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The family of rectangles has the following nice properties:

  1. Every rectangle $R$ can be divided to two disjoint parts, $R_1 \cup R_2 = R$, such that both $R_1$ and $R_2$ are rectangles (i.e. belong to the same family).
  2. Moreover, for every number $p > 0$, it is possible to cut $R$ to two disjoint rectangles $R_1 \cup R_2 = R$, such that the ratio of their areas is $p$.

What other non-trivial families of geometric shapes have one or two of these properties?

  • Triangles have properties 1 and 2 - a triangle can always be divided to two triangles using a ray from one of its vertices.
  • Right-angled triangles have property 1, because they can be divided using a ray from the 90-degree vertex that is perpendicular to the opposite side, and the result will be two right-angled triangles. However, they don't have property 2 because the ratio between the two parts cannot be changed.
  • Squares have none of these properties, as a square cannot be divided to two squares. Ditto for circles.
  • L-shapes seem to have properties 1 and 2, as well as rings.

This raises several questions:

  • Is it possible to give a complete characterization of such families (i.e. find all families with the above properties)?
  • Given a family of shapes, how can we find a minimal containing family that has the above properties? (e.g. for squares, a possible extension is the rectangles; another extension is the L-shapes).
  • (terminology) Is there a formal term for these two properties? If not, what could be a good term? Maybe "closed under division" / "closed under arbitrary proportion division"?

MOTIVATION: When designing algorithms for geometric shapes, if the shapes have property 1 (and possibly property 2), then it is possible to design a recursive algorithm. Otherwise, a recursive algorithm is not possible and we must think of other algorithms.

EDIT: A concept closely related to my property #1 is rep-tile. A rep-tile is a shape that can be divided to two or more smaller copies of itself. This concept was recently generalized to a rep-tile set by Sallows. Specifically, all rep-2 shapes and all rep-tile sets with 2 elements have property #1. But property #1 is more general, as it does not require that the parts will be copies of the original (for example, a rectangle can be divided into two rectangles that are rectangles, although they are not smaller copies of the original).

Property #2 is still open - I haven't found anything similar.

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    $\begingroup$ In many cases, a recursive algorithm based on subdivision is not going to be effective unless the subdivided shapes are actually smaller (in diameter, not just area). From that point of view, rectangles and triangles and L-shapes are nicer than "rings". Note that rings of all shapes will work if you don't impose a diameter condition… circles, ovals, squares, etc. $\endgroup$
    – mjqxxxx
    Oct 31, 2013 at 18:34
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    $\begingroup$ It seems that if the family of shapes consists of shapes with too much symmetry, then it would not satisfy property 1. Squares, circles, equilateral triangles all have radial symmetry. I think dividing any such shape into two cannot yield two shapes of the same type. This may be a possible plan of attack. Of course in the case of rings, this is not quite true, but they have a hole, so this allows us to cut in a circle. We wouldn't be able to do this to a disk since one shape would have a hole. $\endgroup$ Nov 6, 2013 at 8:34
  • $\begingroup$ Just saw this similar question: math.stackexchange.com/questions/545637/… $\endgroup$ Nov 6, 2013 at 10:51
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    $\begingroup$ I just wanted to mention that when I was thinking about this problem yesterday I had an idea. Suppose we cut the shape $R$ with piecewise-smooth curve. We can use the idea of curvature to show that a circle cannot be decomposable into two circles. Simply the curvature of the cut needs to be positive on one side and negative on the other but two circles both have positive curvature. If objects don't contain lines you can show that your family of objects needs to contain objects with positive and negative curvatures. $\endgroup$ Nov 6, 2013 at 18:00
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    $\begingroup$ None-circular rings (as in elliptic rings or just any star-shaped region rings) could work as well as ordinary rings. $\endgroup$
    – Maesumi
    Nov 7, 2013 at 18:10

1 Answer 1

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This is not a complete answer, but here are some thoughts that may be useful.

These families are closed under unions; if families $A$ and $B$ have this property, so does $A\cup B$. (Infinite unions work as well.) This suggests to me that a "nice" characterization of these families may be unrealistic; it would have to include things like "shapes which are either a rectangle or a triangle", which is not a very natural category.

Perhaps you could hope to focus on "prime" families, for which no nonempty strict subset of them is a valid family, but a problem is that there exist families that have no prime subsets; see some of the cases below.

Here are several more families not listed in your post. I'll assume boundedness in all cases, and only specify shapes up to similarity rather than congruence (otherwise we could have examples like "every subset of the following region").

  • All convex shapes.

  • Circular sectors of central angle at most $\theta$, for any $0<\theta\le2\pi$.

  • All parallelograms with an angle of $\theta$ for any $0<\theta<\pi$, and hence also "all parallelograms". Rectangles are the case $\theta=\pi/2$.

  • Triangles whose smallest angle is at most $\theta$, for any $0<\theta<\pi$. Also "strictly less than $\theta$", for which the case $\theta=60^\circ$ gives us the family of non-equilateral triangles.

  • Obtuse triangles.

  • Scalene triangles. (Not positive about this one.)

  • All triangles which are neither equilateral nor right triangles. (Just excluding right triangles doesn't work!)

  • Trapezoids.

  • Quadrilaterals.

  • Non-convex quadrilaterals.

  • Strictly convex quadrilaterals (all internal angles less than $180^\circ$).

  • Convex polygons with at most $n$ sides, for all $n\ge3$ (and including $n=\infty$).

  • Non-triangular convex polygons with at most $n$ sides, for all $n\ge4$ (and including $n=\infty$).

  • Convex shapes whose boundaries are a combination of line segments and circular arcs.

  • Shapes with a line of reflectional symmetry.

  • For any two families $A$ and $B$, the set of disjoint unions of a member of $A$ and a member of $B$. (Similar constructions abound for more than two base families, "two $A$s and three $B$s", etc.)

Some non-examples, and their counterexamples where not obvious:

  • Acute triangles.

  • Isosceles triangles.

  • Rhombuses.

  • Strictly convex pentagons; in fact, no strictly convex pentagon is the union of two others! This applies to strictly convex $n$-gons for all $n\ge 5$, as well.

  • All triangles whose angles are not $(50^\circ,60^\circ,70^\circ)$. (An equilateral triangle needs to use one of these to satisfy property $2$.)

  • All triangles without $10^\circ$ angles. (Again, an equilateral triangle fails here.)

The variety of these examples seems like further evidence to me that a complete classification is probably not feasible.

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