TITLE: Does every ellipse inside a tetrahedron inside a ball fit in a triangle inside the ball?
QUESTION [22 upvotes]: In three-dimensional euclidean space, consider the closed unit ball $B$. Let $T$ be a tetrahedron, and $E$ an ellipse, with $E \subset T \subset B$. Does there necessarily exist a triangle $T'$ with $E \subset T' \subset B$?
Clearly if 1 vertex of the tetrahedron is on one side of the plane of of the ellipse, and the other 3 vertices are on the other side, then intersecting the plane with the tetrahedron gives such a triangle. The interesting case is when 2 vertices of the tetrahedron are on each side of the plane.
I work in quantum information theory, and have come up with a conjecture that, remarkably, is true if and only if the answer to the above question is "yes"! Since this is very far from the sort of thing I normally think about, I don't even know where to begin to look in the mathematics literature, so even just a pointer would be a big help.

REPLY [7 votes]: Mihai-Dorian Vidrighin has suggested the following idea. (He's asked me to post it because he can't post images.)
For simplicity, assume that the setup is "tight" in that $E$ meets every face of $T$ and the vertices of $T$ are on the boundary of $B$. The generalisation should be straightforward.
If any vertices of $T$ lie in the plane of $E$ then the cross-section is already a triangle, so assume otherwise. Pick an arbitrary vertex of $T$ and draw a cone staring from there with $E$ (shown in red) as a base. This cone meets the opposite face of $T$ in a new ellipse $E'$ (shown as a dotted black curve):

$E'$ lies inside a triangle (the face of $T$), which itself lies inside a circle (the cross-section of $B$). By our simplifying assumption, $E'$ touches the edges of the triangle and the triangle's vertices are on the circle. Therefore Poncelet's Porism applies. Hence we can find a different triangle around $E'$ with one edge in the plane of $E$.
Define a new tetrahedron (shown in green) using the chosen vertex of $T$ and the new triangle. By construction it still contains the cone identified before, and in particular it still contains $E$. But the cross-section in the plane of $E$ is now a triangle (shown in thick black).