Since Brahmagupta's formula and Bretschneider's formula we have the inequality:

Any two quardrilaterals $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ with the same sidelengths and $A_1A_2A_3A_4$ is a cyclic but $B_1B_2B_3B_4$ is not a cyclic then area of $A_1A_2A_3A_4$ $\ge$ area of $B_1B_2B_3B_4$

Isoperimetric inequality state that:

For the length L of a closed curve and the area A of the planar region that it encloses, that $L^2 \ge 4\pi .Area$ and that equality holds if and only if the curve is a circle.

I conjecture that:

Let $A_1A_2...A_n$ and $B_1B_2...B_n$ be two n-polygons with the lengths $a_1, a_2,...,a_n$ and $b_1, b_2,...,b_n$ such that with $i \in n$ then exist $j \in n$ such that $a_i=b_j$. If $A_1A_2...A_n$ is a cyclic and $B_1B_2...B_n$ is not a cyclic then area of $A_1A_2...A_n$ $\ge$ area of $B_1B_2...B_n$.

I am looking for the proof of the inequality above.

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  • 1
    $\begingroup$ You can swap neighbouring sides of cyclic polygon. So it is sufficient to prove your conjecture when $a_i=b_i$. $\endgroup$ May 3, 2018 at 14:48

1 Answer 1


Let $B_1B_2\ldots B_n$ be non-cyclic polygon and $B_kB_{k+1}B_{k+2}B_{k+3}$ are four consecutive verteces which not lie on a circle. Making $B_kB_{k+1}B_{k+2}B_{k+3}$ a cyclic polygon and keeping all points excepting $B_{k+1}$ and $B_{k+2}$ at the same places you will get bigger area.

See also Maximum area of a flexible polygon.


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