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Problem:

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.

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.

Community
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user369773
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4 Answers4

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This proof was given here.

Take the polygon and arrange it so that all the vertices lie on a circle: enter image description here

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

enter image description here

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.

Hrhm
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Theorem: For given sidelengths, a cyclic polygon has maximal area.

Proof: Let $\{p_k\}_{k=1}^n$ be the vertices of the polygon.

Set $v_k=p_k-p_{k-1}$ and $m_k=\frac12(p_k+p_{k-1})$, where $p_0=p_n$.

Since the polygon is closed, $$ \sum_{k=1}^nv_k=0\implies\sum_{k=1}^n\delta v_k=0\tag1 $$ Since the $|v_k|$ are given, for $1\le k\le n$, $$ v_k\cdot\delta v_k=0\tag2 $$ The area of the polygon is $$ A=\frac12\sum_{j\lt k}v_j^R\cdot v_k\tag3 $$ where $(x,y)^R=(y,-x)$. Therefore, $$ \begin{align} \delta A &=\frac12\sum_{j\lt k}v_j^R\cdot\delta v_k-\frac12\sum_{j\gt k}v_j^R\cdot\delta v_k\\ &=\sum_{k=1}^n(m_k-p_0)^R\cdot\delta v_k\tag4 \end{align} $$ To maximize $(3)$ under conditions $(1)$ and $(2)$, that is, to guarantee that $(4)$ vanishes when $(1)$ and $(2)$ hold, orthogonality requires that there be constants $\lambda_k$ and a point $p$ so that $$ (m_k-p_0)^R=p+\lambda_kv_k\tag5 $$ In other words, with $c=p_0-p^R$ $$ (m_k-c)\cdot v_k=0\tag6 $$ which translates to $$ |p_k-c|^2=|p_{k-1}-c|^2\tag7 $$ Thus, all $p_k$ lie on a circle with center $c$.

$\large\square$

robjohn
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In general, if $A_1 A_2\ldots A_n$ is a $n$-sided polygon and the lengths $l_1,l_2,\ldots,l_{n-1},l_n$ of $A_1 A_2,A_2 A_3,\ldots,A_{n-1} A_n, A_n A_1$ are fixed, the maximum area is achieved by the cyclic polygon, i.e. the polygon having all its vertices on a circle. You may easily prove this fact through Lagrange multipliers or Steiner's symmetrization method. It is an istance of the isoperimetric inequality for polygons, that can be used to prove the usual isoperimetric inequality in the plane through a limit process.

Jack D'Aurizio
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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

Takahiro Waki
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