3
$\begingroup$

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.

enter image description here

$\endgroup$
  • 1
    $\begingroup$ You can swap neighbouring sides of cyclic polygon. So it is sufficient to prove your conjecture when $a_i=b_i$. $\endgroup$ – Alexey Ustinov May 3 '18 at 14:48
2
$\begingroup$

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.

$\endgroup$

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.