We are given $5$ lines and two circles in a plane. What is the maximum number of possible intersection points among these seven figures ?

My work on the problem: I've considered special cases like what is the maximum number of points of intersection between lines and circles ,between only lines and last between circles.

P.s: Please provide detailed answer so i can follow best.Thanks in advance

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    $\begingroup$ If you think about it, the circles are easy to collocate once you've maxed the intersection betwen the lines. By the way, the answer is 32, try to obtain and prove it $\endgroup$ – Exodd Sep 28 '15 at 20:13

When you only have 5 lines, you can get at most $\binom{5}{2}=10$ intersections. This is the number of distinct pairs of lines among those five. You can draw it on paper if you don't know the binomial coefficient symbol yet. Each of 2 circles can intersect each of 5 lines at 2 points. There can also be 2 circle-circle intersections.

Line-line intersections: $10$

Line-circle intersections: $2\cdot 2 \cdot 5$

Circle-circle intersections: $2$

Add it up to get the answer.

  • $\begingroup$ Pardon, but where is the intuition behind $\binom{5}{2}=10$ using geometrical objects ? i mean i used that method dealing with boxes,people etc.. but in abstract when dealing with geometrical objects what is that combinatorics method doing ? $\endgroup$ – Nameless Sep 28 '15 at 20:28
  • $\begingroup$ Each pair of lines create an intersection (in our maximal case, each pair gives a distinct intersection). Therefore, if we count the number of distinct pairs of lines, we will know the number of intersections. The binomial coefficient $\binom{a}{b}$ means the number of distinct b-subsets of an a-set (I mean a subset with b elements when I say b-subset, etc). The number of 2-element-subsets of a 5-element-set is the same as the number of pairs of lines, when we have 5 lines at our disposal. $\endgroup$ – I want to make games Sep 28 '15 at 20:42
  • $\begingroup$ why do we count distinct pair of lines....why not distinct triad or anything higher than that? $\endgroup$ – Hydrous Caperilla Apr 23 '18 at 13:30
  • $\begingroup$ The base knowledge we're basing the solution on is that a pair of lines can intersect zero times or once (and that we can set up $5$ or any other number of lines such that each pair does in fact intersect, and each of the intersections is different). And so, we think: 'when we have 5 lines, we have 10 distinct pairs of them, and each pair results in an intersection'. $\endgroup$ – I want to make games Apr 24 '18 at 15:37

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