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Given are two, disjoint circles $k_1, k_2$ of equal diameter $d$ in space. One can then choose two points $P_1, P_2$, one on each circle such that the distance between them is greater than $d$.

How can this problem be solved "topologically"?

My thoughts thus far:

One can continuously map $[0,1]$ onto the circle so that $0$ and $1$ are mapped to onto same point. Thus, there is a function $g : [0,1] \times [0,1] \rightarrow k_1 \times k_2 $

Now I am looking at $S = \{ r \in \mathbb{R} | r = d(a,b) \text{ with } a \in k_1, b \in k_2 \}$ with $d(a,b)$ measuring the euclidean distance between the two points $a$ and $b$.

I can now look at a map between $[0,1] \times [0,1] \rightarrow S $ and I should be able to reshape the figure of my square $[0,1] \times [0,1]$, where the edges are glued together, now to a sphere.

As much I feel proud making this connection, as much I feel stuck and stupid. I don't think this gets one anywhere.

How can one salvage this? My 2nd idea revolved around looking at pairs of points on each circle. Each point on a circle can be uniquely identified by its antipodal point. So now I would repeat above's process, however I would define $S' = \{ r \in \mathbb{R} | r = max(d(a,b), d(a', b')) \text{ with } a, a' \in k_1, b, b' \in k_2 \}$ and $a, a'$ and $b, b'$ being antipodal points respectively.

While this all seems nice, I don't see how put everything together. For example, I haven't included the diameter of $d$ of the circles.

Does someone see a better, more constructive way to solve this problem?

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  • $\begingroup$ This is likely not "topological", but draw a line through the centers of the $2$ circles. Where this line crosses each circle on the opposite side of the other circle's center defines $2$ points which are greater than a distance $d$ from each other. If you put into "topological" terms, it might give you an appropriate answer. $\endgroup$ – John Omielan Feb 26 at 20:54
  • $\begingroup$ This only works, if both circles lie a the plane^^ $\endgroup$ – Imago Feb 26 at 20:56
  • $\begingroup$ You are right, it's more complicated than that in $3$ or more dimensions. Good luck with proving the statement in those conditions. $\endgroup$ – John Omielan Feb 26 at 20:58
  • $\begingroup$ What happens if the circles are on an ellipsoid where distance is measured on the surface? $\endgroup$ – William Elliot Feb 26 at 22:06
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A very geometric solution:

In four or more dimensions, it's false. The circles $(\cos\theta,\sin\theta,0,0)$ and $(0,0,\cos\phi,\sin\phi)$ in $\mathbb{R}^4$ each have diameter $2$, and every point in the first circle is at distance exactly $\sqrt{2}$ from every point in the second.

In three dimensions, the circles can be represented as $k_i=\{O_i+v : v\in V_i\text{ and }\|v\|=\frac12d\}$ where $O_i$ is the center and $V_i$ is a vector subspace of $\mathbb{R}^3$ of dimension $2$. Since the sum of the dimensions of $V_1$ and $V_2$ is greater than $3$, they have a nontrivial intersection. Let $w$ be a vector of length $\frac12d$ in this intersection.
Now, the four points $A=O_1+w,B=O_1-w,C=O_2-w,D=O_2+w$ lie on the circles and form a parallelogram. Since the circles are disjoint, they are four different points in $\mathbb{R}^3$. We claim that at least one of the diagonal lengths $\|A-C\|$ and $\|B-D\|$ is strictly greater than $d=\|A-B\|=\|C-D\|$.
Why? Apply the "parallelogram law": $$\|A-C\|^2+\|B-D\|^2 = 2\|A-B\|^2+2\|B-C\|^2 > 2\|A-B\|^2=2d^2$$ With the sum strictly greater than $2d^2$, one of $\|A-C\|^2$ and $\|B-D\|^2$ must be strictly greater than $d^2$. Take the square root, and we're done.

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