In my last post on iso-perimeteric problems I finished with the question "If you have 360 feet of fencing and want to build a pen along a river or barn, the rectangular pen above is NOT the most efficient way to use three straight sections... assume you are limited to four fence posts, so you put two somewhere on the river, and can put two more along the fencing to hold straight sections of fence; how do you place them?
Now extend that problem to four sections of fence (and five fence posts)... and then generalize the results... "
The sweet mathematical idea to make all these types of problems easy is symmetry . Imagine that you had found the perfect solution to the fence problem with three sections and used it on your side of the river. Then the guy on the other side of the river decided to copy your master plan on his side...
NOW make the river really narrow..... no..more narrow... like a line.. Now erase the river completely...
What do you see when you look at your neighbors fence and your fence together? It must be a six sided figure.. and it must contain the most area it could contain for any six sided figure, since both sides hold the maximum possible... you couldn't make the total bigger without making one side or the other bigger...so you must have the most efficient six sided figure possible, a regular hexagon.
So what does your half look like? It must be half of a regular hexagon divided across two opposite vertices.
The solution would be to divide the 360 feet of fencing into three sections, and make the angles between the fence sections 120o and slide the two ends up against the river.
To solve the three sided non-symmetric figure, we used 1/2 of a six sided symmetric figure. And the general solution from there, I hope, is trivial.
Actually with a little handwaving, this blog could have been much shorter..but I remember Pascal once wrote, "I have made this longer than usual, only because I have not had the time to make it shorter." [Lettres Provinciales, No. 16 (1657)]