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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(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    
Rating: 5.0000 (1 votes)

Comments: ( You must be logged in to post comments.)
  Subject Author Date
reviewJohnson Alex2022-06-08 03:03:29
re: comment on part cJohnson Alex2022-06-08 03:02:34
found the error in my irregular quadrilateral programCharlie2007-03-04 21:17:24
Summary and a bonus questionArt M2007-03-04 15:58:00
re: I got the solution right this time!!Art M2007-03-04 14:57:35
re: I got the solution right this time!!Steve Herman2007-03-04 10:35:25
I got the solution right this time!!Brian Smith2007-03-03 20:28:15
I was wrongBrian Smith2007-03-03 11:40:29
re: some progressArt M2007-03-02 19:42:44
Hints/Tipsre(2): Corner Cutting FunctionBrian Smith2007-03-02 11:52:49
re: Corner Cutting FunctionJer2007-03-01 13:34:11
Corner Cutting FunctionBrian Smith2007-03-01 13:03:23
re: interesting comparison: regular vs irregularCharlie2007-02-28 22:39:05
interesting comparison: regular vs irregularCharlie2007-02-28 22:09:39
tabulating F for the triangle, square and regular pentagonCharlie2007-02-28 21:39:56
Some Thoughtssome progressArt M2007-02-28 16:10:06
Some ThoughtsLinks and stuffJer2007-02-28 12:40:52
re: my observationsJer2007-02-28 08:32:57
my observationsArt M2007-02-28 01:50:49
Some Thoughtspart dCharlie2007-02-27 22:25:30
Pick a sideGamer2007-02-27 22:24:32
Some Thoughtscomment on part cCharlie2007-02-27 15:58:44
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