Multiple levels of 'hiding' a convex region with copies

This is based on this post at mathoverflow:

The basic question was how to 'hide' a unit cube which is a source of light with opaque copies of itself such that no light reaches infinity. As was not clear to me when the question was raised, this hiding can be attempted at multiple levels. Let us come down to 2D and try to hide a square.

(1) by fixing 4 opaque squares flush with the faces of the light source square. In this case, vertices of the glowing square can emit light that does reaches infinity but what is visible from infinity are only points, no 1-D subset of the square. Here, if one considers the *total outer boundary of the total layout of 5 squares*, the glowing square has a finite intersection with this boundary - the 4 vertices.

(2) as given on this earlier post here (second figure). Here, we hide the glowing yellow squrare with 5 opaque copies (unshaded). In this case, as pointed out by Erich Friedman, rays can possibly squeeze between opaque squares touching each other but unlike (1), the outer boundary of the total layout has no intersection at all with the glowing square. But yes, there can be rays from the glowing square that need only to pass thru just 1 point on the opaque cordon to get to infintiy.

(3) a still more 'secure' cordon with 6 opaque squares given by Prof. Friedman himself in https://mathoverflow.net/questions/468850/how-many-unit-cubes-are-needed-to-hide-a-unit-cube-fully-in-3d . Difference between this and (2) is that any ray from the glowing square has necessarily to pass thru at some interior points of the opaque squares to get to infinity.

IMO, (2) and (3) are both nontrivial with (3) more so. Going to 3D, things can get a lot more interesting.
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Note: problem 7 of section 2.7 in Research Problems in Discrete Geometry by Moser et al refers to every ray emerging from the glowing body meeting either the interior or boundary of the opaque copies. Here, we have separated the question into 2.
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And a different direction: to hide more than one unit square with opaque unit squares - begin with 2!

On some centers of triangles

This post saves a post deleted from mathoverflow:
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We try to go further from https://mathoverflow.net/questions/484443/on-3-centers-of-triangles and record more questions.

Given any point P in a triangular region T, we denote by $d_1$, $d_2$ and $d_3$ the perpendicular distances from P to the 3 sides of T. We denote by $D_1$, $D_2$, $D_3$, the distances from P to the vertices of T.

It is known that the position of P that minimizes the Arithmetic Mean of the $D_1$, $D_2$, $D_3$ is the Torricelli point. But I don't know answers to the following:

1. Given any T how does one characterize the location of P within T that minimizes the arithmetic mean of $d_1$, $d_2$ and $d_3$?

2. Which position of P inside T maximizes the Geometric Mean of $d_1$, $d_2$ and $d_3$? Which P maximizes the GM of the D's?

3. Which P maximizes the Harmonic Mean of $d_1$, $d_2$ and $d_3$? Which P maximizes the HM of the D's?

Partitioning a Convex Island into Convex Regions

Background:
1. This paper discusses the problem of partitioning a convex polygon C into n convex pieces such that each piece has equal area and equal length of the boundary of C: basically to divide a convex cake into convex pieces of same area and same amount of icing per piece.

2. Our 'fair partition' (spicy chicken) question (2006), a departure from the above, asks for partition of planar convex region C into n convex pieces all of same area and perimeter. It was proved by Avvakumov, Akopyan and Karasev (https://arxiv.org/abs/1804.03057) that such partitions do exist for all C's and all n.

Now, let us record a variant of the question that lies somewhere between the above two:

Given a convex planar region C to divide it into n equal area convex pieces such that each piece has the same length of cut. Basically, to partition a convex island into n convex parts, with equal length of wall needed by each piece - the coast needs no walls or fortification. And some of the portions could be landlocked - not have any coastline.

Remarks: I gather that the existence proof of the fair partition question (mentioned above) also guarantees the existence of a solution to the new question. Initially, there was some doubt as to the case where C is a convex polygonal region - ie not strictly convex but now, it seems clear that the cut length of each piece in a partition will evolve continuously with the piece itself and that the proof can handle it. So, it appears that this new variant is not really much of an issue. Still, I am keeping this post.

