# Variant 4.

1) Let's draw an arbitrary line $y=k-b$, intersecting the parabola at points $A$ and $B$. Then the abscissas of the endpoints of the chord $A$ and $B$ must satisfy the conditions: $k x - b = x^2$, i.e., $x^2 - k x + b = 0$, $x_A = \frac{k - \sqrt{k^2 - 4b}}{2}$, $x_B = \frac{k + \sqrt{k^2 - 4b}}{2}$. Then the abscissa $x_C$ of the midpoint of segment $AB$ is $x_C = \frac{x_A + x_B}{2} = \frac{k}{2}$ and depends only on the slope $k$, but not on $b$. Therefore, if two chords $A_1 B_1$ and $A_2 B_2$ are parallel, then their midpoints $C_1$ and $C_2$ have the same abscissas and the line $C_1 C_2$ is parallel to the $OY$ axis. A chord $A_0 B_0$ perpendicular to this line is parallel to the $OX$ axis and, due to symmetry, its midpoint lies on the $OY$ axis. Therefore, the line through point $C_0$ parallel to $C_1 C_2$ is the $OY$ axis. It intersects the parabola at the vertex $O$, and the line passing through point $O$ perpendicular to $OY$ is the $OX$ axis.
2) Let $t = \frac{1}{x}$; then $x = \frac{1}{t}$. We obtain the equality: $f\left(\frac{1}{t}\right) + \frac{5}{t} f(t) = \frac{3}{t^3}$. Since this equality holds for any $t \neq 0$, it holds for $t = x$. Therefore, we have: $f\left(\frac{1}{x}\right) + \frac{5}{x} f(x) = \frac{3}{x^3}$. Then we get the system of equalities:
\[
\left\{
\begin{array}{l}
f\left(\frac{1}{x}\right) + \frac{5}{x} f(x) = \frac{3}{x^3} \\
f(x) + 5x f\left(\frac{1}{x}\right) = 3x^3
\end{array}
\right.
\Longleftrightarrow
\left\{
\begin{array}{l}
5x f\left(\frac{1}{x}\right) + 25 f(x) = \frac{15}{x^2} \\
5x f\left(\frac{1}{x}\right) + f(x) = 3x^3
\end{array}
\right.
\]
Subtracting the second equation from the first, we get:
\[
24 f(x) = \frac{15}{x^2} - 3x^3 \Rightarrow f(x) = \frac{5}{8x^2} - \frac{x^3}{8}
\]
Answer: $f(x)=\frac{5}{8 x^{2}}-\frac{x^{3}}{8}$.

3) Let

$$
2 x+y-3 z-3=m, x-2 y-4 z-1=n, y+7 z-3 x+7=k
$$

Then we find that $m+n+k=3$. Since by condition $x, y, z$ are integers, then $m, n, k$ are also integers and $m, n, k>0$. Therefore, we get that $m=n=k=1$ and then we have the inequality: $0>z^{2}-9 z+18$. Solving it, we find that $32 \Longleftarrow t0$, i.e., the equality is impossible.

Answer: $a=2$.

Criteria for determining the awardees of the Olympiad

Based on the results of checking each task of the "Sails of Hope" schoolchildren's Olympiad in mathematics, one of the following marks was given (listed in descending order):

«+» the problem is solved completely;

«+/-» the problem is solved with minor flaws that do not affect the overall solution (for example, an arithmetic error at the end of a correct solution is made);

«-/+» the problem is not solved, but there are substantial advances (for example, the problem is solved for a substantial particular case);

«-» the problem is not solved

«0» the participant did not start solving the problem.

When summarizing, only the number of problems solved in general is taken into account, that is, the problems for which the mark «+» or «+/-» is given.

By the decision of the organizing committee of the Olympiad in 2017-2018 academic year, the following criteria for determining the winners and prizewinners of the "Sails of Hope" Olympiad in mathematics were established

[^0]:    q.e.d.