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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?

Edit: The problem in the form that it was stated in the first time was too general, hence I have changed some assumptions.

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    The angles $\alpha_i$ must have the same orientation as the initial angles (i.e. maintain their clockwise/anticlockwise count)?2012-10-27
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    @BeniBogosel, a priori there is no restriction about this, but it is interesting to understand both cases.2012-10-27
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    The question seems interesting, but is way to general to give a precise answer. The idea is that the long segments must make an angle close to $\pi/2$ with the first segment. Have you tried at least the case of 3 segments and 2 angles?2012-10-27

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