Question: > 
> **Problem:**
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> 
> Given a sequence of $n$ lengths, $L\_1,L\_1,...,L\_n$ where each $L\_k$ is the length of the $k$th side, I need to find a sequence of $n$ points, where $p\_k=(x\_k,y\_k)$, such that $\operatorname{dist}(p\_k,k\_{k+1})=L\_k$ and where $\operatorname{dist}(a,b)$ is the Euclidean distance between $a$ and $b$. So they way the sequence goes, $p\_1=(0,0)$ then $p\_2=(0,L\_1)$. The points must correspond to the ordered clockwise vertices of the simple polygon having the maximum possible area for the given side lengths.
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For example, if we were given the lengths $1,1,1$, then obviously this is an equilateral triangle. And the maximum area is given by the coordinates, $(0,0),(0,1),(\sqrt{1-0.25},0.5)$.

Another example, if we were given the lengths $1,2,1,2$, then we would get a rectangle with coordinates $(0,0),(0,1),(2,1),(2,0)$.

How would I work this out for large $n$ or in general for every $n$? Because I cannot think of a way past the case of a square/rectangle.

Answer: 
This proof was given [here](http://www.drking.org.uk/hexagons/misc/polymax.html).

Take the polygon and arrange it so that all the vertices lie on a circle:
[![enter image description here](https://i.stack.imgur.com/U4u4q.png)](https://i.stack.imgur.com/U4u4q.png)

Now, "glue" the red caps onto the side of the polygon:

[![enter image description here](https://i.stack.imgur.com/eUOjA.png)](https://i.stack.imgur.com/eUOjA.png)

Note that the perimeter around the red caps stays constant as I move the vertices. The area of the red shaded sections also stay the same. According to the isoperimetric inequality, the area of the total shape (both the red and white sections) is maximized when the polygon is cyclic. Ergo, the area of the polygon is maximized when the polygon is cyclic.
About convex or concave, it's obviously convex polygon. By repeating this work, we get theorem which area of quadrilateral is max iff four points are on a circle. By using this theorem repeatedly, iff all points are on a circle, we get the area can be max.

When $L\_k $ is on a plane of radius R, S is max.
If angle of chord L\_k is $a\_k$,

$$\displaystyle sin{\frac{a\_k}2}=\frac{L\_k}{2R}$$

$$\displaystyle\sum a\_k=2\sum sin^{-1}{\frac{L\_k}{2R}}=2\pi$$

$\displaystyle\sum \frac{L\_k}{2R}\sqrt{1-(\frac{L\_k}{2R})^2}=2\prod\frac{L\_k}{2R}\tag1$

we get R by solving (1). therefore
$$S=\displaystyle\sum S\_k=\sum \frac{RL\_k}{2\sqrt{1-(\frac{L\_k}{2R})^2}}$$

equation of (1) by <https://in.answers.yahoo.com/question/index?qid=20140526080730AASjOAI>