Problem 10.7. Olya drew $N$ different lines on the plane, any two of which intersect. It turned out that among any 15 lines, there are always two lines such that the angle between them is $60^{\circ}$. What is the largest $N$ for which this is possible?
Answer: 42.

Solution. First, we will show that 42 such lines exist. Draw three lines so that the angle between any two of them is $60^{\circ}$, and take 14 such triplets at different angles:

![](https://cdn.mathpix.com/cropped/2024_05_06_d4986512fdb0330805a0g-37.jpg?height=412&width=422&top_left_y=87&top_left_x=516)

It is clear that when choosing any 15 lines, at least two of them will be in the same triplet, and the angle between them will be $60^{\circ}$. For example, 42 lines passing through one point, with the angle between any two adjacent lines being $\frac{180^{\circ}}{42}$, fit the described construction.

Now we will prove that there cannot be 43 or more lines that meet the condition. Suppose the contrary and consider such 43 lines. Choose an arbitrary point $O$ on the plane; instead of each line, we will consider a line parallel to the given one and passing through $O$ (obviously, the angles between the lines do not change with such parallel translations; since all the lines initially intersected, they cannot coincide after the translation).

Draw an auxiliary line through point $O$ that does not coincide with any of the drawn lines and does not form a $60^{\circ}$ angle with any of them; we will call this line the "horizontal." Choose the "upper" half-plane relative to the horizontal and divide it into three sectors of $60^{\circ}$ each:

![](https://cdn.mathpix.com/cropped/2024_05_06_d4986512fdb0330805a0g-37.jpg?height=218&width=420&top_left_y=1157&top_left_x=517)

It is clear that each of the 43 lines passes through exactly one sector; therefore, by the pigeonhole principle, one of the sectors will contain at least 15 lines. However, all angles between the lines in one sector will be strictly less than $60^{\circ}$, which leads to a contradiction.