3. To determine 11 real numbers such that each of them is equal to the square of the sum of the remaining 10.
Solution. Let the numbers be $x_{1}, x_{2}, \ldots, x_{11}$ and let it hold that:

$$
x_{1} \leq x_{2} \leq \ldots \leq x_{11}
$$

Let the total sum of the numbers be $s$. According to the conditions of the problem, we have:

$$
x_{i}=\left(s-x_{i}\right)^{2}, i=1,2,3, \ldots, 11
$$

from which, according to (1), we get:

$$
\left(s-x_{1}\right)^{2} \leq\left(s-x_{2}\right)^{2} \leq \ldots \leq\left(s-x_{11}\right)^{2}
$$

Since the desired numbers are squares, they are non-negative, so from (1) it follows that $s-x_{1} \leq s-x_{2} \leq \ldots . \leq s-x_{11}$, that is,

$$
\left(s-x_{1}\right)^{2} \geq\left(s-x_{2}\right)^{2} \geq \ldots \geq\left(s-x_{11}\right)^{2}
$$

From (3) and (4) we get $\left(s-x_{1}\right)^{2}=\left(s-x_{2}\right)^{2}=\ldots .=\left(s-x_{11}\right)^{2}$, and due to non-negativity, we have $x_{1}=x_{2}=\ldots=x_{11}$. Then for each of them, $x=(10 x)^{2}$ holds, from which we get that all the desired numbers are 0 or all are equal to $\frac{1}{100}$.