20210905, 20:06  #45 
Jan 2017
2^{3}×3×5 Posts 
I'm not requiring the added surface to match the boundary of the square, so for a small enough added surface area the solution would be a half sphere. However, on a unit square you could fit one with at most radius 0.5, which isn't big enough to use all the 2 available surface area.

20210905, 20:25  #46 
"Vincent"
Apr 2010
Over the rainbow
2·3^{2}·149 Posts 
no comment ;p

20210905, 21:38  #47  
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
10,099 Posts 
Um....
Quote:


20210905, 21:51  #48 
"Vincent"
Apr 2010
Over the rainbow
2×3^{2}×149 Posts 
ah sorry. I should have known that someone already though of this.
Another example would be be mount st michel in france. in that case it is the whole europeanasian plate Last fiddled with by firejuggler on 20210905 at 21:53 
20210905, 21:54  #49 
"Jacob"
Sep 2006
Brussels, Belgium
1753_{10} Posts 

20210905, 21:56  #50 
"Vincent"
Apr 2010
Over the rainbow
2·3^{2}·149 Posts 
St michel island would be, in that case, the 'out' of the fence.
Also, the suez canal cut the AfroEurasia continent Last fiddled with by firejuggler on 20210905 at 21:59 
20210905, 23:24  #51  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×3^{2}×5^{2}×7 Posts 
Vagueness
Quote:
Quote:
And your answer is? 

20210906, 01:37  #52 
Jan 2017
170_{8} Posts 

20210906, 14:03  #53 
Feb 2017
Nowhere
2×5×7×73 Posts 
We assume the lake has radius 1, and the fence is a circular arc of length L which meets the lake at right angles. Let A be the center of the lake, B the center of the circle of which the fence is an arc, and P and Q the points where the fence meets the lake. Let be the acute angle BAP (measured in radians). Then the angle ABP is , and the radius BP has length .
The angle of the arc of fence between BP and BQ is so the length L of fence satisfies the equation The area enclosed by the sector of fence between the radii BP and BQ is The rest of the area enclosed by the fence and the edge of the lake is within the "kiteshaped" quadrilateral APBQA. If F is the intersection of PQ and AB, the quadrilateral is the union of the 4 triangles APF, AQF, BPF, and BQF. These can be rearranged into a rectangle of sides 1 and , so the quadrilateral has area . The circular sector of lake PAQ within the quadrilateral has area , so the area enclosed by the fence is . In the original version, the lake had radius 50 and the fence had length 200 = 4*50, so here we take L = 4, giving the equation . I don't know a closedform solution, so I used Newton's method to obtain a numerical solution. Code:
? { t=Pi/4; print("0 "t); for(i=1,8, f=tan(t)*(Pi+2*t)4; fp=Pi/(cos(t))^2 + 2*tan(t) + 2*t/(cos(t))^2; t = t  f/fp; print(i" "t) ); print(); print("Radius of fence circle is "50*tan(t)); print(); print("Distance between center of lake and center of fence circle is "50/cos(t)); print(); print("Area enclosed by fence is "2500*((tan(t))^2/2*(Pi+2*t) + tan(t)  t)) } 0 0.78539816339744830961566084581987572105 1 0.72304343030496242371104078487746358835 2 0.71815938930962624688096481937315314626 3 0.71813340203617805545425687957399093198 4 0.71813340130850774616059434588664824093 5 0.71813340130850774559009248053397820187 6 0.71813340130850774559009248053397820152 7 0.71813340130850774559009248053397820152 8 0.71813340130850774559009248053397820152 Radius of fence circle is 43.688540881007952979946174346648823038 Distance between center of lake and center of fence circle is 66.397956326317025917614143737943306255 Area enclosed by fence is 4757.9476288799235830166949506623779518 ? 
20210906, 14:15  #54 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2^{2}×3^{2}×5^{2}×7 Posts 
The units are missing, but nevermind. Mrs Fred will be very happy with her new 4757.9... m² rose garden next to the lake. I think Farmer Fred will have a very pleasant evening after erecting the fence.

20210906, 17:31  #55 
"Rashid Naimi"
Oct 2015
Remote to Here/There
2^{2}·3·181 Posts 
I think her name is Wilma. As for the pleasant evening, I wouldn’t count on that. He is probably cooking up a new scheme with Barney Rubble which will eventually get him into trouble.

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