On Elliptical Containers -II

Note: This post reports some investigation done with help from K Sheshadri.

Question(s): Given an ellipse E, what is the least area (least perimeter) convex region for which E is the least area(least perimeter)ellipse containing the convex region?

As can be readily seen there are 4 questions in there - one for each combination: area-area, area-perimeter, perimeter-area, perimeter-perimeter. Out of these, since perimeter of the ellipse has no closed expression, so we first consider the case of least area ellipse container.

Note: The questions above can be generalized to higher dimensions.

Answers

The least area convex region for which a given ellipse E is the least area container is (almost certainly) any triangle for which E is the Steiner circumellipse. Indeed, there are infinitely many such triangles - all of them are inscribed in the ellipse and have their centroid at the center of the ellipse. It is a fairly straightforward calculation to show that the area of all these triangles is equal to 3*sqrt(3)*ab/4 where a and b are the axes of the ellipse. But it would be nice to find deeper arguments (we have none as of now) for "all triangles inscribed in an ellipse with centroid coinciding with the ellipse center have same area". We also find numerically that every point on the boundary of the ellipse is the vertex of one such inscribed triangle.

This appears to suffice: From the Mathworld article on Steiner circumellipse, the area of the Steiner ellipse of a triangle with area Delta is given by: A = (4pi)/(3sqrt(3))* Delta. This implies that for any triangle for which a given ellipse is the Steiner circumellipse should have the same area.

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A side tour:

Guess: Among all 2D convex shapes without 3 fold rotation symmetry, the ellipse is perhaps the only shape such that every point on its boundary is a vertex of an inscribed triangle such that the centroid of the triangle coincides with the center of mass of the container shape. If true, this guess might follow from deeper reasons.

Remark: The guess is incorrect. Numerical checks confirmed that *any* triangle* has that property. Any rectangle too. Indeed, any triangle T appears to have a stronger property: If P is any point in a finite region around the centroid of T, every point on the boundary of T is the vertex of a triangle inscribed in T and with P as its centroid. One can ask if replacing T with any convex region will leave this property unchanged.

Then one sees that the property is not really hard to explain. Take any 2D convex region C and any point P1 on its boundary. Choose any point G within C such that if we extend directed line segment P1-G by a factor of 3/2, the new end point is still in the interior of C (for this G shouldnt be too close to the boundary of C). Let the end point of this extended P1-G be called M. Now, consider all chords of C that pass thru M. There necessarily is one of them such that M is its mid point (obvious). Let the end points of this chord be P2 and P3. It is easy to see that the centroid of triangle P1-P2-P3 (inscribed in C) is identical to G. Reason: G trisects one of its medians by construction and the medians of any triangle are necessarily concurrent.

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The perimeters of triangles inscribed in a given ellipse with centroid at ellipse center show some variation but are quite close. Among them, the triangle with the least perimeter is (numerically), an isosceles triangle to which the major axis of the ellipse is the bisector of the apex angle. This triangle is also almost certainly, the least perimeter convex region for which the given ellipse E is the least area container.

Aside: Given an ellipse, every point on its boundary is also found (numerically) to be the vertex of exactly one triangle such that the *incenter* of the triangle coincides with the ellipse center. In this case, the family of triangles do not all have the same area or perimeter but the values are found to be quite close. One can ask: is there any other container (not ellipse) such that with every point of its boundary as a vertex, we have one triangle such that the incenter of the triangle coincides with the ellipse center? Same question can be asked with any other triangle center.

Moving on to the least area and least perimeter convex regions for which a given ellipse E is the least perimeter elliptical container: These convex regions are almost certainly not triangles - it appears that for any of the inscribed triangles of E for which E is the least area ellipse container, another containing ellipse E' with more area than E but with less perimeter exists.

Note: We can replace 'ellipse' in the basic question set with rectangle, right triangle or a host of shapes and generate several questions from the same ballpark. Some of them might have nontrivial answers. More on that soon.