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:
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Sign up to join this communityUsing 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:
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.
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.
I think the answer is
190380
For 1 circle: 0
For 2: 2
For 3: 6
For n: f(n-1) + (n-1)*2