Five circles and common points

[BMAD]

Given five circles where any group of 4 have a common point.  Does there exist a common point for all 5?

[k-man]

Yes.

Suppose that there isn't a common point shared by all 5 circles. Let A be the point that is shared by circles 1-4, but not by 5. Let B be the point that is shared by circles 1-3 and 5, but not 4. Let C be the point that is shared by circles 1,2,4 and 5, but not 3. Points A, B and C must all be distinct points, but all three are shared by circles 1 and 2. Since two circles can only have at most two points in common, we reached a contradiction.

 

Now, for disks, the answer is the same, although proof is different, but the OP isn't asking about the disks. 

[BMAD]

feel free to extend to disks

[k-man]

The same is true for disks.

 

Let's start the proof by establishing that if any of the five disks is entirely contained by another disk, then all 5 disks must have a common point. The inner disk must have a common point with 3 others (excluding the outer disk), but since every point in the inner disk is also in the outer disk, then that point is common to all 5. So, this excludes the possibility that any of the disks are internally tangent.

 

Now, let's intersect 2 disks in such a way that neither is contained by another. This creates 3 areas

- area exclusive to disk 1 (let's call this area A1)

- area exclusive to disk 2 (A2)

- a common area A12 (intersection of disks 1 and 2). 

 

Similarly to the logic above, we can show that if area A12 is entirely covered by another disk (either 3,4 or 5) then all 5 disks must have a common point. This means that

- area A12 cannot be a single point (e.g. circles 1 and 2 cannot be tangent, but must have 2 points of intersection)

- area A12 is always convex and its "sides" are convex arcs that separate it from areas A1 and A2

- no other disk can cover area A12 completely.

 

What actual areas are created with addition of disk 3 depends on the layout, but after adding disk 3 we need to have the following areas in the layout - A12, A13, A23 and A123. These areas must exist and be larger than a single point to ensure that we can establish common points for every group of 4 disks. This means that circle 3 cannot be tangent to either circle 1 or 2 and must intersect both circles. Similarly to above, it can be shown that area A123 is always convex and borders areas A12, A13 and A23 along its convex sides. It can also be proven (although I will skip this for brevity) that it's not possible to place a disk that would intersect areas A12, A13 and A23, but not cover any part of area A123. 

 

Again, area A123 must not be entirely contained by another disk, so disk 4 must be added to establish the following areas - A123, A124, A134, A234 and A1234. Similarly to above, area A1234 is always convex and surrounded by areas A123, A124, A134 and A234 and it's impossible to place a disk that would intersect with all of A123, A124, A134 and A234, but not cover any part of A1234.