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I have posted this question here without answer. Maybe I can get some light here.

Suppose we are given $n$ segments $l_1,...,l_n$ in $\mathbb{R}^2$ such that $|l_i|=i,\ \forall\ i=1,...,n$, where $|l_i|$ is the length of $l_i$. Let $\alpha_1,...,\alpha_{n-1}$ be $n-1$ angles such that $\alpha_i>0$ and $\sum_{i=1}^{n-1}\alpha_i\leq\frac{\pi}{2}$.

Consider the following operation: (All angles are assumed to be "clockwise").

1) Place $l_{1}$ anywhere in $\mathbb{R}^2$,

2) Place $l_{2}$ at either of the end points of $l_{1}$, making an angle of $\alpha_{1}$,

3) Place $l_{3}$ at the end point of $l_{2}$ not occupied by $l_{1}$, making an angle of $\alpha_{2}$,

...

n-1) Place $l_{n-1}$ at the end point of $l_{n-2}$ not occupied by $l_{n-3}$, making an angle of $\alpha_{n-2}$,

n) Place $l_{n}$ at the end point of $l_{n-1}$ not occupied by $l_{n-2}$, making an angle of $\alpha_{n-1}$.

Now consider arbitrary permutations both of the angles $\alpha_i$ and of the lengths $l_i$. Which ordering of the angles and lengths maximizes the distance between the two endpoints of the chain of segments?

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"Place $l_{2}$ at either of the end points of $l_{1}$, making an angle of $\alpha_{1}$"...

Making an angle $\alpha_{1}$ with respect to which line, measured in which direction? Here is a possible example with $\alpha_i=\pi/8$, but until you remove the ambiguity in your stipulation, it is unclear if this accord with your intention:
  Lengths and Angles
The OP has now made clear that the above fails his definition. So, if $\alpha_i$ is a constant, the distance is entirely determined. For $\alpha_i=\pi/8$, here is the determined shape and so end-to-end distance:
           Lengths and Angles: 2nd version

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  • $\begingroup$ making an angle $\alpha_1$ with respect to $l_1$. All angles are assumed to be "clockwise". $\endgroup$
    – Tomás
    Commented Dec 28, 2012 at 0:39
  • $\begingroup$ Yes you are right. $\endgroup$
    – Tomás
    Commented Dec 28, 2012 at 0:47
  • $\begingroup$ So the above illustration fails your criteria, because the $\pi/8$ angles are not each measured clockwise? Between $l_2$ and $l_1$, the angle is clockwise. But between $l_3$ and $l_2$, it is counterclockwise. Am I correct that my drawing fails your criteria? $\endgroup$
    – Joseph O'Rourke
    Commented Dec 28, 2012 at 0:48
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    $\begingroup$ I think the problem is more tractable and still interesting if the alpha_i are supplementary (or complementary? Anyway, the angles exterior to a convex polygon which add up to 2pi.) angles. Gerhard "Geometry Was So Long Ago" Paseman, 2012.12.27 $\endgroup$
    – Gerhard Paseman
    Commented Dec 28, 2012 at 1:16
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    $\begingroup$ In fact, with all the alphas equal and using my intepretation, it seems clear a length increasing ordering achieves the longest endpoints. From this I think one can also show that a monotonic angle measure ordering is required for optimality when the exterior angles alpha are different. Gerhard "Thinking Outside All The Boxes" Paseman, 2012.12.27 $\endgroup$
    – Gerhard Paseman
    Commented Dec 28, 2012 at 1:25

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