Does every ellipse inside a tetrahedron inside a ball fit in a triangle inside the ball?

Some remarks (2012/10/3):

(0) If the statement is true, then it is tight in the following example:

Draw a circle A, that represents a slice of the ball.

Inscribe inside A a square B, which represents a very thin tetrahedron.

Inscribe inside B a square C joining the midpoints of the sides of B. This square represents the midbase of the tetrahedron.

Inscribe inside C a circle D, which is the ellipsoid.

In this case we have concentric circles with radius(A)=2radius(D), whicch is just enough to fit a triangle between A and D.

This example makes the problem beautiful for me.

(1) If the statement is true, then it is true also for the case in which the outer ball is generalised to an ellipsoid. Why? because by changing the inner product of the space, the ellipsoid turns into a ball.

(2) So the context of our problem is affine geometry of $\mathbb R^3$ (that is, we can drop the inner product). In fact, we can drop even the affine structure and keep only the projective structure.

(3) We can then settle a new inner product so that the ellipse is a circle.

More remarks (2012/10/4):

(4) We have an ellipse inside another one (the intersection between $B$ and the plane of the ellipse, and we are trying to fit a triangle between them.

The problem of finding, given two conic sections, finding an $n$-side polygon circumscribed to the inner conic section and inscribed in the outer conic section is called "Poncelet's second problem" (Santaló, "Geometría proyectiva", p.243-245).

Remarkably, if there exists a triangle that fits between both conic sections, then it is easy to find it: One can choose any point on the outer conic as vertex of the triangle, and the construction will work. See "Poncelet's porism" at http://sbseminar.wordpress.com/2007/07/16/poncelets-porism/ .

I will say that a conic $n$-fits inside another one iff an $n$-sided polygon can be fit between them.

(5) By doing some projective transformations, I think the problem can be reduced to proving the following: Let $0\le a\le b\le 1$. Let $B$ be the cylinder $x^2+y^2\le 1$, $E$ the ellipse in the plane $z=0$ given by $(\frac xa)^2+(\frac yb)^2=1$. If there is a tetrahedron $T$ such that $E\subseteq T\subseteq B$, then there is a triangle $T'$ in the plane $z=0$ such that $E\subseteq T'\subseteq B$.

I don't have a complete proof of this reduction, but I can give more details.

(6) In the above situation, the existence of the triangle $T'$ is equivalent to the fact $a+b\leq 1$. This can be observed by trying to construct a triangle starting with a vertex in the $x$ axis. The starting point is irrelevant by remark (4).

(7) By remarks (5) and (6), we have to prove that if a tetrahedron fits between the cylinder and the ellipse, then $a+b\leq 1$. To fully flatten the problem, we have to be able to recognise if a given quadrilateral surrounding our ellipse is the projection of a tetrahedron that surrounds the ellipse. If we are given the projection $ABCD$ of the vertexes of the tetrahedron, and the quadrilateral $PQRS$ where $T$ intersects the plane $z=0$ (with $P\in[A,B]$, $Q\in[B,C]$, etc), then we can see if the ellipse fits inside $PQRS$. But also, applying Ceva's theorem, it can be shown that $\frac{|P-A| |Q-B| |R-C| |S-D|}{|P-B| |Q-C| |R-D| |S-A|}=1$, and this equation can be used to confirm that the points $A,B,C,D,P,Q,R,S$ of the plane $z=0$ where indeed obtained from a tetrahedron by projecting on and intresecting with the plane $z=0$.

(8) Some experiments that I did with the software GeoGebra suggest that an ellipse $A$ $3$-fits inside another ellipse $C$ iff there is an intermediate ellipse $B$ such that $A$ 4-fits inside $B$ and $B$ 4-fits inside $C$. I think that there is a path of ellipses joining $A$ and $C$, but to define it I would need a notion of $n$-fitting with $n$ non integer.

If we use Marcos Cossarini's reduction, this problem becomes an exercise in chasing cross ratios.

