This is equivalent to asking whether for some scalar D there exists point a s.t. d(a,x)<D/2, d(a,y)<D/2, etc., for all the points that the "Euro" covers.

If d(a,x)<D/2, then if d(a,y)<D/2, then d(x,y)<D. (And vice-versa, of course).

If no point a exists for {w,x,y,z}, then that implies there is at least one of d(w,x), d(w,y)... etc. such that d(-,-)>D. If that is true, then the point a doesn't exist for at least one of {w, x, y}, {w, y, z}, {x, y, z}, {w, x, z}, because collectively those four triplets require d(-,-)<D for all six of the pairs.

Take X = 1 and the points of affixes :
A(0,sqrt(3))
B(-1,0)
C(1,0)
D(0,1-sqrt(3))

You have d(A,B) = d(A,C) = d(A,D)= d(B,C) = 2 and d(B,D) = d(C,D) = sqrt(5-2sqrt(3)) < 2, but you don't have a point a where d(a,-)<1 for each point in {w,x,y,z}. The smaller circle that cover every 4 points has a radius of 2/sqrt(3) > 1 (the 4 points are concyclics)

lets reverse the problem:
put a coin on each of th 4 points:
if the 4 coin overlap, you can center a coin on the overlapping region and it will cover everything.
now lets assume that you have 3 overlapping circle of the same radius R:C1,C2,C3
try to draw a 4th circle o that intersect all three circles but NOT the common region: impossible.

now to prove that mathematically..
lets assume that C1,C2, C3 have a non empty intersection , lets take a point A
lets assume that C1,C2, C4 have a non empty intersection , lets take a point B in it
lets assume that C1,C3, C4 have a non empty intersection , lets take a point C
lets assume that C2,C3, C4 have a non empty intersection , lets take a point D in it

my bet is that the barycenter of A,B,C,D will be in all 4 circles.ie it will be a valid choice.


i guess it is related to the convexity of those intersections: if A and B are in an intesection, the whole [A,B] segment is, and the barycenter of 4 point in on the semgent between all 2 by 2 barycenter.

i need to work a bit (real work that pays the bill) i ll come back

Given 4 points in a plane one of the following is true:
1. One of the points is in or at the triangle formed by the remaining 3 points.
2. The 4 points form a quadrilateral with intersecting diagonals.

please hide your solution for a few days

A convex set is a set S such as if A and B are in S, all points in the segment [A,B] are in S.
-disks are convex/ if K and L are such that d(A,K)<R and d(A,L)<R; d(A,P)<R for each point in [K,L]
that is because on the line K,L you can write:
f(x)=d(x,A)=sqrt(r²+x²)
(here r is the min distance between the (KL) line and A)
the function as a minimum at x=0 but no local maximum;
so for each x,x1,x2 such as x1<x <x2
f(x)< max(f(x1),f(x2));
==>D(A,P)<R if p between K and L

- an intersection of convex sets is a convex (obvious)

i was still thinking about a convexity argument and here is what got:
You have your 4 points A,B,C,D.
suppose you can cover A,B,C with a circle of radius R centered in K
... A,B,D with a circle of radius R centered in L
... A,C,D with a circle of radius R centered in M
B,C,D with a circle of radius R centered in N
1)
The convexity argument is as follow:
as d(A,K) <R AND (A,L)<R , D(A,P)<R for each point between K and L; same thing for d(B,P)
And we have equivalent properties for all 6 segment made of 2 points ou of K,L;M,N

2)now:
lets assume that the segment [K,L] and [M,N] intersects
by the property above the intersection I is such that d(I,A)<R, d(I,B)<R (because I is on [K,L]), and d(I,C)<R and d(I,D)<R (becasue I is on [M,N])
a centre of radius R centerd at I will cover A,B,C and D
same thing if any 2 other segment intersect of course
3)
now what if there is no such intersection?
fo example K=(0,-1);L=(0,+1),M=(2,0) N=( 1,0) : no pair of segment intersects
in that case i think we are still ok:
it means that one of the 4 points is inside the triangle made by the 3 others (i am not sure how to demonstrate that but it seems obvious if you try to draw).
Edit: that is actually exactly cyclops hint so this is a good track
in the example N is inside the triancle K, L, M :
K; L and M are all inside the circle of radius R centered in A, so N is inside the circle too (convexity again):

so N is a good candidate:
d(N,A)<R and by definition of N: d(N,B)<R,d(N,C)<R d(N,D)<R
a centre of radius R centered at N will cover A,B,C and D
there is probably something more beautiful