8.5. Kolya drew 10 segments and marked all their intersection points in red. Counting the red points, he noticed the following property: on each segment, there are three red points. a) Provide an example of the arrangement of 10 segments with this property. b) What is the maximum number of red points for 10 segments with this property?
Answer: b) 15. Solution. a) See the example in the figure. As another example, you can take two copies of the right part of the figure from problem 7.5. b) The example given in the figure shows that 15 red points can be obtained. Let's prove that this is the maximum possible number. Number all 10 segments and write down the numbers of the segments to which each red point belongs near each red point. Since a red point belongs to at least two segments, the total number of recorded numbers is at least $2 N$, where $N$ is the number of red points (the numbers 

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are recorded more than once). But according to the condition, each segment number is recorded at three red points, i.e., a total of exactly 30 numbers are recorded (each of the 10 segment numbers is recorded three times). Therefore, $2 N \leq 30$, i.e., $N \leq 15$. Note. From the proof, it follows that the maximum $N=15$ is achieved only in the case where each red point belongs to exactly two segments (in the second example from part a), where there are points of intersection of three segments, $N=14$).