Non-congruent Tiling - 15

Here is part 14.
Question: For any positive real r1 and r2 with r2 < r1, there are infinitively many triangles with circumradius r1 and inradius r2. Is it possible to tile the plane with mutually non-congruent triangles all of same circumradius and inradius?

Note: I am not sure if we can tile the plane with (a) mutually non-congruent tiles all with same circumradius (b) with same inradius. This less constrained problem might have a "yes" answer.

This question has been put up at mathoverflow .

Further questions:
1. What are the invariants across triangles which have both circumradius and inradius same?
2. Do specifying the Steiner circumellipse and inellipse fully determine a triangle?
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A further mathoverflow post (July 3rd 2022) recorded these questions:

Given an ellipse E:

can we choose from those triangles for which E is the Steiner circumellipse, a pair-wise non-congruent set of triangles that tile the plane?

Can we choose from those triangles for which E is the Steiner inellipse, a pairwise non-congruent set of triangles that tile the plane?

Further query: As was noted in above linked page, given two positive real numbers r1 and r2 with r2 < r1/2. For any such pair, we observe that one can form infinitely many triangles all with circumradius r1 and inradius r2. Under what condition(s) will two specified ellipses be the Steiner circumellipse and inellipse of infinitely many triangles (or a unique triangle)? Basically, how do we characterize those triangles for which two specified ellipses are the Steiner ellipses?
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Let me also record a bunch of variants of the original fair partition question:

Question: Given a positive integer n, will every planar convex region allow partition into n convex pieces all with (a) same smallest containing circle (b) same largest inscribed circle and (c) both smallest containing circle and largest inscribed circle same? Partition into pieces with same largest inscribed circle looks plausible and easier by means of strips.

Here is what I gathered from Prof. Roman Karasev reg the status of the fair partition/spicy chicken problem:
First, the non-prime-power case of the spicy chicken theorem is so far only provable for the situation when one of the quantities being equalized over pieces is area (or something else continuous, zero on degenerate pieces, and additive) and the other value is continuous.
The prime power case works for some cases: If the two quantities we are trying to equalize are continuous and one of them is zero on degenerate pieces then a fair partition is also possible for prime power values of n.
Hence if you inscribe a circle or an ellipse, then its size and area are zero for a degenerate piece, which seems to do the job.