20210830, 11:03  #12  
Jan 2017
120_{10} Posts 
Quote:
Proof: if you claim to have a counterexample  that is, a path between two points points which is not a circle segment and which gives a larger area than a circle segment of equivalent length for the shape formed by the path and the line between the points  then you can take a circle where the corresponding circle segment appears, and switch it with your shape, getting a solution better than a circle for a totally free shape. This proves that a circle segment is optimal, unless you believe a circle is suboptimal for a totally free shape. 

20210830, 11:04  #13 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10954_{10} Posts 
If you can't find a simple expression for the area of an arbitrary shape, part of the boundary of which is a concave circular arc, this sounds like an ideal problem for simulated annealing.

20210830, 13:04  #14  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6282_{10} Posts 
Quote:


20210830, 13:31  #15 
Jan 2017
2^{3}×3×5 Posts 
That is not a counterexample. The second image is wrong and does not have maximal area for that line length. An easy way to see that it must be wrong is that you could move the bases of the straight lines outwards, and for small changes the effect on line length is essentially 0*change (derivative is 0 while it's a right angle) while effect on area is c*change for some c>0.
You can get a larger area by placing a circle segment of suitable curvature against the river. 
20210830, 14:03  #16  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2×3^{2}×349 Posts 
Quote:
What if the lake is really tiny? What if it is larger? What if it impinges just a little bit, or a lot. Do you always get the same result? 

20210830, 14:25  #17 
Jan 2017
2^{3}·3·5 Posts 
You always get circle segments, that's what my proof showed. Compare this to a problem of finding the shortest path between two points that avoids obstacles. Any part of the path in free space must be a straight line  if it weren't, you could make it locally shorter by straightening it. This is similar  any part must be locally a circle segment, or you could get more area locally for the same path length. The curvature must also be the same for all the parts: if you have a pair of points somewhere with distance one and a path of length 1.2 between them, and another pair somewhere with distance also one and path of length 1.1, then you'd get more area for same total length by making both length 1.15 (which gives everything the same curvature).

20210830, 14:59  #18 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
1100010001010_{2} Posts 
I guess it is time to get explicit then.
The attached can't be maximal with a circle or a circle arc, right? The straight lines are better, right? The point being you can't simply ignore the shape of the impinging object. What you say might be true, but is it absolutely always true for the problem as stated? 
20210830, 15:11  #19  
Jan 2017
2^{3}×3×5 Posts 
Quote:
Suppose you have an almostcomplete fence with holes of length 4 and 5. If you have exactly 9 units to complete it, you'll get two straight lines. As the available length of fence increases, you start getting arcs instead, with the same curvature at both places. 

20210830, 15:14  #20 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·3^{2}·349 Posts 
Okay, so what is your solution?

20210830, 17:44  #21 
Jan 2017
2^{3}·3·5 Posts 
For the circular lake case, I get for lake radius=1 and fence length=4:
lake sector angle for fence endpoints = 1.4362668 distance between fence endpoints = 1.3159604 fence circle radius = 0.8737708 fenced area outside lake = 1.90317905 (fence circle area = 2.3985, area of that outside endpoint line = 0.27270, additional area cut off by lake curve = 0.22265) I didn't particularly sanity check the values, could have a simple error somewhere. Last fiddled with by uau on 20210830 at 17:45 
20210830, 22:26  #22 
"Rashid Naimi"
Oct 2015
Remote to Here/There
4147_{8} Posts 
Attached is the maximizationoptimum I get using CAD and successive approximation to more than 1 m resolution.

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