5
$\begingroup$

I am looking for a proof (or a reference) of an inequality related to a rea and the sidelengths of a polygon as follows:

Let $A_1A_2...A_n$ be arbitrary polygon, then:

$$Area(A_1A_2....A_n) \le \frac{1}{4}cotg{\frac{\pi}{n}} \sum_{i=1}^nA_iA_{i+1}^2$$

Geometric meanings: $$Area(A_1A_2\cdots A_n) \le \frac{Area(1)+Area(2)+\cdots +Area(n)}{n}$$

enter image description here

PS: I found this inequality long time ago, that time I thought this is old inequality. But today, I think this is new because I can not see any reference for the inequality.

$\endgroup$
3
$\begingroup$

Presumably the indices $i$ in $A_i$ are taken mod $n$, so "$A_{n+1}$" is to be identified with $A_1$. This must be a known isoperimetric inequality, but it's easier to prove than to find in the literature.

Fix the area $\cal A$ of the $n$-gon. By a standard compactness argument there exists an $n$-gon $A_1 A_2 \ldots A_n$ that minimizes $\sum_{i=1}^n (A_i A_{i+1})^2$. We first show that this polygon is convex (but possibly with $A_{i-1} A_i A_{i+1}$ collinear for some $i$). Indeed if it is not we can replace it by the convex hull, with each side $A_j A_k$ of the convex hull divided into $k-j$ equal subsegments; this both increases the area and decreases $\sum_{i=1}^n (A_i A_{i+1})^2$, so we can shrink the polygon back to area $\cal A$ and make the sum of its sides' squares even smaller.

Given convexity, fix all but one of the vertices, say $A_2$. Then $A_2$ is limited to a line parallel to $A_1 A_3$, and we readily see (as by choosing coordinates that make $A_1,A_3 = (\pm 1, 0)$ ) that $(A_1 A_2)^2 + (A_2 A_3)^2$ is minimized when $(A_1 A_2) = (A_2 A_3)$. Thus the minimizing $n$-gon has all sides equal, say with each $(A_i A_{i+1}) = s$; and then $ns^2$ is minimized when $ns$ is $-$ but $ns$ is the perimeter, and the usual isoperimetric inequality for $n$-gons then finishes the proof that the area-$\cal A$ polygon with the smallest $\sum_{i=1}^n (A_i A_{i+1})^2$ is regular.

$\endgroup$
8
$\begingroup$

Cauchy Schwartz tells you that $$\sum_{i=1}^n |A_iA_{i+1}|^2\geq \frac{P^2}{n}$$ where $P$ is the perimeter of the polygon. Then we need the inequality $$P^2\geq 4n\tan(\pi/n)A$$ which is the classical isoperimetric inequality for polygons with many proofs in the literature, analytic, geometric and algebraic. See this article and its references, for example.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.