3-ча 4. A thousand points are vertices of a convex thousand-sided polygon, inside which there are another five hundred points such that no three of the five hundred lie on the same line. The thousand-sided polygon is cut into triangles in such a way that all the 1500 points are vertices of the triangles and these triangles have no other vertices. How many triangles will result from such a cutting?
Solve 4. Answer: 1998. The sum of all angles of the obtained triangles is equal to the sum of the angles of a 1000-gon and 500 angles of \(360^{\circ}\), corresponding to 500 internal points. Therefore, the sum of the angles of the triangles is \(998 \cdot 180^{\circ} + 500 \cdot 360^{\circ} = (998 + 2 \cdot 500) 180^{\circ}\). Hence, the number of obtained triangles is \(998 + 2 \cdot 500 = 1998\).

3 - part 5. Solve the system

\[
\left\{\begin{array}{l}
x_{1} + 2 x_{2} + 2 x_{3} + 2 x_{4} + 2 x_{5} = 1, \\
x_{1} + 3 x_{2} + 4 x_{3} + 4 x_{4} + 4 x_{5} = 2, \\
x_{1} + 3 x_{2} + 5 x_{3} + 6 x_{4} + 6 x_{5} = 3, \\
x_{1} + 3 x_{2} + 5 x_{3} + 7 x_{4} + 8 x_{5} = 4, \\
x_{1} + 3 x_{2} + 5 x_{3} + 7 x_{4} + 9 x_{5} = 5
\end{array}\right.
\]

Solve 5. Answer: \(x_{1} = x_{3} = x_{5} = 1, x_{2} = x_{4} = -1\).

First, write down the first equation, then the second equation minus the first, then the third equation minus the second, and so on:

\[
\left\{\begin{aligned}
x_{1} + 2 x_{2} + 2 x_{3} + 2 x_{4} + 2 x_{5} & = 1, \\
x_{2} + 2 x_{3} + 2 x_{4} + 2 x_{5} & = 1, \\
x_{3} + 2 x_{4} + 2 x_{5} & = 1, \\
x_{4} + 2 x_{5} & = 1, \\
x_{5} & = 1
\end{aligned}\right.
\]

Now we can sequentially find \(x_{5}, x_{4}, x_{3}, x_{2}, x_{1}\).

\section*{8th Grade}