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Spacy colors (Posted on 2006-08-25) Difficulty: 4 of 5
Every point in 3D-space is colored either red, green or blue. Let R (resp. G and B) be the set of distances between red (resp. green and blue) points. Prove that at least one of R, G, or B, consists of all the non-negative real numbers.

See The Solution Submitted by JLo    
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Solution Solution (Some details required) | Comment 5 of 10 |
 
NOTATION:
  R* = { P in E^3 | P is colored red }
  G* = { P in E^3 | P is colored green }
  B* = { P in E^3 | P is colored blue }
  R = { d(P,Q) | P,Q in R* } 
  G = { d(P,Q) | P,Q in G* }
  B = { d(P,Q) | P,Q in B* }
  X << Y denotes X is a subset of Y
  X ^ Y denotes the intersection of sets X and Y
  X v Y denotes the union of sets Y and Y
  X - Y  denotes the set difference of sets X and Y
  sphere(X,y) denotes the sphere with center X
              and radius y
PROOF (Others can flesh out the details):
  Assume r in [0,inf) - R,
         g in [0,inf) - G, and
         b in [0,inf) - B
  WOLOG let 0 < b <= g <= r.
  Pick a point L in R*. Then sphere(L,r) << G* v B*.
  Case A: sphere(L,r) << G*
          Pick a point M in sphere(L,r). Then
          sphere(M,g) ^ sphere(L,r) << G*
          This is a contradiction since it is a 
          space circle containing colorless points.

  Case B: sphere(L,r) << B*
          Pick a point M in sphere(L,r). Then
          sphere(M,b) ^ sphere(L,r) << B*
          This is a contradiction since it is a 
          space circle containing colorless points.
  Case C: (sphere(L,r) ^ B*) and (sphere(L,r) ^ G*)
          are both non-empty
          
          Pick a point M in sphere(L,r) ^ G*. Then
          sphere(M,g) ^ sphere(L,r) << B*. Pick a
          point N in sphere(M,g) ^ sphere(L,r). Then
          sphere(N,b) ^ sphere(M,g) ^ sphere(L,r) << B*
          This is a contradiction since it is a set 
          containing two colorless points.
  Therefore, at least one of the sets R, G, or B must
  equal [0,inf).
 Oops - Back slashes don't work to well.
To clean up the details:
Case 1: 0 < b <= g <= r
 This is the case discribed above.
Case 2: 0 <= g <= r
 This case is described in Case A above.
Case 3: 0 <= 0 <= r
 No points colored green or blue. Clearly,
 R = [0,inf)
Case 4: 0 <= 0 <= 0
 All points are colorless. Contradicts problem statement.

Edited on August 25, 2006, 9:41 pm

Edited on August 25, 2006, 9:49 pm

Edited on August 27, 2006, 11:54 am
  Posted by Bractals on 2006-08-25 17:23:07

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