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Old 2021-08-31, 00:52   #23
uau
 
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Seems to match the values I calculated (1.90317905 * 50^2 = 4757.947625).
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Old 2021-08-31, 05:41   #24
LaurV
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Quote:
Originally Posted by retina View Post
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.
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(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
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Old 2021-08-31, 05:53   #25
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Quote:
Originally Posted by a1call View Post
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.
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Old 2021-08-31, 11:11   #26
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Quote:
Originally Posted by LaurV View Post
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).
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Old 2021-08-31, 11:28   #27
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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.
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Old 2021-08-31, 15:29   #28
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Quote:
Originally Posted by uau View Post
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.
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Old 2021-08-31, 17:29   #29
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Quote:
Originally Posted by kriesel View Post
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?
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Old 2021-09-01, 00:48   #30
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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.
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Old 2021-09-01, 01:41   #31
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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
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Old 2021-09-01, 12:32   #32
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Quote:
Originally Posted by kriesel View Post
Quote:
Originally Posted by uau View Post
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.
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Old 2021-09-01, 14:42   #33
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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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