.

We will show that $a=x_{k} x_{2 n-k} \leqslant \frac{1}{8(n-1)}$ for $k \leqslant n-1$. Indeed, from the inequalities $x_{k} \leqslant x_{k-1} \leqslant \ldots \leqslant x_{1}$, it follows that $k x_{k} \leqslant x_{1}+x_{2}+\ldots+x_{k}$, so

$$
k a=k x_{k} \cdot x_{2 n-k} \leqslant\left(x_{1}+x_{2}+\ldots+x_{k}\right) x_{2 n-k}
$$

Similarly, from the inequalities

$$
x_{2 n-k} \leqslant x_{2 n-k-1} \leqslant x_{2 n-k-2} \leqslant \ldots \leqslant x_{k+1}
$$

it follows that

$(2 n-2 k) x_{2 n-k} \leqslant x_{2 n-k}+x_{2 n-k-1}+\ldots+x_{k+1} \leqslant$

$$
\leqslant x_{k+1}+x_{k+2}+\ldots+x_{2 n}=1-x_{1}-x_{2}-\ldots-x_{k}
$$

Therefore,

$$
\begin{aligned}
& 2 k(n-k) a \leqslant \\
& \quad \leqslant\left(x_{1}+x_{2}+\ldots+x_{k}\right)\left(1-x_{1}-x_{2}-\ldots-x_{k}\right)=x(1-x)
\end{aligned}
$$

where $x=x_{1}+x_{2}+\ldots+x_{k}$. Since by the inequality of means for two numbers $x(1-x) \leqslant\left(\frac{x+(1-x)}{2}\right)^{2}=\frac{1}{4}$, we get the inequality $x_{k} x_{n-2 k}=a \leqslant \frac{1}{8 k(n-k)}$. It remains to show that $k(n-k) \geqslant n-1$ for $k \leqslant n-1$. But the last inequality can be rewritten as $(k-1)(n-k-1) \geqslant 0$, and both brackets in the last formula are non-negative.

Remark. Only the answer - 0 points.

Only the example of Petya's numbers, for which $A=\frac{1}{8(n-1)}-$ 1 point.

Only the proof that Vasya can always obtain a number $A$ not less than $\frac{1}{8(n-1)}-5$ points.

If in the work it is proven that in Vasya's arrangement there will always be a number not less than $x_{k} x_{100-k}$ for $k=1,2, \ldots, n-1,-$ 1 point is given (this point can be combined with the point for the example).