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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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Comments: ( Back to comment list | You must be logged in to post comments.)
found the error in my irregular quadrilateral program Comment 20 of 20 |
(In reply to re: some progress by Art M)

The program depended on summing the areas of triangles, one vertex of each of which was the "center", while an edge was a segment of the polygon. In the case of the irregular quadrilateral, the "center" was not actually within the figure. Portions that were due to the figure being "scanned" in the opposite direction from the others should have been subtracted out instead of added in as they were.
  Posted by Charlie on 2007-03-04 21:17:24

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