Non Congruent Tilings - 19

Here is the latest episode in this series.



And here is a mathoverflow discussion. The answers show tilings of the plane with mutually similar right triangles of unbounded size. An old instalment of the present series had touched upon the question of tiling the plane with mutually non-congruent but similar triangles.

Question 1: Can we tile the plane with right triangles that are similar to one another but with size bounded?

Question 2: Is there any other shape of triangle, similar and non-congruent copies of it can tile the plane?
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Update - 5th may 2023:
Here is another mathoverflow discussion.

Question: Is it possible to tile the plane with triangles that are (1) mutually similar, (2) pairwise non-congruent and (3)non-right? No other constraints.

Note 1: Reg requirement 3 above: since any right triangle can be cut into 2 pieces similar to itself and non-congruent, we can start with any right triangle and grow a plane-tiling layout outwards with progressively scaled copies of itself.

Note 2: If the question has a "yes" answer, one could consider a more constrained version of the question is got by demanding the tiles to have bounded size. Even if the basic shape of the triangle is a right triangle, this constrained version could be asked.

Another Question on congruent partitions

A few days back, I put up a question at mathoverflow.

Definition: A perfect congruent partition of a planar region C is a partition of it into some finite number n of pieces that are all mutually congruent (any piece can be transformed into another piece by an isometry. We consider only cases where each piece is connected and is bounded by a simple curve) and with no part of C left out (not part of any piece).

Question: If from a convex planar region C, 2 connected, mutually congruent pieces (pieces need not be convex) can be cut such that no point of the boundary of C is left out, then will C always allow a perfect congruent partition into 2 pieces (with no point on the entire C left out)? I can't think of a counterexample or proof.

Note: The question can be generalized from 2 pieces to n pieces.

Partial Answer: For n >=5, the answer is "not necessarily".

Example for n = 5 (Obviously generalizable to higher values of n). Consider a square S and a slightly smaller square S' got by insetting S. C is a convex region got by replacing each edge of S with an outward bulging arc of very large radius - ie very close to a line segment.

Now, consider cutting C into 5 mutually congruent pieces. It is pretty much obvious that no perfect congruent partition of C into 5 pieces is possible. It is also easily seen that the entire boundary of C can be 'used' by dividing that part of C between its boundary and S' (S' obviously is in the interior of C) into 4 mutually congruent thin pieces. What remains of C after these 4 pieces are cut is the interior of S'. There is more than enough space there to cut another thin piece congruent to the 4 already cut. So, we can have 5 mutually congruent pieces that can be cut from C that together use the entire boundary of C (in fact, only 4 of the 5 can do it) but we cannot cut C entirely into 5 mutually congruent pieces.

This simple construction generalizes to n =6 (just start with a regular pentagon instead of square) but does not seem to give a clue for n =4,3,2.

Oriented Containers (again) - Biconvex Lenses

We add one more link to the chain of Oriented containers, the latest of which was here .
And there are two arix preprints on this topic - both done some years ago: this and this

Definition: A biconvex lens is the intersection of two circular disks - not necessarily of the same radius. It is an oriented convex region - the orientation given by the line joining the centers of the intersecting disks or its perp
endicular.

Question: To find the smallest (least area or least perimeter) biconvex lens that contains a set of n points on the plane.

A preliminary approach:
- Find convex hull of the set of n points.
- Find circles formed by 2 triplets of points on this hull and find their intersection - a biconvex lens. For each intersection check if all input points are on or within the lens. If not, reject this intersection.
- From among the lenses found in previous step that do contain all n points, pick the smallest.


The algo is polynomial but seems to have complexity O(n^7) in the worst case. Am reminded of David Mount's beautiful treatment of finding the smallest containing disk for a set of n points - a linear(!) time algo is possible there. Such a randomized incremental algorithm might be possible here as well with massive reduction in complecity.

There must be much faster algorithms!

And a constrained version of above problem - both arcs forming the lens reqd to be of equal radius - could also be interesting.

Note: Just like the biconvex lens, one can think of a crescent which is also the (concave) intersection of the interior of one disk with the exterior of another disk.

Yet another thought: Among all the oriented containers listed in all those pages and preprints - isosceles triangles, right triangles, rectangles, semicircles, ellipses, sectors, circle segments, trapezoids and now lenses - only the semicircle certainly has the property: "for any spread of points, the least perimeter container and least area container have the same orientation." (ie if "container" means semicircle"); but then, maybe the circle segment and sector too have this property. Perhaps even lenses.

And of course for every container question, there could b a containee - largest oriented convex regions that can be drawn inside convex regions!

Note added on May 28th 2023: Recall from this earlier post that most, if not all, above questions can have inside out versions. For example: Given a circle segment S, find the smallest (largest) convex region form which S is the smallest containing (largest embedded) circular segment. Smallest could be in terms of area of perimeter.

Non-regular tilings of Hyperbolic plane

We add a bit to https://mathoverflow.net/questions/398191/which-polygons-tessellate-the-hyperbolic-plane.

Background: It is known that there is a tiling of the hyperbolic plane by regular hyperbolic n-gons (n-gons with all angles and sides equal) with k polygons meeting at each vertex iff 1/n + 1/k < 2. So, for any n, there are regular n-gons that can tile the hyperbolic plane. Further, if the internal angles of a hyperbolic n-gon are equal, then the lengths of its sides are also equal.

**Question:** Given a number n, how does one find some non-regular hyperbolic n-gon that tiles the hyperbolic plane? One obvious way is to find any regular (2n-4)-gon that tiles the plane and cut it in half. What one would want to ask is if there are non-regular hyperbolic tiles which cannot be got in this way. For example, is there any hyperbolic quadrilateral tile with all edge lengths different?