Problem 4. Three lines intersect to form 12 angles, and $n$ of them turn out to be equal. What is the maximum possible value of $n$?
Answer: 6.

Solution: Three lines limit a certain triangle. If this triangle is equilateral, then out of twelve angles, six are $60^{\circ}$, and the other six are $120^{\circ}$.

Can any external angle of the triangle be equal to its internal angle? It is equal to the sum of the non-adjacent internal angles, so it is greater than each of the non-adjacent ones. Therefore, it can only be equal to the adjacent one. Then each of them is $90^{\circ}$, and there are 4 such angles (they share a common vertex). There are no other right angles in this construction. But then the other external angles are not equal to the internal ones. Equal can only be the internal ones (and the vertical ones to them) or only the external ones (and the vertical ones to them). Then there will be no more than 4 equal angles of each type.

Criteria: 3 points for an example with six equal angles, 3 points for comparing external and internal angles. 1 point for an answer without justification.