# Infinitely simple polygon solipsism

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.)

• Why a simple insets of polygons will not work. The relation and removal of polygons need to be better defined. – Moti Feb 20 '17 at 5:40
• You could just have insets of similar polygons... and remove the largest – Moti Feb 20 '17 at 6:07
• That would leave a hole. Or did I understand? – humn Feb 20 '17 at 6:07
• 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. – Moti Feb 21 '17 at 4:51
• @humn This paper seems to show that your intended right-angled solution is optimal because no others are possible. – xnor Feb 21 '17 at 23:48

A solution with ratio $\varphi^\frac12\approx1.27201965$, where $\varphi$ is the golden ratio:
Set $x=\varphi^{-\frac12}$. Then: