Oriented containers - kites

This continues an old thread on oriented containers with latest episode here

How does one find the least area convex kite that contains a given convex polygon P? What about the least perimeter kite container?Isosceles triangles may be treated as degenerate kites

Which is the convex shape that maximises the difference in orientation (measured by angle between the lines of symmetry of the two containers) between the smallest area and smallest perimeter kite containers?

Note 1: Unlike in the case of the smallest area containing rectangle for a convex P, the smallest enclosing kite might not share an edge with the polygon - if P is a thin rectangle, a containing kite that shares an edge with it appears suboptimal. well not sure!

Note 2: one can also ask about the largest area/perimeter kite contained inside a given convex polygon.



if the smallest area kite container too shares an edge with P, one can ask if there is any type of container where the optimal container need not share an edge with P.

One more non congruent tiling thought

Question:

Is there any set of polygons that have the same angle set with each member individually incapable of tiling the plane but together can manage to tile?

Obviously, the polygons in the set should be pentagons or hexagons...

A further constraint would be: elements of the polygon set should not only possess the same angle set but be pairwise non-congruent.

And what if we constrain "angle set" above to "angle sequence"?

Packing - some more thoughts

1. Which convex hexagon is worst for packing (ie leaves largest fraction of plane open)?

2. it has been conjectured that the regular heptagon is the worst convex region for packing. Is it known if regularity is significant? Indeed, for hexagons, regularity gives a perfect pack.

So, one can ask: among all convex heptagons, is the regular one the worst for packing?

Further, is there any reason to believe that the worst packing region has SOME rotational symmetry?

And what is the connection, if any, between a body being bad for packing and bad for COVERING?

Stretching Fair Partitions - 2

We add another very speculative claim to those in the last post.

"If a convex planar region C has the property for any positive integer n that at least one convex fair partition of C into n convex pieces has at least 2 (or maybe 3, say) of the pieces mutually congruent, then for any n, C allows a convex fair partition into n pieces that are all mutually congruent - and further, C is either a sector of a disk or parallelogram."

Stretching the Fair Partition question

Can one make claims of the following type?
"If some convex region C allows partition into n convex pieces all of equal area, perimeter and one more quantity, say diameter or least width for all values of n (or infinitely many values of n), then all pieces are necessarily congruent."

If the above is true, one can stretch things a bit and guess: "If all pieces are congruent for all n, C is necessarily a sector of a disk (with the full disk as a limiting case) or a parallelogram (including rectangles). If 'for all n' is relaxed to 'infinitely many values of n', one also has the case of C being a triangle." This latter guess was once posted at mathoverflow.

Note: perimeter and diameter can be nonzero even when a polygon is degenerate but not area or least width. Basically the question/claim is about 3 quantities being equal among pieces (with 2 of the quatities being like perimeter and one like area or vice versa). If 3 quantities being equal isnt enough for the congruence claim to hold, consider 4!

Smallest quadrilateral containing a set of points

I am not sure how to find the least area/ least perimeter triangles that contain a set of points on the plane. The guess for both would be that the triangle has an edge coincident with at least one side of the convex hull of points so a plane sweep kind of approach would work.

An apparently more difficult question: How does one find the least area quadrilateral containing a set of points? The quadrilateral is allowed to be non-convex. If the quad is required to be convex, we might be able to manage with a plane sweep type of algo.

Non-congruent Tilings - 22

Some more tiling questions occured recently. Simply recording them. Hopefully they don't feature in earlier episodes of this series, the most recent being this and this.
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1. Can the plane be tiled by mutually non-congruent triangles all of equal area and with all three edges unique? If possible, one could demand the edge lengths to have an upper bound.
(This question was raised at this earlier episode) and here is another discussion that appears closely related but for the equal area requirement.

2. Can the plane be tiled by mutually non-congruent triangles all of same area and having one angle common?

3. Can the plane be tiled by mutually non-congruent triangles all having two sides common? No equal area constraint here.

4. Can the plane be tiled by mutually non-congruent triangles all with one side and one angle common? No equal area constraint.

And so on...there seems to be no end to possibilities...