Partitioning the Perimeter of Convex Polygons

This post was inspired by 'Mathematical Omnibus' by Fuchs and Tabachnikov (FT).

Questions:

1. 'Fair bisectors' of convex planar shapes are those lines which divide both the area and perimeter of the shape equally. What can one say about the totality of fair bisectors of a convex region? How do their intersections reflect the shape of the polygon itself?

For any centrally symmetric C, all fair bisectors pass thru its center. So the issue is whether all fair bisectors of any C are concurrent. Numerical calculations show that they are not in general - intersections among the fair bisectors split the interior of C into several pieces. Let us call the totality of those pieces which do not touch the boundary of C as the core of C.

So one could ask which convex region C gives the largest such 'core area' as a fraction of the area of C.

2. Generalizing a bit, what about lines that break off the same fraction t of the area and outer boundary length of C? For a circular disk, it appears that only for t=1/2, we have such lines (any diameter). Are there C's for which such lines exist for several (maybe even arbitrarily many) different values of t? Notes: Ellipses should have more than one such lines. Needs checking.

These questions have obvious higher dimensional analogs.

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Now let me go on to record some experimental observations based on a reading of lecture 11 from FT titled: 'Segments of Equal Area'.

FT says that the envelope of a set of lines that cut off the same area from a plane wedge is a hyperbola.

Here is something I am not clear about: For which polygon is the area bounded inside the envelope of area bisectors the highest fraction of the area of the polygon? Note: some properties of this envelope are given in FT.

Let us now go from area to perimeter:
Definition: A perimeter bisector of a convex polygon is a line that divides its boundary into two segments of equal length. Length of the line intercepted within the polygon is immaterial. One can also imagine lines that cut off a specific fraction of the boundary.

A Basic Question: What is the envelope of lines that cut off from the arms of a wedge, pieces of equal edge length sum ?

(Note: this edge length sum does not include the length of the cutting line; for ex: if the wedge were of 90 degrees, we are looking for the envelope of lines which pass through (t, 0) and (0, 1-t) for varying t). Moreover, this envelope is not infinite like the hyperbola, for obvious reasons)

The curve is the segment of a parabola (assuming I did the algebra okay!).

We make some guesses based on further experiments..

For an equilateral triangle, the envelope of its perimeter bisectors (dividing the outer boundary into 2 equal pieces) is as shown below; it consists of 3 parabola segments with cusps at points dividing each median in the ratio 1:2.


For an equilateral triangle, the envelope of lines that divide the perimeter in the ratio 1:2 appears to be formed of 6 parabolic segments and three discontinuities. The parabolic segments meet each edge of the triangle at 2 points which trisect that edge (shown as red dots in the schematic diagram below) and the envelope has a discontinuity between each point pair on each edge.



Remarks: To my knowledge, no work has been done on perimeter partitioning lines on the lines of what FT describe for area partitioning lines. May be there aren't nice results for envelopes of perimeter partitioning lines. These envelopes are discontinuous - those for lines which segment the area of polygons are never discontinuous. Further, the envelope of area partitioning lines (for any area fraction cut off) is the locus of mid points of the partitioning lines and I couldn't see any such nice property for perimeter partitioning lines.

On 'Inside Out' Dissections of Polygons

This post attempts to add a little bit to this discussion initiated by Prof. Joseph O'Rourke.

There, an 'inside out dissection' of a polygon P is a defined as a dissection of P into another polygon P' such that (1) P' is congruent to P and the perimeter of P is internal to P' - ie the perimeter of P' is composed of the internal cuts made to P. It is also very nicely proved there that any triangle has an inside out dissection via 4 pieces, thus:

Let us consider doing an inside-out dissection of a polygon such that the total length of cuts is minimized. From the construction given on the above linked page, for any general triangle T, there is an inside out dissection with total cut length = half the perimeter of T itself; it is a 'perfect' dissection.

