Bogdanov I.I.

Each segment with endpoints at the vertices of a regular 100-gon is painted red if there is an even number of vertices between its endpoints, and blue otherwise (in particular, all sides of the 100-gon are red). Numbers are placed at the vertices such that the sum of their squares equals 1, and on the segments - the products of the numbers at the endpoints. Then, the sum of the numbers on the red segments is subtracted from the sum of the numbers on the blue segments. What is the largest number that could result?

#
Let the numbers at the vertices be denoted as $x_{1}, x_{2}, \ldots, x_{100}$. Then the sum of the numbers on the red segments (denoted as $R$) is the sum of all pairwise products of numbers at positions with different parity:

$$
\begin{gathered}
R=x_{1} x_{2}+x_{1} x_{4}+\ldots+x_{1} x_{100}+x_{3} x_{2}+x_{3} x_{4}+\ldots+x_{3} x_{100}+\ldots \\
\ldots+x_{99} x_{100}=x_{1}\left(x_{2}+x_{4}+\ldots+x_{100}\right)+ \\
+x_{3}\left(x_{2}+x_{4}+\ldots+x_{100}\right)+\ldots+x_{99}\left(x_{2}+x_{4}+\ldots+x_{100}\right)= \\
=\left(x_{1}+x_{3}+\ldots+x_{99}\right)\left(x_{2}+x_{4}+\ldots+x_{100}\right)=P Q
\end{gathered}
$$

where $P=x_{1}+x_{3}+\ldots+x_{99}$, and $Q=x_{2}+x_{4}+\ldots+x_{100}$. The sum on the blue segments (denoted as $B$) is the sum of all pairwise products of numbers at positions with the same parity: $B=x_{1} x_{3}+\ldots+x_{97} x_{99}+x_{2} x_{4}+\ldots+$ $x_{98} x_{100}$. Considering that $x_{1}^{2}+x_{2}^{2}+\ldots+x_{100}^{2}=1$, we can write that
$2 B+1=x_{1}^{2}+x_{3}^{2}+\ldots+x_{99}^{2}+2 x_{1} x_{3}+\ldots+2 x_{97} x_{99}+$

$+x_{2}^{2}+x_{4}^{2}+\ldots+x_{100}^{2}+2 x_{2} x_{4}+\ldots+2 x_{98} x_{100}=$

$=\left(x_{1}+x_{3}+\ldots+x_{99}\right)^{2}+\left(x_{2}+x_{4}+\ldots+x_{100}\right)^{2}=P^{2}+Q^{2}$.

The desired difference $R-B=P Q-\frac{P^{2}+Q^{2}-1}{2}=\frac{1-(P-Q)^{2}}{2} \leqslant \frac{1}{2}$.

This estimate is achieved, for example, when $x_{1}=x_{2}=\ldots=x_{100}=1 / 10$ or when

$x_{1}=x_{2}=\frac{1}{\sqrt{2}}, x_{3}=x_{4}=\ldots=x_{100}=0$.

## Answer

$1 / 2$.