20210831, 00:52  #23 
Jan 2017
2^{3}×3×5 Posts 
Seems to match the values I calculated (1.90317905 * 50^2 = 4757.947625).

20210831, 05:41  #24  
Romulan Interpreter
Jun 2011
Thailand
2^{2}·3·5·163 Posts 
Quote:
(or one single big ark, depending on the available fence, which at the extreme, yes, it is a line at the end of the parcels) Last fiddled with by LaurV on 20210831 at 05:50 

20210831, 05:53  #25  
Romulan Interpreter
Jun 2011
Thailand
2634_{16} Posts 
Quote:
Nice work btw. 

20210831, 11:11  #26 
Jan 2017
2^{3}·3·5 Posts 
If by "angle at the center of the lake" you mean the angle of the lake sector between the fence endpoints, that was in my earlier post: it's 1.4362668, which is less than 90 degrees.
There is a relevant 90 degree angle involved however: I think the fence should always meet the lake boundary at a 90 degree angle (with sharp corners counting as all directions they turn over). 
20210831, 11:28  #27 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
10009_{10} Posts 
https://en.wikipedia.org/wiki/Oxbow_lake
This may have an impact on how much land a 200 meter fence and a lake or river can encompass. 
20210831, 15:29  #28 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
3^{2}·643 Posts 

20210831, 17:29  #29 
Jan 2017
1111000_{2} Posts 
What's your point? This was a question about a circular lake. Yes, the answer is smaller than in the straight line case (a half circle is the optimal answer there). So what?

20210901, 00:48  #30 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
7×1,367 Posts 
It seems that the visitors of this thread, after this trivial warmup, are now ripe to admire the real beauty.
Behold!  The couch problem Hammersley's solution had an attractive beauty to it  look at the value  it is , but it was not the limit. Joseph Gerver holds the champion title since 1992 but rigorous proof has been elusive. 
20210901, 01:41  #31 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2151_{10} Posts 
There seems to be similarities to the Trammel of Archimedes dating back to at least 5th century:
The Sofa with hinged armrests. https://en.wikipedia.org/wiki/Trammel_of_Archimedes https://youtu.be/Hfw0yYur5S4 Last fiddled with by a1call on 20210901 at 01:45 
20210901, 12:32  #32  
Feb 2017
Nowhere
2^{2}×29×43 Posts 
Quote:
Assuming the "meets at right angles" idea is right, one of the boundaries is a circle of radius 1, and the fencing is a circular arc, then the radii of the two cicles that meet at the fence posts meet at right angles, and each is tangent to the other circle. So if is the angle between the boundary radius to a fence post and the line between the centers of the two circles, the radius of the other circle is . The angle between a fencing radius to a fence post and the line of centers is . This leads to fairly simple formulas for the radius of the fencing circle and the area it encloses. As to proving the "right angles" idea correct (I'm not sure in what generality), I'm not sure how to do that short of invoking the "calculus of variations." An analog model could probably be devised with wire, thread, and a soap film. 

20210901, 14:42  #33 
"Rashid Naimi"
Oct 2015
Remote to Here/There
3^{2}·239 Posts 
FTR, FWIW, I won’t be able to put any time on this puzzle before the weekend. It should be easy to verify things with a parametric CAD model. If possible I will attach a file and folks can verify for themselves. In the meantime there is 30 days free trial of SolidWorks online.

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