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.

  • $\begingroup$ A narrower question that may be easier to answer is this: Is a triangle $T_s \supset E$ such that $E$ is a Steiner ellipse of $T_s$ itself in the ball: $T_s \subset B$? The Steiner ellipse of a triangle is the unique ellipse inscribed in the triangle touching at the side midpoints. $\endgroup$ – Joseph O'Rourke Sep 29 '12 at 2:09
  • $\begingroup$ It looks probable that one may assume (by applying appropriate projective transformation) that the ellipse is a circle. Then the condition that it may not be put in the triangle is $OI^2>R^2-2Rr$, where $I$, $r$ denote center, radius of the circle, $O$, $R$ of the section of ball by the plane of this circle. This may help at least in computational brute force approach. $\endgroup$ – Fedor Petrov Sep 29 '12 at 8:18
  • $\begingroup$ Please add the tag "projective-geometry". $\endgroup$ – Marcos Cossarini Oct 5 '12 at 3:59

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.

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  • $\begingroup$ Right now I have the following question: Assume we have an ellipse $E$ fitting tightly inside a tetrahedron $T$ (i.e., there is no tetrahedron containing $E$ that is strictly contained in $T$) so that $T$ fits tightly inside $B$ (i.e., there is no tetrahedron contained in $B$ that strictly contains $T$). Is it possible that we find another tetrahedron contained in $B$ and containing $E$ for which the fit is not tight? $\endgroup$ – Marcos Cossarini Oct 4 '12 at 23:46

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):

Tetrahedron with ellipse and cone

$E'$ 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).

alt text

  • $\begingroup$ This is like my second idea, but better expressed and with more rigor. I'm glad someone with the geometry chops has realized it. Gerhard "Ask Me About Pork Chops" Paseman, 2012.10.08 $\endgroup$ – Gerhard Paseman Oct 8 '12 at 15:29
  • $\begingroup$ That's beautiful - and it also provides a visual explanation of the equality case for the cylinder version of the problem. $\endgroup$ – zeb Oct 8 '12 at 15:51
  • $\begingroup$ Mihai has pointed out that his proof includes a partial version of a Poncolet's Porism for ellipsoids inside tetrahedron inside ellipsoids, it might be interesting to see how far that can go. My original question could also be generalised to simplices in higher dimensions. But I think my original problem is solved (the best way to reduce the none-tight case is perhaps to shrink the base of the tetrahedron to make it tight, then shrink the cross-section of the Bloch sphere to a tight ellipse round that.) Thanks for all the fantastic suggestions, and in particular for the name of the Porism. $\endgroup$ – Matt Pusey Oct 10 '12 at 17:20

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.

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  • $\begingroup$ This is a very pretty argument. $\endgroup$ – Gjergji Zaimi Oct 8 '12 at 8:36
  • $\begingroup$ Good job! I don't understand the projection through $C$ from $\omega$ to $\Omega$ (in the comment between brackets in the last paragraph), but anyway, from the equality of ratios, it is easy to see that the fourth point of intersection $C′$ should satisfy $(C'A,C'C;C'X,C'Y)=(C'M,C'C,C'X,C'Y)$, so it should be on the line $AM$. In case anyone is wondering, the expression "cross ratio of the four points $A$, $C$, $X$, $Y$ with respect to $\omega$" is justified because given four points E,F,G,H and a number $\lambda$, the locus of the points T such that $(TE:TF:TG:TH)=\lambda$ is a conic section. $\endgroup$ – Marcos Cossarini Oct 9 '12 at 1:01

Here is a bit of Mathematica code that rather supports Joseph's conclusion.

Tet[phi_] := {{Sin[phi], 0, Cos[phi]}, {-Sin[phi], 0, Cos[phi]}, {0, 
   Sin[phi], -Cos[phi]}, {0, -Sin[phi], -Cos[phi]}};

Rect[a_, phi_] := Module[{x, y, u, w},
  {x, y, u, w} = Tet[phi]; 
  Polygon[{a x + (1 - a) u, a x + (1 - a) w, a y + (1 - a) w, 
    a y + (1 - a) u}]];

v = Subsets[Range[4], {3}];

 Graphics3D[{Opacity[0.2], Sphere[{0, 0, 0}, 1], Opacity[0.3], 
   GraphicsComplex[Tet[phi], Polygon[v]], Opacity[0.8], 
   Rect[a, phi]}], {phi, 0, Pi/2}, {a, 0, 1}]

Using parameters like a=0.9, phi=1.4 one obtains an elongated rectangle inscribed a flat tetrahedron, close to an equatorial plane. The maximal inscribed ellipse in this rectangle hardly is contained in any triangle that fits in the unit ball.

    rectangle in tetrahedron in sphere

Edit (2):

The following improved version uses an analytic solution for tetrahedra $ABCD$ with $[AB]$ perpendicular to $[CD]$, and the maximal ellipse $E$ in a plane cutting $ABCD$ parallel to $[AB]$ and $[CD]$. In this case the maximal distance of the vertices of the smallest enclosing triangle $T'$ from the origin is

    $1 - 2 (1 - a) a (1 + \cos(\phi - \psi))\leq 1$ for $0 < a < 1 $,

where $A=(\sin \phi, 0, \cos\phi)$, $B=(-\sin \phi, 0, \cos\phi)$ , $C=(0, \sin \psi, -\cos\psi)$, $D=(0, - \sin \psi, -\cos\psi)$. Thus in all these cases the triangle $T'$ lies inside $B$.

I even think that all other (nontrivial) cases of $E\subset T$ can be reduced to one of the above cases by aligning two edges of the enclosing tetrahedron $T$ to lie along the principal axes of $E$, while keeping $E$ inside. But this I cannot prove yet.


   psi_] := {{Sin[phi], 0, Cos[phi]}, {-Sin[phi], 0, Cos[phi]}, {0, 
    Sin[psi], -Cos[psi]}, {0, -Sin[psi], -Cos[psi]}};
Rect[a_, phi_, psi_] := Module[{x, y, u, w},
   {x, y, u, w} = Tet[phi, psi]; 
   Polygon[{a x + (1 - a) u, a x + (1 - a) w, a y + (1 - a) w, 
     a y + (1 - a) u}]];
v = Subsets[Range[4], {3}];
ParPl[a_, phi_, psi_] := 
  ParametricPlot3D[{a Sin[phi] Cos[t], -(-1 + a) Sin[psi] Sin[t], 
    a Cos[phi] + (-1 + a) Cos[psi]}, {t, 0, 2 Pi}, Boxed -> False, 
   Axes -> False];
Tangent[a_, phi_, psi_, t_, 
   lam_] := {a Sin[phi] Cos[t], -(-1 + a) Sin[psi] Sin[t], 
    a Cos[phi] + (-1 + a) Cos[psi]} + 
   lam {-a Sin[phi] Sin[t], (1 - a) Cos[t] Sin[psi], 0};

 bt = 2 ArcTan[
    1 - (a Csc[psi] Sin[phi])/(-1 + a) - Sqrt[(
     a Csc[psi] Sin[phi] (2 - 2 a + a Csc[psi] Sin[phi]))/(-1 + a)^2]];
 p1 = Tangent[a, phi, psi, bt, Cot[bt]];
 p2 = Tangent[a, phi, psi, bt, -Sec[bt] - Tan[bt]];
 p3 = {-1, 1, 1}*p2;
 Show[Graphics3D[{Opacity[0.2], Sphere[{0, 0, 0}, 1], Opacity[0.3], 
    GraphicsComplex[Tet[phi, psi], Polygon[v]], Opacity[0.8], 
    Rect[a, phi, psi], Green, Polygon[{p1, p2, p3}]}], 
  ParPl[a, phi, psi]], {{phi, Pi/4}, 0, Pi/2}, {{psi, Pi/4}, 0, 
  Pi}, {{a, 0.7}, 0, 1}]


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  • $\begingroup$ @Karl: I get a slew of Mma errors, starting with: ReplaceAll::reps: "{rule} is neither a list of replacement rules nor a valid dispatch table, and so cannot be used for replacing." $\endgroup$ – Joseph O'Rourke Oct 2 '12 at 23:58
  • $\begingroup$ @Joseph : Thanks, that is corrected now. $\endgroup$ – Karl Fabian Oct 3 '12 at 5:26

A possible counterexample?:
The tetrahedron is nearly a flat rectangle (red), and the ellipse $E$ nearly fills it. Then I don't see how to enclose $E$ in a triangle that remains in the ball. Not a proof, I know...

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  • 4
    $\begingroup$ Thanks. I agree that that ellipse looks like it won't fit in a triangle, but I don't agree that it fits inside the tetrahedron. If you try to make a thin tetrahedron like that then a typical cross-section through it will be a diamond which your ellipse wouldn't fit into. If you have Mathematica, the code at pastebin.com/UCAYTPpc shows the sort of tetrahedron I think your talking about and lets you look at cross-sections through it. $\endgroup$ – Matt Pusey Sep 28 '12 at 14:33
  • 1
    $\begingroup$ Turn it into a (dis-?)proof by embedding an ellipse into a cross-section of the tetrahedron. I think this is a beautifully misleading picture, because it suggests that a tetrahedral cross section can be almost as big as two of the faces, which I believe is never the case for tetrahedra of positive volume with faces of roughly equal area. Gerhard "Almost Like Sixty-Five Equals Sixty-Four" Paseman, 2012.09.28 $\endgroup$ – Gerhard Paseman Sep 28 '12 at 15:03
  • $\begingroup$ This is a nice picture, but I think there is a discontinuity in the set of ellipses you can inscribe in a tetrahedron when you degenerate to a rectangle. By the way, a square would work as well. $\endgroup$ – Douglas Zare Sep 28 '12 at 15:06
  • $\begingroup$ Thanks everyone---I stand corrected! I will leave my "answer" intact as an instructional error. :-) $\endgroup$ – Joseph O'Rourke Sep 28 '12 at 15:36

I suspect the answer is yes. Let me paint a word picture, which won't be as beautiful as Joseph O'Rourke's contribution, but may be as useful.

Consider all planes passing through the tetrahedron which are parallel to the plane of the ellipse. As Matt points out, the interesting case is when the ellipse lies in the cross-section of the tetrahedron that is not a triangle. I think one can use some Cavalieri type estimates to bound the size and shape of the rectangle to get an affirmative answer. Another idea is to reflect the tetrahedron across an angle bisecting plane which swaps two adjacent edges of the enclosing quadrilateral, and use that to construct a triangle which encloses the ellipse. Any such triangle which extends outside the original sphere says something about the tetrahedron, in which case it may be provable that another pair of edges can be used to construct the desired triangle.

Another idea is that the vertex and two incident edges opposite a certain vertex of the bounding quadrilateral can represent a point on one of the tetrahedral edges and two lines on different faces, and that by "rotating" the tetrahedral edge, one can construct a bounding triangle tha stays within the bounding sphere.

I hope these words are as suggestive as the picture, which I hope Joseph leaves up.

Gerhard "Ask Me About Suggestive Writing" Paseman, 2012.09.28

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  • $\begingroup$ @Suggestive Gerhard: Can you expand upon "some Cavalieri type estimates"? I am intrigued. $\endgroup$ – Joseph O'Rourke Sep 28 '12 at 23:48
  • $\begingroup$ I may have used the wrong term. I am thinking of his theorem where volumes are equal if the slices are equal, combined with (Simpson?) type quadratic estimates that give area at distance t if you know values at 0 and 1 (and maybe 1/2). Gerhard "Ask Me Not About Names" Paseman, 2012.09.28 $\endgroup$ – Gerhard Paseman Sep 29 '12 at 1:59

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