TECH-MUSINGS

Thoughts On Algorithms, Geometry etc...

Monday, March 30, 2020

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.

-------------

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.

0 Comments:

Post a Comment

<< Home