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The Olympic symbol has 5 rings that intersect at 8 points:

enter image description here

What is the most number of intersection points can you achieve by moving the rings?

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    $\begingroup$ Why "Olympic"? It is just about five rings, right? $\endgroup$ – Florian F Oct 2 '20 at 13:23
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There is an easy upper bound:

Any two distinct circles can intersect in at most two points. If every pair of circles intersects, then you get $\binom{5}{2}*2=20$ intersections.

This upper bound can be reached in this way:

enter image description here
Two distinct circles intersect if and only if there is a region that lies inside both circles. To let all circles intersect, arrange them so that they enclose a central region, and such that none of the intersections coincide. This obviously generalises to any number of circles.

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    $\begingroup$ Well done! That's a very neat solution. $\endgroup$ – Dmitry Kamenetsky Oct 2 '20 at 5:38
  • $\begingroup$ Quite possibly this also maximises the total number of produced regions, but I need to think about it. Perhaps this could be a new puzzle. $\endgroup$ – Dmitry Kamenetsky Oct 2 '20 at 12:26
  • $\begingroup$ @DmitryKamenetsky Maximising intersections also maximises the number of regions. Adding the $n$-th circle adds a number of intersections with the previous circles ($2(n-1)$ but that is not important). The circle consists of the same number of arcs between those intersections, and each arc splits a previous region in two. So the $n$-th circle adds the same number of intersections as regions, at least for $n>1$. The first circle does not add any intersections, but does leave you with two regions. The total number of regions is therefore equal to the number of intersections, plus $2$. $\endgroup$ – Jaap Scherphuis Oct 2 '20 at 13:06
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    $\begingroup$ That art fails to take into account the proper proportion of the rings. Just to demonstrate that it works with the ring proportion, here's an alternative image...imgur.com/mSDFq9h $\endgroup$ – Strawberry Oct 2 '20 at 16:10
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As is shown in my drawing, the number of intersections of the five rings is 18.rings

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    $\begingroup$ very nice! I reckon you can get more intersections by bringing the top left and the bottom right circles closer together. $\endgroup$ – Dmitry Kamenetsky Oct 2 '20 at 2:16
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Can this be solved with the "Handshake" formula?

Let's say, 5 rings can "shake hands" with 4 other rings. So there are $\frac{5 * (5-1)}{2}=10$ possible "handshakes". Rings always "shake hands" with other rings at two points though, so the answer is twice that amount: $20$

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