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Sphere Intersections (Posted on 2015-11-11) Difficulty: 3 of 5

  
SPHERE INTERSECTIONS IN En (n≥1)

   K = H(P1,a1) ∩ H(P2,a2) ∩ ... ∩ H(Pn,an)

               where

        ai ≥ ai+1 > 0        for i = 1, 2, ... , n-1.

               and

        P1En  and

        PiH(P1,a1) ∩ H(P2,a2) ∩ ... ∩ H(Pi-1,ai-1)
                                   for i = 2, 3, ... , n.

DEFINITIONS

     • En denotes Euclidean n-space. The set of n-tuples of
             real numbers.

     • If P ∈ En and r is a real number greater than zero, then
                      H(P,r) = { Q ∈ En | δ(P,Q) = r }.
       A "sphere" with center P and radius r.

     • δ(P,Q) denotes the distance between points P and Q in En.

QUESTION

     Is K empty ?
  

No Solution Yet Submitted by Bractals    
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analytical solution | Comment 1 of 4
I am going to assume that the question is if there are values of n for which K is empty for all possible selections of the a's and P's.  I assume this because for n>=3 if you make the radius of the second "sphere" larger than twice the radius of the first then they have no intersection and thus K is empty.

if n=1 then we have just a single "sphere" namely two points on the number line for K.

If n=2, then the first and second circles intersect in one point if a2=2*a1, 2 points if a2<2*a1 and zero points if a2>2*a1.  Thus K is not always empty.

Now assume n>=3.  Imagine we start building the set K but starting with K initially being the entire first sphere.  We then pick a point in K and a radius and then redefine K as the intersection of the current K with the new "sphere".  Now each time we do this K remains a "sphere" just simply being of lower dimension (with the empty set having zero dimension), thus by the time we add the last circle we need K to be at least 1 dimensional.  Now at each step, we can minimize the amount of dimension reduction by keeping the radius constant.  Thus if we say that a1=a2=...=an=r then each redefinition of K creates a "sphere" of 1 less dimension and reduces the radius by a factor of sqrt(3)/2.  However, if at any point the reduced radius of K is less than r/2 then the next iteration of K will be empty.

Thus we need r*(sqrt(3)/2)^(n-2)>=r/2
(sqrt(3)/2)^(n-2)>=1/2
which translates to n<=6

thus for n>=7 K is empty for all possible selections

  Posted by Daniel on 2015-11-11 14:20:46
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