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Feb 11, 2021 at 22:07 comment added WhatsUp @HerbertKociemba This is not surprising, as the sum of $\cos \theta$ and $\cos (\theta \pm \frac{2\pi}3)$ is zero, for any $\theta$.
Feb 11, 2021 at 21:06 comment added Zomulgustar Phrased another way, the sum of the squares of your three red lines must equal the sum of the squares of the edge lengths of the two triangles.
Feb 11, 2021 at 19:37 comment added Herbert Kociemba Btw., from your equations (*) we get the remarkable relation u^2 + v^2 + w^2 = a^2 + 1, but I do not see in the moment how this could be exploited.
Feb 11, 2021 at 2:28 comment added Laska @WhatsUp thanks sorry I deleted my comment about a before I saw you'd swiftly replied. I had realized we don't gain any more solutions by looking for only the 3 inter-triangle lengths to be rational. Using any two of the quadratics in a we can eliminate a^2 to see a is rational.
Feb 11, 2021 at 2:09 comment added WhatsUp $a$ is, by definition, the side length of one of the given triangles, hence it belongs to "distance between any two of the $6$ vertices".
Feb 10, 2021 at 14:00 comment added Herbert Kociemba The solution I found using integers and not rationals nevertheless has distances small enough and can be found just by brute force in an appropriate way.
Feb 10, 2021 at 2:14 comment added Herbert Kociemba Thank you very much for this analysis which gives me new insights to the problem. I can confirm your set of diophantine equations and have transformed my solution to a rational tuple (a,t) which indeed then gives rational u, v and w, which are the lengths of the red lines above if the outer triangle is scaled to edge length 1. I agree that it would be very interesting to know something more about the solution space in general. My knowledge about elliptic curves is more or less zero.
Feb 9, 2021 at 22:17 history edited WhatsUp CC BY-SA 4.0
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Feb 9, 2021 at 22:12 history answered WhatsUp CC BY-SA 4.0