Protasov V.Yu.

Find all positive numbers $x_{1}, x_{2}, \ldots, x_{10}$, satisfying for all $k=1,2, \ldots, 10$ the condition ( $x_{1}+\ldots+$ $\left.x_{k}\right)\left(x_{k}+\ldots+x_{10}\right)=1$.
Let $x_{1}+x_{2}+\ldots+x_{10}=s$. By setting $s$, we can express $x_{1}+x_{2}+\ldots+x_{k}=s_{k}$ for all $k=1,2, \ldots, 10$:

$$
s_{1}=x_{1}=\frac{1}{s}, s_{2}=x_{1}+x_{2}=\frac{1}{s-s_{1}}, s_{3}=\frac{1}{s-s_{2}}, \ldots, s_{10}=\frac{1}{s-s_{9}}=s
$$

From this, it is clear that the equations can only be satisfied for one value of $s$. Indeed, if for some $s$ they are true, increasing $s$ will decrease $s_{1}, s_{2}, \ldots, s_{10}$ (and vice versa); in this case, the last equation will be violated.

Since all unknowns can be expressed in terms of $s$, the system has no more than one solution.

Construct an isosceles triangle $A_{0} O A_{10}$ with lateral sides $A_{0} O=O A_{10}=1$ and angles of $15^{\circ}$ at the base; mark points $A_{1}, A_{2}, \ldots, A_{9}$ on the base $A_{0} A_{10}$ such that segments $A_{0} A_{1}, A_{1} A_{2}, \ldots, A_{9} A_{10}$ are seen from the vertex $O$ under equal angles $\alpha$ (the angle $A_{0} O A_{10}$ is $150^{\circ}=10 \alpha$). We will prove that the lengths of these segments $x_{1}, x_{2}, \ldots, x_{10}$ satisfy our system.

Draw another ray $O P$ through point $O$, parallel to $A_{0} A_{10}$. Triangles $A_{0} O A_{k}$ and $A_{10} A_{k-1} O$ are obviously similar, with sides $A_{0} O=1$ and

$A_{0} A_{k}=x_{1}+\ldots+x_{k}$ of the first corresponding to sides $A_{k-1} A_{10}=x_{k}+\ldots+x_{10}$ and $O A_{10}=1$ of the second. From this, we obtain the required equation

$\left(x_{1}+\ldots+x_{k}\right)\left(x_{k}+\ldots+x_{10}\right)=1$

Using the Law of Sines, from triangles $O A_{k-1} A_{k}$ and $O A_{0} A_{k}$, we get $\frac{x_{k}}{\sin \alpha}=\frac{O A_{k}}{\sin k \alpha}, \frac{O A_{k}}{\sin \alpha}=$

$\frac{1}{\sin (k+1) \alpha}$, from which

$x_{k}=\frac{\sin ^{2} \alpha}{\sin k \alpha \sin (k+1) \alpha}(k=1,2, \ldots, 10)$

The values of $\sin k \alpha$ for $k=1,2, \ldots, 6$ are respectively

$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{3}-1}{2 \sqrt{2}}, \frac{1}{2}, \frac{\sqrt{2}}{2}, \frac{\sqrt{3}}{2}, \frac{\sqrt{2+\sqrt{3}}}{2}=\frac{\sqrt{3}+1}{2 \sqrt{2}}, 1$. Substituting these values into the general formula, we get the answer.

## Answer

$x_{1}=x_{10}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{2}^{2}}, x_{2}=x_{9}=\sqrt{2}-\frac{\sqrt{6}}{2}, x_{3}=x_{8}=\frac{2 \sqrt{6}-3 \sqrt{2}}{6}$,

$x_{4}=x_{6}=\frac{9 \sqrt{2}-5 \sqrt{6}}{6}, x_{5}=x_{6}=\frac{3 \sqrt{6}-5 \sqrt{2}}{4}$.

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