I came across this from the Longlists page, so as it appears, it is the problem 71 of the IMO Longlist 1983. Let 3 dimansional space is partitioned into 3 disjoint subsets $M_1, M_2, M_3$. We will call the pair $(x,r),\, x \in \mathbb{R}^3,\ r \in \mathbb{R}^+$, a good pair, if for every $r_1,\, 0<r_1 \leq r$, there exists $x_1 \in \mathbb{R}^3$, such that $|x-x_1| = r_1$ and $x_1$ belongs to the same subset as $x$. Let denote with $S(x,R)$ the sphere with center $x$ and radii $R$, i.e. $S(x,R) = \{y\in \mathbb{R}^3 |\, |y-x|=R \}$. Lemma 1. If every point of the sphere $S(x,R)$ belongs to one of the two sets e.g. $M_1, M_2$, then there exists a good pair $(y, \sqrt{2} R)$. Proof. Let consider some point $y \in S(x,R)$. WLOG let assume that $y \in M_1$. If $(y, \sqrt{2}R)$ is a good pair we are done. If not, there exists $r \leq \sqrt{2}R$ such that (1) $S(y,r) \cap S(x,R) \subset M_2$. Let $x_1 \in S(y,r) \cap S(x,R)$. From (1) we get that $(x_1, r_1),\,r_1=\sqrt{2}r$ is a good pair. If $\sqrt{2}R \leq r_1$ we are done. If not, let assume that $(x_1, \sqrt{2}R)$ is not a good pair. Than there exists $r,\, r_1 < r \leq \sqrt{2}R$, such that: (2) $S(x_1,r) \cap S(x,R) \subset M_1$. let $x_2 \in S(x_1,r) \cap S(x,R)$, then $(x_2,r_2)$ is a good pair, $r_2=\sqrt{2}r > \sqrt{2}r_1$ . If $\sqrt{2}R \leq r_2$ we are done, if not, the same argument gives us a good pair $(x_3, r_3), r_3 > \sqrt{2}r_2$ ... Thus, we get a chain of good pairs $(x_1, r_1), (x_2,r_2),\cdots, r_{i+1} > \sqrt{2}r_i$ and in the end, for some $k,\, r_k \geq \sqrt{2}R$ which means $(x_k, \sqrt{2}R)$ is a good pair. $\square$. Let consider some $x\in \mathbb{R}^3$ and assume WLOG $x \in M_3$. If for every $R \in \mathbb{R}^+$, there exists $y \in \mathbb{R}^3,\, |y-x|=R$, we are done. If not, there exists $R \in \mathbb{R}^+, S(x,R) \subset M_1\cup M_2$. Using Lemma 1, we get a good pair $(x_1, r_1),\, r_1=\sqrt{2}R$. WLOG let $x_1\in M_1$. If assume that for every $r,\, (x_1,r)$ is a good pair, we are done. If not, $\exists r > r_1, \, S(x_1,r) \subset M_2\cup M_3$. Using Lemma 2 again, we get a good pair $(x_2, r_2),\, r_2 > \sqrt{2}r_1$. In this way we get an infinite chain of good pairs: (3) $(x_i, r_i), r_{i+1} > \sqrt{2}r_i,\,i=1,2,\cdots$ are good pairs Hence one of the sets $M_1, M_2, M_3$, contains infinite many $x_i$. Let WLOG it be $M_1$. (4) $(x_{i_s}, r_{i_s} ), \, x_{i_s}\in M_1,\, r_{i_s} \to \infty,\, s=1,2,\cdots$ are good pairs From (4) it follows that $M_1$ contains all possible distances.

Call the three sets of points $A$, $B$ and $C$. Lemma. If there is a sphere $\mathcal{S}$ with radius $r$ that contains points from only two of the sets (say $A$ and $B$), then either $A$ or $B$ contains every distance in $(0;r]$. Proof. Suppose that neither $A$ nor $B$ contains every distance in $(0;r]$, then for some $d\le r$, there are no two points in $B$ on $\mathcal{S}$ with distance $d$. If $\mathcal{S}$ contains only points from $A$, we are done. Otherwise, we can find a point $P\in\mathcal{S}\cap B$, and then the circle $\{X\in \mathcal{S}:\, |XP|=d\}$ only has points from $A$. The diameter of this circle is $2d\sin\left(\cos^{-1}\frac{d}{2r}\right) \ge d$ because $\cos^{-1}\frac{d}{2r}\ge \cos^{-1}\frac12=60^\circ>30^\circ$ ($d\le r$). Hence we have found in $A$ every distance from $(0;d]$. Now if for some $d^*\in(d;r]$, we cannot find two points in $A$ with distance $d^*$, then a circle $\{X\in\mathcal{S}:\, |XP|=d^*\}$ must be in $B$, while containing two points with distance $d$, contradiction. This means that every distance in $(0;r]$ occurs in $A$, proving our claim. $\blacksquare$ Suppose for a contradiction that there is an $a,b,c\in \mathbb{R}^+$ such that there are no two points from $A,B,C$ with distance $a,b,c$, respectively. If one of $A,B,C$ is empty, reassign finitely many points from one of the infinite sets to the sets which are empty. Every sphere centered around a point in, say $C$, with radius $c$, will contain only points from $A$ or $B$. By the Lemma, either $A$ or $B$ will contain every distance in $(0;c]$, hence either $a>c$ or $b>c$, or equivalently, $c<\max\{a,b,c\}$. We can say the same for $a$ and $b$ as well, yielding that neither of $a,b,c$ can be the largest of them, which is absurd. This completes the proof.