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 Cutting Corners (Posted on 2007-02-27)
Take a polygon with area S1 and pick a number r in [0,1/2]. Take vertex A that connects sides AB and AC and add points M and N on these sides so that AM/AB=AN/AC=r. Cut corner A along MN. Cut all other corners the same way.

After repeating these steps infinite times we will get a figure with an area S2. Let's F(r)=S2/S1. It's clear that F(0)=1 and F(½)=0.

Questions:

(a) What is this function for square?

(b) What is this function for equilateral triangle?

(c) Is it possible to get a circle from a square or from an equilateral triangle this way?

(d) Is it possible that this function is universal for all triangles, or for all rectangles, or for all polygons?

 No Solution Yet Submitted by Art M No Rating

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 re: some progress | Comment 14 of 20 |
(In reply to some progress by Art M)

Using the two last formulas in my previous post we can link F(r) for triangles and quadrilaterals:

F4(r)=(1+2*F3(r))/3.

Calculating F4(r) from here using Charlie's numerical F3(r) values, we get for F4(r):

0.01             0.999796002
0.03             0.998092412
0.05             0.994505495
0.07             0.988858572
0.09             0.980994838
0.11             0.970787059
0.13             0.958147598
0.15             0.943037975
0.17             0.92547705
0.19             0.905546834
0.21             0.883395029
0.23             0.859233635
0.25             0.833333334
0.27             0.806013838
0.29             0.777630884
0.31             0.748560963
0.33             0.719185147
0.35             0.689873418
0.37             0.660970778
0.39             0.632786094
0.41             0.605584233
0.43             0.579581628
0.45             0.554945055
0.47             0.531793133
0.49             0.510199918

which agrees with Charlie's calculations for F4(r).

The approximation for the function F0(r) is (from Charlie's F3 values):

0.01             0.999592003
0.03             0.996184824
0.05             0.989010989
0.07             0.977717144
0.09             0.961989676
0.11             0.941574119
0.13             0.916295196
0.15             0.886075949
0.17             0.8509541
0.19             0.811093669
0.21             0.766790059
0.23             0.718467271
0.25             0.666666668
0.27             0.612027676
0.29             0.555261767
0.31             0.497121927
0.33             0.438370294
0.35             0.379746836
0.37             0.321941555
0.39             0.265572187
0.41             0.211168466
0.43             0.159163256
0.45             0.10989011
0.47             0.063586265
0.49             0.020399837

For regular pentagon we get F5(r)=sin(54)^2+cos(54)^2*F0(r):

0.01             0.999859041
0.03             0.998681889
0.05             0.99620339
0.07             0.992301463
0.09             0.986867756
0.11             0.979814354
0.13             0.971080701
0.15             0.960640209
0.17             0.948505908
0.19             0.934734468
0.21             0.919427947
0.23             0.902732834
0.25             0.884836166
0.27             0.865958859
0.29             0.84634672
0.31             0.826259899
0.33             0.805961709
0.35             0.785707802
0.37             0.765736569
0.39             0.746261431
0.41             0.727465408
0.43             0.70949805
0.45             0.692474596
0.47             0.676477011
0.49             0.661556467

agrees with calculations for regular pentagon.

According to my post, F4(r) is the same for all quadrilaterals, so either I'm wrong  or Charlie's calculations for irregular quadrilaterals.

1-F0(r) function is the same as the function f(r) in Brian Smith's post but the values don't agree with the numerical values above.

 Posted by Art M on 2007-03-02 19:42:44

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