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the diagramReference image ^^^

Ok so I think I might have found a new theorem or maybe rediscovered an old one. Also please note that I am just a 13 year old so I don't really understand super complex notation so please try to keep the language as simple as possible. Here is how the theorem goes:

In a $3D$ plane if you have $2$ circles who share a common diameter and the circles look like perpendicular lines from the top angle, then if you make $2$ chords per circle (that makes it $4$ chords in total) that are parallel to the diameter and $\frac{1}{\sqrt{3}}$ times the diameter then upon joining the endpoints of all $4$ chords together you would always get a cube. I don't really know how I got the number $\sqrt{3}$ as I worked on it a while ago and just forgot it. I checked with chatgpt and it said it was able to find some things about this but it wasn't able to find the formal name for this theorem.top view here is the top view from which the circles look perpendicular^^^

I just wanted to make sure if this theorem is already discovered or something new which chatgpt was heavily implying it is a well known theorem but I cannot find the exact name. Thanks for your time have a nice day

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    $\begingroup$ Your terminology has issues; for example, a plane is 2D. I think what you mean is if two circles in orthogonal planes in 3D space share a diameter and you draw in each circle two chords parallel to that diameter with $1/\sqrt{3}$ times its length their eight endpoints are a cube's vertices. Except that can't be right, because a cube's vertices don't fit in two orthogonal planes. $\endgroup$ Commented 18 hours ago
  • $\begingroup$ Hi, thanks for the feedback! Yes I did mess up a bit there. The chords are parallel to and 1/✓3 times smaller than the diameter. I will be shortly attaching an image in a some time, thanks! $\endgroup$ Commented 17 hours ago
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    $\begingroup$ On second thoughts, we can get a cube's vertices in two orthogonal planes if each has two edges that aren't from the same face. In that case, we can prove the $\sqrt{3}$ factor with repeated applications of the Pythagorean theorem. $\endgroup$ Commented 17 hours ago
  • $\begingroup$ Ty, just added reference images. $\endgroup$ Commented 17 hours ago
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    $\begingroup$ You probably won't find this result as a known theorem because is just a simple consequence of the construction. May be this could be found as an exercise in an Analytic Geometry book. $\endgroup$ Commented 17 hours ago

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OP has “rediscovered an old one.”

In Elements, XIII, 15 Euclid constructs a cube, comprehends it in a sphere, and proves that “the square on the diameter of the sphere is triple of the square on the side of the cube” (Heath translation).

Hence$$\frac{side}{diam}=\frac{1}{\sqrt 3}$$

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