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Suppose you're given a cuboid centered at the origin, and with its edges parallel to the coordinate axes. Suppose its measure is known to be $2 a \times 2 b \times 2 c $ along the $x, y, z$ directions, respectively.

Further suppose you have a plane with unit normal $n$, that cuts through the faces of the cuboid. Assume that this cutting plane cuts through all six faces of the cuboid.

Question: Now, I'd like to find the parametric equation of the ellipse that lies in the cutting plane, centered at the origin, that is tangent to all the faces of the cuboid.

My thoughts:

The intersection of the cutting plane with the faces of the cuboid results in an irregular hexagon, but one that has parallel opposite sides. The ellipse is obviously centered at the origin, so it is determined by $3$ parameters. Using tangency to three consecutive sides of the hexagon of intersection, these parameters can be determined.

  • is it obvious that such an ellipse exists? – user619894 Mar 31 '24 at 15:11
  • May be it is not obvious, but one can prove its existence by construction. –  Mar 31 '24 at 15:14
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    From your hexagon $ABCDB'C'$ you can construct a pentagon $ABCDE$, where $E$ is the intersection of lines $DB'$ and $AC'$. That pentagon is also tangent to the ellipse, hence you can easily find its tangency points as described here: https://math.stackexchange.com/questions/4330755/determining-center-of-inscribed-ellipse-of-a-pentagon/4330822#4330822 – Intelligenti pauca Apr 02 '24 at 22:18

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I've implemented the method outlined in the question statement, for a cuboid of dimensions $6 \times 8 \times 10 $ and a normal to the plane of the ellipse given by $N = (\sin 60^\circ \cos 40^\circ, \sin 60^\circ \sin 40^\circ, \cos 60^\circ) $

The result is shown below

enter image description here