Area of Generalized Koch Snowflake

I asked on the Math Stack Exchange here how I could find the area of a "generalized Koch snowflake". An $$n$$th generalized Koch snowflake, in my case, is formed almost the same as the Koch snowflake - but instead of starting with an equilateral triangle, the start is a regular $$n$$-gon and the next iteration is formed by placing a regular $$n$$-gon with a side length $$1/3$$ of the previous side length on each side of the previous iteration.

How can I find or at least approximate this area given $$n$$ and the starting side length? I think maybe using polygons is the way to go, but I am new to Mathematica and don't really know how to use them.

• It might be possible to do this with GeometricScene, although I haven't put in the time to create it. Perhaps something useful to be gleaned at this answer (and the question in general for other methods of generating scenes) – Carl Lange May 12 at 19:51
• You may also find KochCurve interesting, and there are many Demonstrations relating to Koch curves and snowflakes. – Carl Lange May 12 at 19:59 Here is a start. KochCurve transforms a unit vector {{0,0},{0,1}}. This unit vector can be geometrically transformed via FindGeometricTransform into any polygon side. So given a side of a polygon you 1st FindGeometricTransform of unit vector into it and then transform with it KochCurve. The following function can transform any line into KochCurve:

ClearAll[lineKoch];
lineKoch[n_][vec_]:=
Last[FindGeometricTransform[vec,{{0,0},{1,0}}]][
ReflectionTransform[{0,1}][First[KochCurve[n]]]]

And the following function using lineKoch will transform any set of polygon vertex points into a n-Koch MeshRegion:

ClearAll[polyKoch];
polyKoch[n_][pt_]:=
MeshRegion[#,Polygon[Range[Length[#]]]]&@
Flatten[lineKoch[n]/@Partition[pt,2,1,1],1]

You need MeshRegion because Area or RegionMeasure can get the approximate area. You can see it at work, by applying it to some random point-sets:

Labeled[#,Area[#]]&[polyKoch[#]]&/@
Table[(#[[Last[FindShortestTour[#]]]]&@
RandomReal[1,{RandomInteger[{3,7}],2}]),10]

which will produce the image at the top of this post. You can also design an interactive app to change your polygons:

Manipulate[
Column[{Area[#],Show[#,PlotRange->2]}]&@polyKoch[n][pt],
{{n,2},Range[2,5]},
{{pt,CirclePoints},Locator,LocatorAutoCreate->True}] UPDATE: other Koch shapes

To change Koch substitution shape use proper syntax of KochCurve function:

shape={{1,0},{1,90°},{1,-90°},{2,-90°},{1,90°},{1,90°},{1,-90°}};

lineKoch[n_][vec_]:=
Last[FindGeometricTransform[vec,{{0,0},{1,0}}]][
ReflectionTransform[{0,1}][First[KochCurve[n,shape]]]]

polyKoch[n_][pt_]:=
MeshRegion[#,Polygon[Range[Length[#]]]]&@
Flatten[lineKoch[n]/@Partition[pt,2,1,1],1]

Labeled[#,Area[#]]&[polyKoch[#]]&/@
Table[(#[[Last[FindShortestTour[#]]]]&@
RandomReal[1,{RandomInteger[{3,7}],2}]),10] • How can I change it from adding triangles to adding other polygons? – automaticallyGenerated May 15 at 23:22
• @automaticallyGenerated I do not understand your question. In this post you see demos with arbitrary polygons. What do you mean? – Vitaliy Kaurov May 15 at 23:24
• I see that it starts off with arbitrary polygons, but how can I change it so that it adds a square/regular pentagon, hexagon, etc in the middle of each side instead of an equilateral triangle? Sorry if this is a basic question - I'm new to Mathematica. – automaticallyGenerated May 15 at 23:26
• @automaticallyGenerated I added example at the end. Please look through examples in docs on KochCurve --- there is syntax of this function to make arbitrary-shape replacements . – Vitaliy Kaurov May 15 at 23:38