geometry problem ( that nagged me )

[perceval]

maybe i got it:

hint:

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)

proposed solution/

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

[cyclops]

maybe i got it:

Congratulations. You are too clever for my problems. I doubt there is a more elegant solution. It is the solution of the book.
For the missing link see:

http://en.wikipedia.org/wiki/Helly%27s_theorem

Edit: oops, corrected two terrible grammatical errors. Shame on me.