For example, I believe the following arrangement solves both questions above for N=4. (I'll let you determine the distances.)
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Source: Original. I've limited N to 6 to keep the problem simple to approach, but I'd be happy to accept and publish results for higher N.
Hereafter the distances that I have found according to the two cases: N case 1 case 2 1 1/2 1 2 4/3 2 3 9/4 4 4 16/5 61/10 5 191/42 17/2 6 85/14 183/16 7 61/8 1171/80 The attached JPG file 011003sol.jpg (750K) gives the corresponding diagrams. The scales of the rods have been respected.
For N = 1 or 2 there is no real choice.
For N = 1 you hang the rod horizontally at its middle.
Longest horizontal distance from the loop is 1/2.
Longest total horizontal distance is 1.
For N = 2, you have the 2 horizontally at a point so that it balances
with the 1 at its end. Say it is hung X units from the end with the 1.
Then (2-X)(2-X)/2 = X*X/2 + 1*X,
2 - 2X + X^2/2 = X^2/2 + X,
2 - 2X = X
2 = 3X
X = 2/3.
The distance from the loop is thus 4/3, and the total distance is 2.
For N = 3, you maximize both distances by hanging the 3 horizontal,
hanging the 1 from its end, and hanging the 2 horizontal from the bottom
of the 1. The 2 must be balanced at its middle, and the 3 is balanced
against the 3 units of weight hanging at its end, X units from that end.
(3-X)(3-X)/2 = X*X/2 + 3X
9/2 - 3X = 3X
9/2 = 6X
X = 3/4.
So the 3 extends 3/4 unit toward the other rods and 9/4 units the other
way. The 2 extends an additional 1 unit beyond the end of the 3, for a
total of 7/4 on that side of the original loop. Thus, the max distance
from the loop is 9/4 and the max total distance is 4.
For N = 4, the diagram given in the puzzle is best.
The balance of the 4 rod is as follows, with X units on the short end:
(4-X)(4-X)/2 = X*X/2 + 6X
8 - 4X = 6X
8 = 10X
X = 4/5.
4/5 unit of the rod hangs on one side, with 3 and 1/5 units protruding on
the other side. Let X represent the short end of the 3 rod:
(3-X)(3-X)/2 = X*X/2 + 2X
9/2 - 3X = 2X
9/2 = 5X
X = 9/10.
So 2 and 1/10 units of the 3 rod protrudes beyond the short end of the 4 rod.
That makes 2 and 9/10 on that side of the main loop, 3 and 2/10 on the other
side, and 6 and 1/10 total width.
By this point I've solved the same algebra 4 times for different numbers.
Let's find a general solution. For a rod of length L, with a weight W
hanging from one end, the distance X from the weight where the rod is
balanced is:
(L-X)(L-X)/2 = X*X/2 + WX
(L^2)/2 - LX = WX
(L^2)/2 = (L+W)X
X = (L^2)/[2(L+W)]
For the case of putting the long rod at the top and hanging all the weight
off one of its ends, L+W = the sum of 1 to L, or L*(L+1)/2. This reduces
to X = L/(L+1) in this case, explaining the pattern which is evident in
the preceding results.
For 5 rods I think this is the best arrangement:
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We know now that 5/6 unit of the 5 rod will be on the short side, and thus
4 and 1/6 units on the long side. For the 4 rod, L=4 and W=5; X=16/18 = 8/9.
The 3 and 1/9 units of the 4 rod which sticks out here beats the 1.5 + 8/9 on
the other side of the 1, and beats the 2 + a very small bit of the 3 rod
you could get from hanging the rods in the order 5 2 3 1 4.
The total horizontal distance is 8 and 1/9; the longest horizontal distance
is that of the 5 rod, 4 and 1/6.
For 6 rods, the similar arrangement (with the 6 and 5 horizontal, and the
4 hanging as dead weight off the 3) gives 5 and 1/7 horizontal extension
on the 6 rod. The short end of the 5 rod will be given by L = 5, W = 9,
X = 25/28. So the 5 rod sticks out 4 and 3/28 past the short end of the 6,
making the total width 10 and 3/28.
However, there are several competing arrangements:
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In this arrangement, the length of the short end of the 4 rod is given by
L=4, W=9, X = (L^2)/[2(L+W)] = 16/26 = 8/13.
The length of the short end of the 5 rod is given by L=5, W=3, X = 25/16.
This gives the total extension to the right of the 6 rod to be
8/13 + 5 - 25/16 = 4 + (8*16 - 9*13)/208 = 4 and 11/208.
Not as good as the original.
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In this case we have weights hanging from both ends of the 4.
This expanded equation looks like (using Z as the weight on the long end):
(L-X)(L-X)/2 + Z(L-X) = X*X/2 + WX
(L^2)/2 - LX + ZL - ZX = WX
(L^2)/2 + ZL = (W+L+Z)X
X = L*(L/2 + Z)/(W+L+Z)
So with L=4, W=6, Z=3, we get X = 20/13.
The end of the 5 rod is thus 2.5 + 20/13 = 4 and 1/26 units beyond the
end of the 6, still less than the first case.
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In this arrangement, the length of the short end of the 3 rod is given by
L=3, W=10, X = (L^2)/[2(L+W)] = 9/26.
The length of the short end of the 5 rod is given by L=5, W=4, X = 25/18.
This gives the total extension to the right of the 6 rod to be
9/26 + 5 - 25/18 = 4 + (9*18 - 7*26)/468 = 4 - 20/468 < 4.
There are some other arrangements, but each is clearly inferior to
one of these. So we still have 10 and 3/28 total width and
5 and 1/7 horizontal extension as the best for N=6.
However, for longer N, maybe as soon as for N=7, there will be enough
extra dead weight around to make some of these other arrangements useful.
Eventually they should be able to pass the 2N+1 width limit that
arrangements in the pattern of these solutions for 4, 5, and 6 rods
are approaching.