On Reflection Properties of Convex Regions

It is well known that any ray of light passing thru a focus of an ellipse will pass thru the other focus after a single reflection from the ellipse boundary. If A and B are the foci of an ellipse, this property of rays holds both ways (those passing thru A meet at B and vice versa). Two basic queries on this reflection property:

1. Is there a closed convex region C with the property: there exists a pair of points A and B within C such that all rays thru A will reflect once on C and pass thru B but not all rays thru B will pass thru A after one reflection from C?

2. Is there a closed convex region C such that: there is a pair of points A and B in the interior such that all rays thru A pass thru B after exactly 2 reflections from C? Note: This question can have 'one-way' (convergence only of rays thru A at B) and 'both-ways' variants

Update (June 28th, 2020):
See this page for a nice discussion on the above questions:
https://mathoverflow.net/questions/364134/on-reflection-properties-of-convex-regions

Non-congruent Tilings - 8

This post brings to the present (2020) - and would try to build on - the most recent update to this post begun in June 2016 (basically the post has had too many updates over the years so it would be better to bring the latest update to the present and take it forward from here):
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Here is a nice discussion, initiated by Prof. O'Rourke and dating hack to 2015, on the question of tiling the plane with non-congruent isosceles triangles (a question also raised in the above linked old post of mine in a November 29th 2017 update):
https://mathoverflow.net/questions/221431/tiling-the-plane-with-incongruent-isosceles-triangles

This mathoverflow page describes 2 constructions, both showing what appear to be tilings of the plane with pairwise non-congruent isosceles triangles with no upper bound on the size of the tiles. So, one naturally wonders if such a tiling with such an upper bound can be worked out. And of course, area and perimeter requirements on the tiles can be added!

Indeed, we can tile plane with non-congruent isosceles triangles with bounds only on edge lengths: perturb each vertex of an equilateral triangle tiling infinitesimally in a unique direction, to get a tiling with non-congruent acute triangles with every side unique and with both upper and lower bounds on max edge length.. Then, divide each triangle into isosceles triangles by joining its circumcenter with the vertices. That should be it.
However, it looks like further constraints such as equality of area/perimeter would make the problem harder
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A further variant: to tile the plane with mutually non-congruent acute isosceles triangles (without or with area/perimeter constraints).

Indeed, one can relax 'isoscelesness' and go back to the basic non-congruent general triangle tiling question - starting with "area equal among tiles and perimeter unconstrained" (see here):
http://stanwagon.com/potw/2015/p1199.html - and add acuteness of tiles as an additional requirement to generate a new sequence of questions.

Observations: Tiling with mutually non-congruent triangles of equal area and unbounded perimeter that are also all constrained to be strictly obtuse is pretty easy. In the construction shown in the page linked just above, instead of dividing the plane into 4 quadrants, begin with 3 wedges of 120 degrees each and do pretty much the same construction. However, with the tiles required to be strictly acute, things appear much harder.

Non Congruent Tilings - 7

This post continues the train of thought in the non-congruent tiling series. The following questions arose from considering how non-congruence of tiles could go over into Elliptic and Hyperbolic geometry (part 6 of the series - http://nandacumar.blogspot.com/2020/04/non-congruent-tilings-6.html). Note: here, we do not always insist on mutual non-congruence of tiles.

Question 1: Is it possible to partition a spherical surface (simply 'sphere' hereafter) into N equal area spherical triangles for any value of N?

Answer: For even N, one can cut a sphere into N identical triangular pieces, the northern hemisphere into N/2 and southern into N/2.
For odd N, there is no vertex to vertex triangulation into N pieces (indeed, for a vertex to vertex triangulation, we have e = 3f/2 => (by Euler formula) v+f = 3f/2 +2. => v = f/2 +2. The last condition cannot be satisfied if f is odd - v has to be an integer).

Question 1a: If the vertex to vertex restriction is relaxed, can we have a triangulation of a sphere into N triangles for any N? Further, can we have triangulations with all pieces having equal area? And equal perimeter? And yes, mutually non-congruent?

