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 2021-08-31, 00:52 #23 uau   Jan 2017 23·3·5 Posts Seems to match the values I calculated (1.90317905 * 50^2 = 4757.947625).
2021-08-31, 05:41   #24
LaurV
Romulan Interpreter

Jun 2011
Thailand

24·13·47 Posts

Quote:
 Originally Posted by retina The attached can't be maximal with a circle or a circle arc, right? The straight lines are better, right?
Wrong. The max area there still has only circle fences.

(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 2021-08-31 at 05:50

2021-08-31, 05:53   #25
LaurV
Romulan Interpreter

Jun 2011
Thailand

100110001100002 Posts

Quote:
 Originally Posted by a1call Attached is the maximization-optimum I get using CAD and successive approximation to more than 1 m resolution.
Ha! I wanted to say 90°, but I was a bit scared, that's why I said "between 60 and 120". Can you confirm that is exactly 90? How does it vary if the proportion of radius of the lake to available fence changes?

Nice work btw.

2021-08-31, 11:11   #26
uau

Jan 2017

23×3×5 Posts

Quote:
 Originally Posted by LaurV Can you confirm that is exactly 90?
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).

 2021-08-31, 11:28 #27 Uncwilly 6809 > 6502     """"""""""""""""""" Aug 2003 101×103 Posts 5·1,999 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.
2021-08-31, 15:29   #28
kriesel

"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest

168E16 Posts

Quote:
 Originally Posted by uau Seems to match the values I calculated (1.90317905 * 50^2 = 4757.947625).
Which is less area than the 3-sides-fenced rectangle adjacent to a straight river section, 50 x 100 m2.

2021-08-31, 17:29   #29
uau

Jan 2017

23·3·5 Posts

Quote:
 Originally Posted by kriesel Which is less area than the 3-sides-fenced rectangle adjacent to a straight river section, 50 x 100 m2.
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?

 2021-09-01, 00:48 #30 Batalov     "Serge" Mar 2008 Phi(4,2^7658614+1)/2 2·7·683 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 ${\pi \over 2} + {2 \over \pi}$, but it was not the limit. Joseph Gerver holds the champion title since 1992 but rigorous proof has been elusive.
 2021-09-01, 01:41 #31 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 2,141 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 2021-09-01 at 01:45
2021-09-01, 12:32   #32
Dr Sardonicus

Feb 2017
Nowhere

3·1,657 Posts

Quote:
Originally Posted by kriesel
Quote:
 Originally Posted by uau Seems to match the values I calculated (1.90317905 * 50^2 = 4757.947625).
Which is less area than the 3-sides-fenced rectangle adjacent to a straight river section, 50 x 100 m2.
... which is less than the 6366.1977+ m2 area of a semicircle of perimeter 200 m - the semicircle being the circular arc that meets a straight line at right angles.

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 $\theta$ 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 $\tan(\theta)$. The angle between a fencing radius to a fence post and the line of centers is $\frac{\pi}{2}\;-\;\theta$. 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.

 2021-09-01, 14:42 #33 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 214110 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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