How many intersections can we can get from $6$ lines and $3$ circles?
I'm kind of a beginner in combinatorics and only know counting and multiplication rule so if you use any other method it's a request do elaborate please, thank you!
How many intersections can we can get from $6$ lines and $3$ circles?
I'm kind of a beginner in combinatorics and only know counting and multiplication rule so if you use any other method it's a request do elaborate please, thank you!
Let's suppose no lines are equal and no circles are equal.
Each circle has maximal 2 intersections with a line or with a circle. Each line can have maximal 1 intersection with another line and 2 with a circle. Thus the maximal amount of intersections of 3 circles with one line is $3 \cdot 2=6$. We have $6$ lines so we get $36$ intersections of 3 circles with $6$ lines. Now we also have to count the maximal intersections of circles with each other. Each pair of circles can have maximally 2 intersections. When you have 3 circles, you have 3 different pairs (order of the pairs doesn't matter). This means we get $3 \cdot 2=6$ intersections of circles with circles. At last we calculate the intersections of lines with lines. Each pair of lines intersect in maximally 1 point. We get $5+4+3+2+1=15$ pairs of lines, hence $15$ intersections of lines with each other.
Summing this all together we get $36+6+15=57$.
To show that it is geometrically possible I also drew a configuration of 3 circles and 6 lines maximally intersecting:
