# Convex curves with many inscribed triangles maximizing perimeter

A classical nice result of Euclidean geometry states that the triangles maximizing the perimeter among all inscribed triangles of a given ellipse constitue a one-parameter family. Precisely, for each point on the ellipse there is exactly one such a maximizing triangle with vertex on that point, which can also be viewed as a period-3 bouncing trajectory.

I am wondering is this property is characteristic of the ellipses:

Let $C$ be a compact convex subset of the plane such that for any point $P\in\partial C$ there is a triangle maximizing the perimeter among all inscribed triangles, with $P$ as a vertex. Is it true that $\partial C$ is necessarily an ellipse?

• Any reference fot the "classical result"? Oct 14 '11 at 19:45
• @I.Rivin: the "classical result" follows from the easy (or at least easier) special case of a circle via linear transformation. Oct 14 '11 at 20:11
• I believe Pietro is referring to a result of Michel Chasles; the triangles are all "billiard triangles" as Pietro says. Oct 14 '11 at 20:17
• @Noam: why so? Linear transformations change lengths of non-parallel segments quite differently... Oct 14 '11 at 21:54
• @Fedor: You're right, sorry. I misread the question as asking for maximal area. The argument does not work for perimeter. Oct 15 '11 at 18:29

N.B. This is an edit of my original post, confirming the guess that I made originally.

The answer is no, i.e., such curves are not forced to be ellipses.

Here is a sketch of the argument. (The details will take a while to type in, and that will have to wait. Also, I'm not sure that there will be that many people interested in seeing the details, so I might not put them in unless I get a request.)

Such a curve $C$, if it is strictly convex (which I am assuming from now on) defines a closed curve in the $5$-dimensional space $T$ of triangles in the plane with fixed perimeter $P>0$. (Note that $T$ is an $S_3$-quotient of a hypersurface in the space of triples of non-collinear points in the plane.) These triangles of perimeter $P$ are so-called 'billiard triangles', i.e., the angle of incidence and reflection of the sides of each vertex of the triangle with the curve $C$ are equal.

This means that such a curve in $T$ is tangent to a certain rank $3$ plane field $D\subset TM$, and, conversely, a closed embedded curve in $T$ that is tangent everywhere to this plane field $D$ represents a closed curve of triangles of perimeter $P$ moving in such a way that the velocity of each vertex is perpendicular to the triangle's angle bisector at that vertex. Thus, the curves we want to study are solutions of an underdetermined system of ODE.

The $D$-integral curves represented by the ellipses (which do give solutions, by Chasles' Theorem) are 'regular' in the sense of control theory (this is what I had to check by looking at the structure equations of $D$), so that one can make an arbitrary functions' worth of perturbations of any such closed $D$-integral curve (say, one given by an ellipse) to get other nearby $D$-integral curves. Thus, there will be lots of such closed convex curves, near ellipses but not ellipses, that have a circle of inscribed triangles of maximum perimeter.

It might be interesting to construct some explicit closed integral curves that don't come from ellipses, but I don't see any easy way to do that right now.

Remark: The plane field $D$ is interesting. It is bracket-generating and does not contain an integrable $2$-plane field, so that it belongs to the type studied by Élie Cartan in his famous "Five Variables" paper of 1910. It is not 'flat' in Cartan's sense, i.e., the group of symmetries of the plane field is not of dimension $14$, but rather of dimension $3$ (very small). Still, the general theory says that the $D$-integral curves near the ellipses (in a suitable sense) are regular curves, so that they are freely deformable.

• This is an impressive analysis! It would make a nice paper. I assume these curves $C$ are smooth? Or, for example, could they be composed of pieces of ellipses joined together? Oct 19 '11 at 11:00
• Absolutely beautiful! Though one thing I don't see yet: a closed integral curve in $T$ could in principle give rise to 3 different closed curves in the plane, one for each vertex. Are there examples for this? Could a perturbation of an ellipse split up like this? Plus I'd be grateful if you can add a reference (or definition) of "regular" integral curve. Oct 19 '11 at 14:45
• @Michael: If you weren't careful, yes, they could split, into, say, 3 circles. However, I am working on the quotient space $T$ of triangles (actually, it turns out to be easier to work with the oriented triangles, a $\mathbb{Z}_3$ quotient), not the space of 'marked' triangles, which is an open subset of three copies of the plane. When you work on this space, this splitting problem doesn't arise for deformations. 'Regular' is in the usual sense of nonholonomic systems (which this is). Bliss is a good reference, but you can also consult Bryant-Hsu, Inventiones M. 114 (1993), 435–461. Oct 19 '11 at 15:32
• @Joseph: Thanks. The curves don't have to be smooth (but I like them better that way). I guess you could, in principle, construct ones that are merely $C^1$ by using $C^1$-but-not-$C^2$ constraints to generate the solutions, but I don't know whether I could rig it to be pieces of ellipses. Maybe it would be worth writing a short note about this problem, because it is a somewhat interesting application of a technique that appears not to be well-known. Oct 19 '11 at 15:37
• @Robert: thanks for the comment and the references. I have to think more about what you wrote. Oct 19 '11 at 16:23

