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Using 20 Circles, what is the maximum number of intersecting point that can be obtained?

For example, if there were 3 circles, the answer would be $6$ as shown below:

enter image description here

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    $\begingroup$ 20 non-concentric circles? $\endgroup$ – nikki Aug 15 '18 at 20:00
  • $\begingroup$ I'd assume so (there's a facetious answer otherwise). Is it also a safe assumption that they all be uniformly sized? $\endgroup$ – El-Guest Aug 15 '18 at 20:01
  • $\begingroup$ @nikki :) then there would be infinite intersections. $\endgroup$ – Oray Aug 15 '18 at 20:01
  • $\begingroup$ @El-Guest not necessarily, you can use any size you would like $\endgroup$ – Oray Aug 15 '18 at 20:01
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    $\begingroup$ @Oray interesting, thanks for the clarification :) $\endgroup$ – El-Guest Aug 15 '18 at 20:02
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I think the best way to go about this is to see that

any two circles will have maximum two intersection points between them (think of a Venn diagram for this).

I'm not able to program/draw this out at the moment (especially not with 20 circles),

but I feel that the maximum can be obtained when each pair of circles intersects twice.
Now, how many pairs of circles are there? Since order doesn't matter, using choose notation we can say that there are $C(20,2)$, ie. 20 choose 2 pairs of circles. This evaluates to $C(20,2) = \frac{20!}{18!2!} = \frac{20(19)}{2} = 190$. Since there are 190 pairs of circles and 2 intersection points per pair, the maximum number of intersection points should be $190 \times 2 = 380$ intersection points.

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I'd say $2 \times \frac{n!}{(n-2)! 2!}$ which works out to

380

for n=20.

Every circle can intersect every other circle at exactly two points. From 20 circles, there are 190 distinct ways to choose a pair.

To show that such a construction is possible,

Have each circle be the same size. Then choose a point somewhere, and draw the circles so that the point is inside every one of them. Two same sized circles will intersect if (and only if) there is some point that's inside both circles.

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  • $\begingroup$ That's a pretty complex way of writing $n^2-n$ $\endgroup$ – Kruga Aug 16 '18 at 8:13
  • $\begingroup$ @Kruga it sure is! $\endgroup$ – Bass Aug 16 '18 at 9:51
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I think the answer is

190 380
For 1 circle: 0
For 2: 2
For 3: 6
For n: f(n-1) + (n-1)*2

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  • $\begingroup$ I think your formula makes sense and looks correct; what I can't figure out is how you got that particular answer with it. :-) $\endgroup$ – Bass Aug 15 '18 at 20:48
  • $\begingroup$ Oops, looks like I made a mistake in my calculation. I mentioned I was busy when I was answering. :P I was trying to do 2.1 - 2 + 2.2 - 2 + 2.3 -2 + ... + 2.20 - 2 = 2.[(1+2+3+ ... +20) - 20] = 380 Looks like I mentally only calculated 190, and then forgot to multiply the whole thing by 2 :P $\endgroup$ – nikki Aug 16 '18 at 3:52

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