Step 1: It's enough to show that if $A,B,C,D$ are points on a circle $\omega$, and if $P,Q,R,S$ are points on the segments $AB,BC,CD,DA$ respectively such that the lines $PQ$, $RS$, and $AC$ meet at a point - call it $M$ - outside the circle, then any ellipse contained in quadrilateral $PQRS$ is contained in a triangle contained in $\omega$. (The way I visualize this reduction is: start by employing a projective transformation to assume the ellipse is contained in a plane passing through the center of the sphere, then without loss of generality two of the vertices of the tetrahedron are below the plane of the ellipse and two are above. Then push the vertices out further until they are lying on the boundary of the cylinder perpendicular to the plane of the ellipse, and project them down. The point $M$ is the intersection of the line containing the vertices projecting onto $A$ and $C$ with the plane of the ellipse.)

Step 2: By Poncelet's Porism, we may as well pick any line tangent to the ellipse as one of the lines of our triangle. So, I pick the line $PS$, and let $X$ and $Y$ be its intersections with the circle $\omega$, with $X$ closer to $P$ and $Y$ closer to $S$. Let $Z$ be the intersection of the other two tangents from $X$ and $Y$ to the ellipse. We just need to show that $Z$ is inside $\omega$, so we start by trying to find the locus of such $Z$.

Let $U$ be the intersection of $XZ$ and $PQ$, let $V$ be the intersection of of $YZ$ and $SR$, and let $W$ be the intersection of $XR$ and $YQ$. Then the hexagon $XUQRVY$ has all six sides tangent to our ellipse, so by Brianchon's Theorem we see that the point $W$ is on the line $UV$. Now, the points $U$, $V$, $W$ are the intersections of the opposite sides of the hexagon $XZYQMR$, so by the converse to Pascal's Theorem the points $X$, $Y$, $Z$, $M$, $Q$, $R$ lie on a conic section, so the locus of $Z$s is an arc in a conic section.

Step 3: I claim that the conic passing through $M,X,Y,Q,R$ - call it $\Omega$ - is tangent to the circle $\omega$ at the point $C$. Note that if this holds, then since two conics can intersect in at most four points, the arc $XCY$ of this conic section will be contained inside the circle $\omega$, so the claim finishes the problem. (In fact, the claim also gives us a construction for an equality case: take the inscribed ellipse of $PQRS$ that is also tangent to the line $XC$.)

To see that the point $C$ is on the conic section $\Omega$, it suffices to verify that the cross ratio of the four lines $CX,CY,CQ,CR$ is the same as the cross ratio of the four lines $MX,MY,MQ,MR$. Intersecting with the conic $\omega$ passing through $C$, we see that the cross ratio of the four lines $CX,CY,CQ,CR$ is the cross ratio of the four points $X,Y,B,D$ with respect to the conic $\omega$. Projecting through the point $A$ lying on the conic $\omega$ onto the line $XY$, we see that this cross ratio is the same as the cross ratio between the four points $X,Y,P,S$. This cross ratio is the same as the cross ratio of the four lines $MX,MY,MP,MS$, which by definition are the four lines $MX,MY,MQ,MR$. We have demonstrated that $C$ is on the conic $\Omega$.

Next, to see that $\Omega$ is tangent to the circle $\omega$ at $C$, it suffices to check that the cross ratio of the four points $A,C,X,Y$ with respect to $\omega$ is the same as the cross ratio of the four points $M,C,X,Y$ with respect to $\Omega$ (since when we project through the point $C$ which is on both conics, $A$ goes to $M$, $X$ and $Y$ go to themselves, and $C$ goes to the second intersection between $\Omega$ and the tangent to $C$ with respect to $\omega$). Let $I$ be the intersection of $CD$ and $XY$. Then projecting the points $A,C,X,Y$ through the point $D$ on the conic $\omega$ onto the line $XY$, we see that their cross ratio with respect to $\omega$ is the cross ratio of $S,I,X,Y$. Projecting through the point $R$ of the conic $\Omega$, we see that this is the same as the cross ratio of $M,C,X,Y$ with respect to the conic $\Omega$, so we are done.

Step 4: Pray that there are no configuration issues in the above argument.

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).