Likewise for rectangles or regular hexagons, we have inside out dissections where the total length of cuts is half the perimeter. Intuitively, it is obvious that half the perimeter of P is a tight lower bound on the total cut length.

Question 1: Which convex polygon P gives an upper bound on the length of cuts (compared to perimeter of P) for an inside out dissection? Basically, we are asking for polygons which need much more total cut length than half its perimeter for any inside out dissection.

Note: We observe that every convex polygon P can be dissected inside out with total length of cuts arbitrarily close to P (which is thus a loose upper bound on the total cut length). Proof: Form another polygon P' in the interior of P by insetting P by a small distance d. It is clear that the region between P and P' can be broken into parallelograms and triangles; dissect all these triangles inside out (as given in above picture) and reflect the parallelograms by 180 degrees and we have an inside out dissection of P. And as d goes very small, the total cut length arbitrarily gets close to P.
But we guess, the dissection just described must be suboptimal for any P and that the perimeter of P must be a very loose upper bound for total cut length. So, the question is which polygon approaches this upper bound closest for its "least cut length inside out dissection".

Question 2: For convex P, how will such an upper bound change if we allow P' to be non-congruent to P - insisting only that the perimeter of P should become totally interior to P'? It seems that allowing P' alone to be non-convex might not have an impact.

Question 3: Going to 3D, it is easy to see that a cube can be turned inside out by first cutting it into 8 cubes with three mutually perpendicular cutting planes - and the total area of the cuts is only half the surface area of the cube. It is not immediately obvious if for tetrahedra, a simple generalization of the triangle inside out dissection will go through.

Question 4: What happens to inside-out dissections in non-Euclidean geometry? Even the inside out triangle dissection doesn't seem to generalize obviously to hyperbolic geometry.
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Note: Regarding question 2 above, if P is allowed to be non convex and if P' is allowed to be non-congruent to P, the length of the cuts can be much less than half the perimeter of P. A simple example is shown below:



P is the non convex polygon to the left. It consists of the big yellow non concave region AND the 4 small white triangles at its fringes. We could do the following inside out dissection of it: as described in the above linked page, first dissect only each of the 4 white triangles inside out and keep them at the same place relative to the full polygon P. Then cut P along the blue 'equatorial' line and reassemble the 2 pieces to get the new polygon to the right (P'). Now, no portion of the perimeter of P is exposed. It can also be seen that keeping the 4 small white triangles fixed, if the number of 'teeth' in P are arbitrarily increased, the perimeter of P' is much less than that of P and the total length of cuts for our inside out dissection can be arbitrarily small compared to perimeter of P.

Thinnest Rigid Packings of the Plane - Contd.

Here we elaborate a bit on this one year old post.

Let us recall the definition: A packing with copies of any shape is called rigid (or "stable") if every unit is fixed by its neighbors, i.e., no unit can be translated without disturbing others in the packing. We are interested in thin and rigid packings of the plane, those that leave the largest possible fraction of the plane uncovered.

1. How does one form the thinnest rigid packing of the plane with unit squares? The layout may well be non-lattice! Note: As was noted in above linked page, even for unit circles, the thinnest rigid pack of the plane with them is not known for sure.

2. For a thin enough rectangle as unit, the configuration given in the above linked post

seems a good candidate for the thinnest rigid packing. Of course, as the rectangle unit get thinner, the coverage of the plane can be brought arbitrarily close to 0.

However, for fat rectangles (where length and breadth are close) tending to squares, this doesn't appear to be the way to thinnest packing.

3. For a general triangle, we can always have a rigid packing that leaves 1/2 of the plane uncovered:

But for thin enough triangles, it is better to form parallelograms with pairs of them and then to form analogs of above layout with rectangles - thus achieving arbitrarily low coverage of the plane.

4. Which is the convex shape for which the densest and thinnest rigid packs of the plane show least difference in coverage? Note: As noted above, for a thin rectangle or thin triangle as unit, the densest and thinnest rigid packs of the plane can show arbitrarily large difference in coverage.