Ellipse 3-partition: same area and perimeter

Inspired by the question, "How to partition area of an ellipse into odd number of regions?," I ask for a partition an ellipse into three convex pieces, each of which has the same area and the same perimeter. The perimeter includes both arcs of the ellipse and whatever cuts are used.

This is known to be possible by recent results of Aronov and Hubbard, "Convex Equipartitions of volume and surface area," and by Karasev, "Equipartition of several measures," but perhaps the general techniques in these papers (which I have not studied) need not be used in this special case. Perhaps there is a natural construction?

Update. The two papers I cited above are both difficult for me to penetrate. The special case of equipartition into three parts was achieved earlier in a paper by Imre Bárány, Pavle Blagojevićc, and András Szűcsd, "Equipartitioning by a convex 3-fan," which I cannot access at the moment. But as you can infer from the title, the partition is accomplished via a convex 3-fan: a point with three rays emanating from that point.

• Going by the linked papers, I assume you want the three pieces to be convex as well? If so, you may want to mention that in the question; otherwise there are some pretty simple constructions that work. – Rahul Sep 7 '11 at 15:43
• Thank you, Rahul! Corrected. – Joseph O'Rourke Sep 16 '11 at 20:20
• One possibility at least worth a try is three outward rays from the origin given by angles $\alpha,\beta,\gamma$. The areas can be found with $\frac{1}{2}\int r^2d\theta$ and the perimeters using elliptic integrals of the second kind. – anon Sep 16 '11 at 20:40
• @Peter: I had the pleasure of attending a talk by Steiger on this nice result :-) Good idea, to use the uniqueness of the 6-partition! That would reduce to search to just one variable, a point on $S^1$. – Joseph O'Rourke Sep 18 '11 at 0:18
• If you want convex pieces, I think you cannot do much else then paritition by a fan with straight lines, or else possibly cut up the ellipse by two non-intersecting straight lines. If the boundary between two pieces has nonzero curvature (and I think even more so if it is not smooth) then one of the two separated pieces will fail to be convex. – Marc van Leeuwen Oct 3 '12 at 6:10

Let $V_1 V_2$ be the major axis of our ellipse, with area $\Delta$. Take a point $P$ between $V_1$ and $V_2$ such that $PV_1=x$, and two points $Q_1,Q_2$ such that $Q_1 Q_2$ is perpendicular to $V_1 V_2$ and the elliptic sector $E_1$ delimited by the rays $PQ_1,PQ_2$ has area $\Delta/3$. Let $E_2$ be the elliptic sector delimited by the rays $PQ_1,PV_2$ and $p_j(x)$ the perimeter of $E_j$. The function
$$f(x) = p_1(x) - p_2(x)$$
is clearly continous, so, if we find two points $x_0,y_0$ such that $f(x_0)f(y_0)<0$, we are sure that $f$ has at least a zero $z$ and we have done, since:
$$\Delta(E_2) = \frac{\Delta-\Delta(E_1)}{2} = \Delta(E_1).$$
• However, to ensure the convexity of $E_1$ we must have $x\geq 0.367534\dots V_1V_2$. – Jack D'Aurizio Oct 14 '12 at 11:53