On Some Infinite Planar Arrangments with Triangles

Background:
Given a convex region C. One can think of a graph corresponding to any planar arrangement of copies of C - each unit is a node and an edge connects it to another node if the two share some finite length of boundary. Such a graph is necessarily planar. As is well known, the average degree of a planar graph can at most be 6.

Now, given any triangle, one can tile the plane with it: join a pair of the triangle into a parallelogram and form infinite strips with these parallelograms and arrange these strips side by side. While putting two strips together one can slide one length-wise a bit with respect to the other. The graph corresponding to this tiling (with each pair of adjacent strips shifted slightly lengthwise) will have the same degree = 4 for every node. One notes a gap between this 4 and the maximum average degree of a planar graph (6).

Question: Can one have any plane-filling arrangement (not necessarily a tiling) with any triangle where the average degree of the corresponding graph is between 4 and 6?

Note: With squares, it is easy to form a tiling with corresponding graph having degree 6 at every node.

Generalization: Given any convex 2D shape C, not necessarily one that tiles. Assume the arrangement(s) with infinite copies of C that maximizes the average degree of the corresponding graph is known. Are these arrangements also good at achieving max packing density?

Largest Regular Polygons that Fit into a Polygon

It is well known that for any 2D region P (not necessarily convex), the center of the largest circle that can be drawn in P necessarily lies on the medial axis of P - the medial axis of P is the set of all points in its interior with more than one closest point on the boundary of P.

Question 1: What if we are looking for the largest equilateral triangle or largest square or in general the largest regular N-gon that can be drawn inside P/

Observation: The center of the largest N-gon need not lie on the medial axis of P. Here is an example.



The figure shows a circle and its largest contained square (the latter shown with broken boundary). The slightly larger red square touches the circle at two points and two of its corners poke out of the circle. The centers of the two squares are also marked. Now consider P to be the union of the circular region and the big square. Obviously, the largest square that can be drawn inside P is the red square. Its center has only one closest point on the boundary of P (the other end of the blue broken line) and so is not on the medial axis of P.

Note: Even if P is restricted to be convex, the center of the largest square inside it need not be on the medial axis. Indeed, take a square and inflate three of its sides very slightly outwards, keeping the boundary convex with all vertices fixed and leaving the 4th side untouched. If this object is taken as P, its largest contained square is the one we started with. The center of this square has only a single closest point on the boundary of P - the mid point of the untouched 4th side of the original square.

The question of the largest contained regular N-gon should be a well studied area. I shall link references as I find them.
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This is more speculative...
Question 2: Given a convex region C and a positive integer n>=3. Let A(n) be the area of the largest n-gon that can be drawn inside C. What could one say in general about the behavior of A(n) as n varies?

For example, if C is an equilateral triangle, A(n) appears to steadily decrease with n from the area of C itself for n =3 until the incircle is reached for n = infinity. If C is a circle, A(n) appears to steadily increase from the inscribed equilateral triangle (n =3) to C itself for infinite n. So, do we have something like: "For any C, there is a unique n for which A(n) is a maximum and it is monotonically increases or decreases on either side of that n"?
Or one can wonder: Can one construct a convex C - or even a non-convex region - such that the derivative of A(n) changes sign arbitrarily many times?
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Question 3: Given a general convex region C, to find the largest regular polygon that is totally contained in it. Basically, one needs to find that particular value of n for which a regular n-gon contained in C has the largest area.
A specific case: Given a specific value of n, can one find some triangle such that the largest regular polygon inscribed in the triangle has exactly n sides? One guess is that n cannot be arbitrarily large if C is restricted to be some triangle. iow, the largest contained regular n-gon cannot be the incircle for any triangle.

Remark: This set of questions suggests a way to classify all convex regions - not only polygonal ones - into a countable number of sets given by the value of n such that the regular n-gon inscribed in a polygon is maximum.

Partitioning into Isosceles Triangles

Basic Question: To partition a given polygon (not necessarily convex) with N sides into the least number of isosceles triangles.

All polygons do admit such a partition. And some N-gons might need at least 4(N-2) isosceles triangles.
Indeed, every N-gon has a triangulation into N-2 triangles. Any triangle can be cut into two right triangles and further, any right triangle admits a partition into 2 isosceles triangles (for this, simply join its circumcenter that lies on the hypotenuse to the opposite vertex). So, we have an obvious upper bound of 4(N-2) for the number of isosceles triangles that result from partitioning any given N-gon. And this bound appears tight. For, there are polygons like this:

where any triangulation appears to yield only obtuse triangles. A general obtuse triangle appears NOT to allow partition into less than 4 isosceles triangles.

Further question: What if a given N-gon is to be cut into the least number of acute isosceles triangles?

Here is a discussion on how few acute (not necessarily isosceles) triangles can result when an obtuse triangle is partitioned. An argument is given that says at least 7 acute triangle pieces may result from a single obtuse one.

So to cut any given N-gon into acute isosceles triangles, one can

1. triangulate the N-gon into ~ N triangles.
2. Divide each triangle into 2 right triangles (yielding a total of 2N right triangles) and then partition each right triangle into 2 isosceles triangles - note that except when the right triangle is itself isosceles, one of these 2 isosceles triangles is acute and the other is obtuse. So, at the end of this step, we have 2N acute isosceles triangles and 2N obtuse isosceles triangles.
3. Now, partition each of the 2N obtuse isosceles triangles into 7 acute pieces as described in the above linked page. By the symmetry of the input obtuse isosceles triangle, it is seen that each of the 7 acute triangles is also isosceles. Thus we have a total of 2N + 7 X 2N = ~16 N acute isosceles triangles.

Remaining Question: Is 16N a tight lower bound on the least number of acute isosceles pieces that can team up to recreate any given N-gon?

On Tiling with Isosceles Triangles

A few years back, I wrote this post .
Therein I had wondered in passing: what if the tiles are to be *isosceles and non-congruent* (with various possibilities for area and perimeter)?
Quite recently, I found this very interesting discussion :

A particularly nice observation there, from Noam Elkies: Any acute non-isosceles triangle can be tiled by three pairwise incongruent isosceles triangles, by connecting each vertex to the circumcenter.

An elegant construction follows where a tiling of the plane into pairwise non-congruent isosceles triangles is built up from the above observation.

Remark: One wonders if the construction given there leads to arbitrarily large isosceles triangles being used. Looks like a way out (tiling with non-congruent, bounded size isosceles triangles) can be found based on the construction given at the top of that very page:

Choose any acute and non-isosceles triangle T and tile the plane with copies of it. Now, put a random point in the interior of a unit T to divide it into 3 small acute triangles. Since there are infinitely many ways by which such a point can be put inside T, we can put a point inside each T unit in the tiling to create a tiling of plane with pair-wise non-congruent acute non-isosceles triangles. Now, using Elkies's observation, we could connect the vertices of each triangle in our latest tiling to its circumcenter to create a fresh tiling of the plane with isosceles triangles which appear pair-wise non-congruent.

Note: To my knowledge, no discussion has happened on what happens if in addition to pair-wise non-congruence, equal area requirement is also put on the isosceles tiles. And same appears to be the scene if we replace isosceles triangles with right triangles (It appears that a square can be cut into 4 mutually non-congruent right triangles in infinitely many ways; so to achieve only pair-wise non-congruence with right triangle tiles, we could tile the plane with squares and cut each square in a different way into 4 mutually non-congruent triangles).