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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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 comment on part c | Comment 1 of 20

If the final shape is to be a circle, it would seem that each step of the process should be a regular polygon. If that is in fact the case, then it would be impossible to form a circle, as the same x cannot be used at each step and maintain regularity. The reasoning is as follows:

If the side is taken as 1, the remaining portion of the side will be 1 - 2*x. The chord will be 2 * x * sin(alpha/2), where alpha is the angle of the polygon, alpha = 180 - 360/n degrees. We need to equate the chord with the remaining portion of the side. This program evaluates this for various sided regular polygons:

DEFDBL A-Z
pi = ATN(1) * 4

FOR n = 3 TO 30
alpha = 180 - 360 / n
x = 1 / 3
DO
ppx = px
px = x
x = .5 - x * SIN((alpha / 2) * pi / 180)
x = (x + px) / 2
LOOP UNTIL ppx = x
PRINT USING "## ###  ##.##########"; n; 2 * n; x
NEXT

The table resulting is:

`n1  n2        x 3   6   0.3333333333 4   8   0.2928932188 5  10   0.2763932023 6  12   0.2679491924 7  14   0.2630237709 8  16   0.2598915325 9  18   0.257772801010  20   0.256271407711  22   0.255168049512  24   0.254333095013  26   0.253685824514  28   0.253173797815  30   0.252761725116  32   0.252425139117  34   0.252146637818  36   0.251913566619  38   0.251716542820  40   0.251548489721  42   0.251403987422  44   0.251278828123  46   0.251169704124  48   0.251073986425  50   0.250989563826  52   0.250914725427  54   0.250848072528  56   0.250788452629  58   0.250734909430  60   0.2506866438`

where n1 is the number of sides of the starting regular polygon and n2 is twice that, the number of sides of the resulting regular polygon. So for example, to transform an equilateral triangle into a regular hexagon, x needs to be 1/3. But to change a regular hexagon into a regular dodecagon requires x to be 0.2679491924. And then to change that to a regular 24-gon, requires x to be 0.2543330950.  The above table results are monotonically decreasing, so no matter what regular polygon you start with, the series will require ever decreasing x values.

 Posted by Charlie on 2007-02-27 15:58:44

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