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I have a simple claim in spherical geometry that has come out of my research into the so-called "perspective 3-point (pose) problem." Here it is:

Fix three (distinct) great circles on the unit sphere, say a red one, a green one and a blue one. Also, on a movable great circle, put three pairs of antipodal dots (six dots), one pair being red, one pair being green, and one pair being blue. Call this movable circle (with attached dots) "the ring." Going around the ring (in either direction), assume that the length of each arc connecting consecutive dots never exceeds $\pi/2$. I claim that it is always possible to move the ring, rotating it on the sphere, so as to place the red dots on the red circle, the green dots on the green circle, and the blue dots on the blue circle.

I would be interested in any reference related to this claim, any simple proof of it (I believe I have a not-so-simple proof), a counterexample (if I'm wrong), or any helpful thoughts at all. Note that this claim can be recast in terms of the projective plane instead of the unit sphere.

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    $\begingroup$ If you replace the circles by their axes, the claim becomes: For any non-zero vectors $a,b,c$ and any $p,q\in[0,1]$, there is a non-zero vector $d$ with $$((a\times d)\cdot(b\times d))^2=p |a\times d|^2 |b\times d|^2$$ $$((a\times d)\cdot(c\times d))^2=q |a\times d|^2 |c\times d|^2$$ These equations are then quartic and homogeneous in the coefficients of $d$. $\endgroup$
    – user44143
    Commented Nov 3, 2022 at 18:58
  • $\begingroup$ That is a nice way to look at it, Matt, and squarely in the "why didn't I think of that?" department. However, I'm not sure about this capturing my constraints. Let me write R for the unit vector from the origin to one of my red points, so basically my R is your (a x d) / |a x d|. Similar for G and B. So, you say that (R.G)^2 = p and (R.B)^2 = q for known values of p and q, which is correct. However, identifying my six points with the six vectors, R, -R, G, -G, B, -B, and going around the ring, I require that the dot product of two consecutive vectors/points be positive. $\endgroup$
    – Michael Rieck
    Commented Nov 3, 2022 at 22:51
  • $\begingroup$ I think there is a similar version faithful to those sign considerations: for any non-zero vectors $a,b,c$, and for any $p,q,r$ in $[0,1]$ with $\arccos \sqrt p + \arccos \sqrt q + \arccos \sqrt r =\pi$, there is a non-zero vector $d$ with the two equations above and a third parallel equation in $b,c,d,r$. $\endgroup$
    – user44143
    Commented Nov 4, 2022 at 1:24
  • $\begingroup$ Okay, thanks. That would do it. If you have a reference for that, or even vaguely remember where to look, I'd appreciate that information. Cheers. $\endgroup$
    – Michael Rieck
    Commented Nov 4, 2022 at 8:26
  • $\begingroup$ I just meant that this would be a more faithful statement — I don’t know a nice analysis of the quartics. $\endgroup$
    – user44143
    Commented Nov 4, 2022 at 11:00

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