Question 2: If we go for spherical quadrilaterals (or pentagons...) instead of triangles, what happens?

Question 3: If we put no restrictions on the number of sides on the pieces: if I understood the experts right, we can have a fair partition of a sphere into N pieces for any N (all N pieces have same area and perimeter) - the proof (by Akopyan, Avvakumov and Karasev) for the existence of a fair partition into N pieces for convex planar regions goes thru for spherical surfaces as well. Can we also achieve piece wise non-congruence?

Question 4: From sphere, one can go over to torus, cube and so on and ask the same questions as above - relaxing the vertex to vertex condition, equal area pieces, equal perimeter pieces, pieces with both area and perimeter equal, then mutually noncongruence added to these conditions.
Note: For a torus, Euler's formula becomes v+f = e => f cannot be odd - because, for an edge to edge triangulation, v = f/2.

Question 5: For what values of N are there N-faced polyhedrons such that all faces have the same area and same perimeter (and perhaps same number of sides, but that may be too strong a constraint) and are mutually non-congruent? Will allowing candidate polyhedrons to be non-convex make the question easier (guess: probably not)? What about relaxing the equal perimeter (or equal area) constraint alone among the faces?

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Some Basic Results on question 5:
1. It is easy to see that there is no tetrahedron where the 4 faces are triangles of the same area and perimeter but mutually non-congruent. Indeed, each pair of tetrahedral faces share an edge and that plus the faces having same area and perimeter will demand them to be congruent.

2. With mutually non-congruent faces required to have only equal perimeter (areas free), there is no tetrahedron that achieves it.

Proof: Let such a tetrahedron exist and let a, b, c, d, e, f be the lengths of its 6 edges {a,b,c}, {a,d,e}, {b,e,f} and {c,d,f} being the 4 faces as in this picture.

Since all faces should have equal perimeter, we have a+b+c = a+d+e = b+e+f = c+d+f
Taking these equations pairwise, we have the 6 equations:
b+c = d+e
a+d = b+f
b+e = c+d
a+c = e+f
a+b = d+f
b+e = c+f

Now, a face of the tet can have at most 1 edge length in common with another face (if two triangles with equal perimeter also have two sides each with same lengths, then, the third edge will have the same length - due to the equal perimeter constraint. And if two faces have the sets of edge lengths identical, then the faces are congruent). Faces {a,d,e} and {a,b,c} share a. So, we infer that {d,e} has no value in common with {b,c}. Same is the case with the following pairs of edge pairs:
{e,f} and {a,c}
{d,f} and {a,b}
{a,e} and {c,f}
{b,f} and {a,d}
{d,c} and {b,e}
From this property, we infer that a,b,c,d,e,f have to be all different.

From two of the set of 6 equations we noted earlier, select 2:
b+c = d+e
b+e = c+d
Together these 2 imply that e-c = c-e => c=e, which contradicts the above requirement that a,b,c,d,e,f are all different. Done.

3. Guess: Tetrahedrons exist such that faces are pairwise non-congruent and all have same area.

Justification: Two faces with same area plus two side lengths common will be congruent. So, the 6 relations with edge pairs given above for equal perimeter faces case apply here as well. ie. each of the following pairs of edge pairs have no value in common.
{d,e} and {b,c}
{e,f} and {a,c}
{d,f} and {a,b}
{a,e} and {c,f}
{b,f} and {a,d}
{d,c} and {b,e}
So, all edge lengths: a,b,c,d,e,f are different.

Now, Consider a base edge of the tet, say a. Consider the set of triangles with a as base and some specified area A. The third vertex of this triangle can be any point on an infinitely long cylinder with a as axis. Since a,b,c are different, if we form triangles of same area A with each of a,b,c as base, we have three cylinders with a,b,c as axis and with each cylinder having a unique suitable value of radius. Now, it looks pretty obvious that for some value(s) of the area, the radii of the 3 cylinders will be such that the three cylinders all pass through a common point. That's what we need.

Remark: For number of faces above 4, these questions seem much harder.