Fair Bisectors of Triangles - Observations

We recall the old definition: A fair bisector of any convex polygon is a line cutting it into 2 pieces of equal area and perimeter.

The following are some numerical observations exclusively on triangles.

- A triangle has either 1 or 3 fair bisectors. That the number of fair bisectors should be odd is easy to show by simple continuity arguments. That it cannot be greater than 3 for triangles looks obvious although I can't quite nail a proof.

- If any triangle has 3 fair bisectors, they are concurrent - I have no theoretical justification yet for this observation.

Assuming the fair bisectors of any triangle with 3 of them to be concurrent, one can ask if this results in a new center of a triangle or if it is coincident with some already well understood center.

- And one can always ask about concurrency properties of fair bisectors of convex regions with more than 3 sides - and the max number of fbs of convex polygons with no pair of sides parallel.
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Update ( 6th May 2020): That a triangle cannot have more than 3 fair bisectors is fairly easy to show. Indeed, if a pair of edges {a,b} or a triangle support a pair of fair bisectors (that they are at most a pair follows from each fair bisector being determined by a root of a simple quadratic equation), then, it is seen that the third edge c has to be shorter than both a and b. Then, it readily follows that neither of the edge pairs {a,c} or {b,c} can support a pair of fbs. This plus the requirement that the number of fbs is odd yields the limit as 3.
Thanks to K Sheshadri for pointing out that fbs can be found by solving a quadratic equation.
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Update (9th May 2020): K Sheshadri and self did a lot of calculations and programming with a whole range of triangles. For each triangle, the fair bisectors were found to concur - with coordinate precision up to 14 decimal places in our latest attempt - and also trying to find a proof that they do. And then, yesterday, I stumbled upon this article by Shailesh Shirali: http://publications.azimpremjifoundation.org/1655/1/3_Equalizers%20Of%20A%20Triangle.pdf

The lines I have been calling 'fair bisectors' were, in fact, given the name equalizers about 10 years back and it has been well established that they do concur - at the incenter of a triangle!

Let me sign off with another observation: For convex polygons with more sides, the fair bisectors do not seem to concur. This is based only on rough estimates! Maybe one can link their intersection points to some structure such as the medial axis.

Isosceles Triangle Containers of Triangles - 4

This post answers a specific question raised in this recent post :

The question: To find the least area isosceles triangle that contains a given convex polygon P, Will this work: 1)first find the least area general triangle that contains P and then 2) find the smallest area isosceles triangle that contains this least area general triangle?
Note: The same question can be asked for least perimeter isosceles triangle container of P with the same method proposed -perimeter simply replaces area. Also note that algorithms have already been found to find both smallest general triangle containers of convex polygons (above linked page has pointers) and also for least area isosceles triangle containers of general triangles.

The answer is negative. This picture illustrates an example:



The construction goes thus: Take a flat obtuse isosceles triangle ABC (AC = BC). Extend AC a little to D and join D to E on AB with segment DE crossing BC at F such that the area of triangle CDF is slightly less than that of triangle BFE.

Consider the grey quadrilateral AEFC as the convex region for which the least area isosceles triangle is to be found. The smallest area general triangle container containing AEFC is obviously not ABC and is not entirely contained within ABC either. However, it is also clear that ABC is the smallest area isosceles triangle that contains AEFC. Since ABC does not contain the whole of the smallest area general triangle containing AEFC, the method proposed above will fail.

Note 1: The above basic question and the proposed method can be restated with perimeter replacing area. It can be seen that if we find the least perimeter triangle container of AEFC, it will not be entirely inside ABC which will however be the least perimeter isosceles container of AEFC (note that CD and EB are quite small and almost equal in length). So, the basic method of first finding the general smallest triangle container and then finding the smallest isosceles container of this general container will fail for both area and perimeter cases.

Note 2: It can also be shown that the basic method fails to find the smallest area right triangle containing a convex polygon as well, as shown in this figure:

The construction is made with area marked '1' slightly less than area '2'. Then for the grey region as the input convex region, the red triangle will be the smallest area general triangle container and the black right triangle will be the smallest area right triangle container and the latter does not contain the former. The same construction should work as a counter to the method for min perimeter right triangle containers as well.

So we could conclude that the question of finding the min area (and min per) isosceles triangle containers of a convex polygon may well be different from finding the min area (min per) general triangle containers of the convex polygon.

Non-Congruent Tilings - 6

This post continues the 'broken chain' of posts on the non-congruent tiling front:

Here are links to the first 5 installments (am keeping them here for reference):

1. part 1
2. part 2
3. part 3
4. part 4
5. part 5
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Far more importantly, here are some of the major expert findings in this ongoing story (I am far behind the story!):

- The Euclidean 2D plane cannot be tiled with mutually noncongruent triangles all of the same area and perimeter. (Kupaavski, Pach, Tardos https://arxiv.org/pdf/1711.04504.pdf)
- non congruent tiling is possible if the triangles have equal area and bounded perimeter. (Frettloh https://arxiv.org/abs/1603.09132)
- the Euclidean plane can be dissected into mutually incongruent convex quadrangles of the same area and the same perimeter. (Dirk Frettloh and Christian Richter https://arxiv.org/abs/2004.01034)

Remarks: I just gathered from an expert that the same question (or set of questions) could have very different behavior and answers on the Hyperbolic plane. One hopes experts would look into those possibilities.

The hyperbolic possibility was mentioned in post number 3 listed above (an update made in 2018). However, despite a lot of effort (not focussed but substantial), I am yet to gain sufficient command over Hyperbolic geometry to attack these questions.

In elliptic geometry, since there is roughly speaking less space around each point than hyperbolic or even Euclidean, one suspects non congruent tilings to be harder than in Euclidean plane. So, impossibility results may be easier to prove.

On comparing Planar Convex Regions

This post is based on this old post: https://nandacumar.blogspot.com/2012/11/maximizing-and-minimizing-diameter-ii.html

Given two planar convex regions C1 and C2 both with unit perimeter, we define the difference between C1 and C2 as the least value of Hausdorff distance between C1 and C2 can have when the regions are placed above one another and adjusted to minimise the Hausdorff distance between them

Questions:

1. What are the specific pair of unit perimeter regions {C1,C2} with some equal specified area such that the difference between C1 and C2 is maximum?

2. What are the specific pair of unit perimeter regions {C1,C2} with equal specified area and equal specified diameter (diameter of a region is the greatest distance between any two points in the region). such that the difference between C1 and C2 is maximum?

These questions have been posted at https://mathoverflow.net/questions/357124/on-comparing-planar-convex-regions-of-equal-perimeter-and-area