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Cutting Corners (Posted on 2007-02-27) Difficulty: 5 of 5
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()=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    
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re: my observations | Comment 5 of 20 |
(In reply to my observations by Art M)

I did the calculation for the midpoint of the segment after the second iteration.  The distance is

sqrt((13/4)-(17/8)sqrt(2)) which is about .49477,

so a circle is impossible from a square.

As far as finding a function goes, this looks very difficult.  I was able to make a plot of r vs F(r) and the result resembles a portion of a bell curve (normal curve.)  I feel this resemblance is probably superficial.  I will post a picture and link later today.

If anyone has geometer's sketchpad and would like the file I used, I'd be happy to share that too.


  Posted by Jer on 2007-02-28 08:32:57
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