# Most intersections with Olympic rings

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

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

• Why "Olympic"? It is just about five rings, right? Oct 2, 2020 at 13:23

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:

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.

• Well done! That's a very neat solution. Oct 2, 2020 at 5:38
• 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. Oct 2, 2020 at 12:26
• @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 $2$. Oct 2, 2020 at 13:06
• 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 Oct 2, 2020 at 16:10

As is shown in my drawing, the number of intersections of the five rings is 18.

• very nice! I reckon you can get more intersections by bringing the top left and the bottom right circles closer together. Oct 2, 2020 at 2:16

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: $$20$$