4
$\begingroup$


           Solipsism — The self is all that can be known to exist.


Above is a simple polygonal region divided into infinitely many different-sized copies of itself.   Each copy is √2 = 1.414... times as large as the next smaller one (in terms of linear scale, not area).   If the largest copy is removed, the remaining polygonal region is a scaled-down version of the original.


Can you find another simple polygon that has 4 or more sides and can be divided into infinitely many different-sized copies of itself, where the original polygonal region is geometrically similar to what remains if the largest component copy is removed?

The open-ended goal is a maximum successive-size ratio as close as possible to 1.


Reflection is allowed.   Each copy size occurs only once.   Polygons in this puzzle have _finitely_ many vertices.   Note that the goal is to minimize the _maximum_, not average or smallest, ratio between any two successively sized copies.   The large composite polygon is not included in these ratios.   A neat solution with only right angles and a maximum ratio less than 1.3 is known at pose time.

(This puzzle is similar, but with different conditions, to Unreflected Infinitely simple polygon reflexivity.)

$\endgroup$
8
  • $\begingroup$ Why a simple insets of polygons will not work. The relation and removal of polygons need to be better defined. $\endgroup$
    – Moti
    Feb 20, 2017 at 5:40
  • $\begingroup$ You could just have insets of similar polygons... and remove the largest $\endgroup$
    – Moti
    Feb 20, 2017 at 6:07
  • $\begingroup$ That would leave a hole. Or did I understand? $\endgroup$
    – humn
    Feb 20, 2017 at 6:07
  • 1
    $\begingroup$ You require that every time a nested rectangle will be left - each rectangle, based on your drawing - share a side with another rectangle and all similar - means that the ratio of the sides of the rectangle is $sqrt\{2}$. There are infinite such families of 4 - all parallelograms that are resulting from selecting any desired angle between the sides. $\endgroup$
    – Moti
    Feb 21, 2017 at 4:51
  • 1
    $\begingroup$ @humn This paper seems to show that your intended right-angled solution is optimal because no others are possible. $\endgroup$
    – xnor
    Feb 21, 2017 at 23:48

1 Answer 1

8
$\begingroup$

A solution with ratio $\varphi^\frac12\approx1.27201965$, where $\varphi$ is the golden ratio:

Set $x=\varphi^{-\frac12}$. Then:

The numbers work out due to how the golden ratio works. The outer numbers define the dimensions for the entire hexagon. Note that I did not come up with this polygon; it is called the Golden Bee and was originally devised by Karl Scherer in his 1987 book. This is probably optimal.

$\endgroup$
1
  • $\begingroup$ Right on! (Incidentally, Golden Bee was originally published, not devised, in the mid/late 1980s. I called it PP in 1983 and assumed that others had found it too, as turned out to be the case.) $\endgroup$
    – humn
    Apr 6, 2017 at 18:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.