Marking out Geodesically convex regions on surfaces of polyhedrons

We refer to any subset S of the surface of any convex solid C as being geodesically convex if for any two points on S, the shortest line on the surface of C that joins them is contained entirely in S - and if there are more than one such shortest path, at least one is entirely contained in S.

Question: Consider any convex 3d solid. We need to mark out its surface into n regions all of equal area such that each region is geodesically convex. First of all one needs to decide if a given convex polyhedron's surface can be so marked out for a specified n - an algorithmic decision question.

It appears that the surfaces of all convex solids of revolution about an axis can be marked out into n such regions for any n.

It seems unlikely that the surface of a regular tetrahedron can be marked out into 5 geodesically convex equal area regions. For even a cube, it is not clear if such a marking out of the surface can be done for any n.

Note: This question was posted at mathoverflow a few days back and simply disappeared!

To rig coins and cubes

Sometime back, I put up a couple of questions at mathoverflow and they got deleted. I believe the questions deserve to survive.

1. Given a fraction α > 1/2, how does one make a coin for which probability of landing heads in a fair toss is α ?

Basically, we need to 'rig' a coin to a specified extent. By coin, we mean a cylinder with low but non-zero height (compared to radius). Obviously, the density should vary within the body of the coin. The radius to height ratio could be chosen suitably so as to achieve the required α . I am not sure what assumptions should be made of the ground on which the coin lands.

2. How does one 'rig a cube' such that the six faces have some specified set of six probabilities on a toss - for example, 5 of the faces could have equal probability but the sixth has a slightly greater probability. This seems a clearer question in that there is no confusion about the dimensions of the object.
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The questions (espcially the one on coins) had elicited some comments:

- "Fair coins do not really give probability 1/2 for heads. They actually tend to land on the side they started with."
(but as far as I can make out, if one performs the same number of tosses with heads and tails up, this bias could get cancelled).

- "People commonly expect that coin tossing would be mainly affected by the weighting (perhaps just the location of the mass center). Not so; E. T. Jaynes wrote about how the method of tossing is more crucial. He devised three different tossing methods that lead to quite different heads probabilities for the same coin."
(but again, for a fair coin, these biases all seem cancellable by following whatever is the 'complement' of each biased tossing method).

On the whole, my feeling is that for a uniform and equally weighted coin, there are indeed ways of tossing to ensure that as the total number of tosses goes to infinity, the probability of heads does get to 1/2. Then question 1 could be rephrased: how does one alter a coin in such a way that the very manner of tossing that yields 1/2 for heads for a fair coin would yield α probability for heads with this new coin?

An addition to a mathoverflow post

We add to - generalize - this post at mathoverflow and add a couple of questions. The mathoverflow post is on central symmetry. We here talk about any symmetry:

Note: in what follows, instead of "mirror symmetry", one can think of any other symmetry.

1. Which convex solids have the property: all planar sections have mirror symmetry?

2. If a convex solid has the property: all its maximal area planar sections - sections of max area with normal in any specified direction - have mirror symmetry, what can be said about the solid?

Note: in both above questions, we can replace max area planar sections with max perimeter planar sections. One can also consider shadows instead of planar sections; and solids for which both shadows and planar sections have some specified symmetry.

Non-congruent Tilings - 21

Although plenty has been written on this thread, a pretty obvious question seems to have escaped one's attention (at least, looking through all the earlier posts one didn't spot any trace thereof!). Here goes.
The main theorem proved by Kupaavski, Pach and Tardos on the basic question:

Theorem: There is no tiling of the plane with pairwise noncongruent triangles all of the same area and the same perimeter

Question: Can the plane be tiled by triangles all of same area and perimeter that are pair-wise non-congruent except that exactly 2 of them are mutually congruent? Of course, '2' can be generalized to 'finitely many mutually congruent triangles' or 'finitely many pairs of congruent triangles' etc.

This has been put up at mathoverflow: https://mathoverflow.net/questions/469784/trying-to-extend-a-theorem-on-tiling-with-mutually-non-congruent-triangles