This 1988 paper of Innami gives a construction of a convex curve, all of whose points are vertices of billiard triangles and therefore also maximal-perimeter inscribed triangles. In Innami's construction, all triangles are isosceles.

• That's a nice paper. In fact, though, you can impose just about any extra single condition on the triangles and there will be non-circular (and non-elliptical) solutions. For example, instead of requiring that two of the sides be of equal length, you could require that one of the sides always be of length equal to $\tfrac13$ of the total (fixed) perimeter of the triangle, and so on. The point is that the set of such triangles in the plane will be of dimension $4$, and the differential equation that constrains the motion of the vertices will still be an underdetermined ODE with many solutions. Dec 6 '12 at 18:10
• That's interesting. Is it also possible to impose that the curve be 2-fold symmetric about the origin? I am actually interested in this origin-symmetric case for outer, instead of inner, billiards. Dec 6 '12 at 18:49
• By '2-fold symmetric about the origin', do you mean 'invariant under reflection about a line through the origin' or 'invariant under the $x\mapsto -x$ involution centered at the origin'? I don't know the answer, but I'll think about it. Also, what do you mean by 'outer billiards'? If the curve is convex, a single bounce of a ball coming in from infinity will go back out to infinity, so I'm not sure what you mean in this case. Dec 6 '12 at 19:10
• I mean the latter ($x\mapsto -x$). In each iteration of outer billiards you start with a point outside of the billiards table, draw the tangent line from the point to the table (in the counterclockwise direction), and reflect the point about the point of tangency to get the starting point for the next iteration. Outer billiard triangles are circumscribing triangles whose midpoints lie on the boundary of the table. Dec 6 '12 at 19:16
• Ah! I see. However, this really isn't a metric construction. If $V_i$ ($1\le i\le3$) are the vertices of the 'outer triangle', then the points of tangency of the sides are $\tfrac12(V_i{+}V_j)=T_k$ (where $(i,j,k)$ are distinct in $\\{1,2,3\\}$), so the side $T_iT_j$ of the inscribed triangle is parallel to the tangent to the curve at $T_k$. In other words, the triangle $T_1T_2T_3$ is a critical point for the area for inscribed triangles in the curve. Of course, this can be reversed, and there are many curves symmetric about the origin that have a circle of max-area inscribed triangles. Dec 6 '12 at 21:38

Please see a relevant paper Yu. Baryshnikov, V. Zharnitsky, Sub-Riemannian geometry and periodic orbits in classical billiards, Math. Res. Lett., 13 (2006), 587–598.

As far as I know, the following question is open: are there curves, rather than ellipses, that possess circles of p- and q-periodic billiard trajectories for different p and q.

Dear Colleagues, the question that was originally posed is slightly different:

Let C be a compact convex subset of the plane such that for any point P∈∂C there is a triangle maximizing the perimeter among all inscribed triangles, with P as a vertex. If the perimeter of maximizing triangle at P∈∂C is CONSTANT, independent of P, is it true that ∂C is necessarily an ellipse?

Sorry for misunderstanding

• @Vladimir: Actually, that is exactly how I understood the question, and the answer is the same. The examples of convex curves $C$ (not all of them ellipses, by the way) that I indicate have the property that they have a circle of inscribed triangles of perimeter $L$, where $L$ the maximum value of the perimeter of any inscribed triangle in $C$. Moreover, for each point $P\in \partial C$, there is one inscribed triangle of perimeter $L$ passing through $P$. I don't see how this is different from what you are asking. Oct 20 '11 at 20:11