A square piece of paper was punctured at 1965 points. From the puncture points and the vertices of the square, no three lie on the same straight line. Then, several straight, non-intersecting cuts were made, each starting and ending only at the punctured points or the vertices of the square. It turned out that the square was cut into triangles, none of which contained any punctures inside. How many cuts were made and how many triangles were formed?
Answer: 5896 cuts, 3932 triangles. Let's solve the problem in the general case when the square is pierced at $n$ points. Let the number of resulting triangles be $x$. On the one hand, the sum of the angles of all these triangles is $x \cdot 180^{\circ}$. On the other hand, it is equal to $360^{\circ} + n \cdot 360^{\circ}$ (the sum of the angles of the square and the sum of the angles $360^{\circ}$ at $n$ vertices). Therefore, $x = 2n + 2$. Let the number of cuts made be $y$. On the one hand, the number of sides of the resulting triangles is $3x$. On the other hand, it is equal to $4 + 2y$ (each side of the square is counted once, and each cut is counted twice).

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