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Every point of three-dimensional space is colored red, green or blue. Prove that one of the colors attains all distances, meaning that any positive real number represents the distance between two points of this color.

Let's suppose that no color attains all distances, meaning that exist real numbers $r \ge g \ge b$ such that no two red points have distance $r$, the same for g and green and b and blue.

Then consider a sphere of radius $r$ centered at a red point. Clearly its surface has green and blue points only.

Taking a look at the solution, they affirm that:

1) "since $g,b \le r$ the surface of the sphere must contain both green and blue points"

It doesn't seem clear to me and I don't know how to prove it. I think I must show that the points on the surface of the sphere of radius $r$ attains all distances between $0$ and $2r$ (and it would clearly imply that these points attains all distances between $0$ and $g$ or $b$).

2) "Let $M$ be a green point in the surface. There exist two points $P$ and $Q$ on the sphere such that $MP = MQ = g$ and $PQ = b$"

I also can't see how they can affirm that such points exist. With that two facts the problem is done.

I know that I should suppose the contrary to prove those things but even so I'm getting nowhere.

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  • $\begingroup$ By $MP$ I mean the lenght of the line segment from $M$ to $P$. $\endgroup$ Jul 6, 2017 at 18:25
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    $\begingroup$ For 1. you could suppose that the surface is all green, then take a green point $G$ on it and consider a sphere of radius $g$ and center $G$. $\endgroup$
    – krirkrirk
    Jul 6, 2017 at 18:27

2 Answers 2

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I think you've got the ordering of the radii for each color mixed up.

First, if only one color is used, it is trivial to show that it attains all distances.

If exactly two colors are used, say green and blue, but neither attains all distances, then there are distances $g \ge b$ such that no two green points are $g$ apart, and no two blue points are $b$ apart. Pick a green point $P_g$; then the sphere $S_g$ around $P_g$ with radius $g$ will be entirely blue. Now pick a blue point $P_b$ on $S_g$. The sphere $S_b$ around $P_b$ with radius $b$ necessarily intersects $S_g$ since $b \le g$. But this contradicts the assumption, since $S_g$ is entirely blue. Indeed, we can see this argument holds even in the plane, using circles instead of spheres.

Therefore, three colors must be used. The idea outlined above is extended. If $r \ge g \ge b$, then we start by choosing a red point $P_r$. The sphere $S_r$ about $P_r$ with radius $r$ is entirely colored with blue and green. (It cannot be monochromatic since this would collapse into the previous 2-color argument.) So now pick a green point $P_g$, and consider the sphere $S_g$ around $P_g$ with radius $g \le r$, which intersects $S_r$ in a circle $C_g$ that must be entirely blue. Now pick a point $P_b$ on $C_g$, and draw the sphere $S_b$ around $P_b$, which intersects circle $C_g$ in at least one point since $b \le g$. It follows that the distance between these two points is $b$, but you have no more colors to use, as $C_g$ is entirely blue, contradicting the assumption that there exist distances $r \ge g \ge b$ such that no two red points are $r$ apart, no two green points are $g$ apart, and no two points are $b$ apart. Therefore, at least one of the colors attains all distances.

There is an important subtlety in the above argument: the circle $C_g$ has a radius $\rho$ that is strictly less than $g$, being the intersection of a sphere with radius $r$ and a sphere of radius $g$. But it cannot be smaller than $g \sqrt{3}/2$, which is attained only when $g = r$. In other words, we can't make $\rho$ so small that when picking $P_b$ on $C_g$, $b$ exceeds $2\rho$ and we get away with $C_g$ being all blue. This can't happen since $2\rho \ge g \sqrt{3} > g \ge b$.

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  • $\begingroup$ Great answer; I'm guessing this is still true with $n$ colors ? $\endgroup$
    – krirkrirk
    Jul 6, 2017 at 18:37
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    $\begingroup$ @krirkrirk It may seem intuitively true (and it may actually hold for arbitrary many dimensions), but the $n$-ball argument we are employing here is not so trivial, as evidenced by the "subtlety" I mentioned in the last paragraph. $\endgroup$
    – heropup
    Jul 6, 2017 at 18:45
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I, too, had difficulty understanding the solution in the text. If I may, I'll explain the text's solution since @heropup explained an alternative approach.

For 1), I think the reasoning is clear from @heropup's answer.

For 2), it should be clear that such a green point $M_g$ is guaranteed to exist. We can then choose any color points $P'$ and $Q'$ along the same circle, $C_s$, of the sphere such that $M_gP'=M_gQ'=g$. We can then adjust our choice of points by moving along the circle $C_s$ until $P'Q'=b$.

In the pictures that follow, $C_s$ is outlined in red, along with the lines $M_gP',M_gQ'$ that are to be varied. The Mathematica code is also enclosed in the large likelihood someone else can code better than I!

enter image description here

enter image description here

Addendum:

g1 = ParametricPlot3D[{Cos[u] Sin[v], Cos[u] Cos[v], 
      Sin[u]}, {u, -\[Pi], \[Pi]}, {v, -\[Pi]/2, \[Pi]/2}, Mesh -> None, 
      PlotStyle -> Opacity[.40, Blue], PlotPoints -> 80, 
      MaxRecursion -> 4, Exclusions -> {Cos[u] Cos[v] == .7}, 
      ExclusionsStyle -> ({Directive[Opacity[1], Thick, Red]})]
 g2 = Graphics3D[{Thick, Red, Line[{{0, 1, 0}, {0, 0.7, 0.7}}]}]
 g3 = Graphics3D[{Thick, Red, Line[{{0, 1, 0}, {0.35, 0.7, 0.606}}]}]
 Show[g1, g2, g3]
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