Mathematics Asked by Ak2399 on December 9, 2020

Given a quadrilateral *ABCD* with a fixed perimeter and the side *AD* having a fixed length

how can we prove that quadrilateral with maximum area satisfying the conditions above must also satisfy:

*$angle$ABC = $angle$BCD**AB = BC = CD*

For the second condition its straight forward. U can easily see that for any triangle. The triangles having the maximum are is the one where it’s isosceles (using ellipses)(or equilateral if you allowed all the sides to change accordingly) . Applying this for the points A , B , C and D its easy to see that the lengths of the sides of the quadrilateral must be equal.

The problem now boils down to the first condition

Any ideas on how proving it?

Let $ABCD$ be the quadrilateral ($AB=BC=CD$) in its "equal angles" configuration (i.e. $BCparallel AD$) and consider a small displacement $AB'C'D$ from that configuration $(AB=AB'=B'C'=C'D)$: we want to prove that the area of $ABCD$ is greater than the area of $AB'C'D$. Let's suppose $B'$ and $C'$ are the result of a counterclockwise rotation of $B$ and $C$ about $A$ and $D$ respectively, as in figure below.

As $BB'to 0$, segments $BB'$ and $CC'$ form the same angle with line $BC$; moreover, their projections on $BC$ must be equal (to ensure $B'C'=BC$). This implies that $BB'=CC'$ and that triangle $ABB'$ is congruent to triangle $DCC'$.

Hence the difference of the areas of quadrilaterals $ABCD$ and $AB'C'D$ is the same as the difference of the areas of triangles $MBB'$ and $MCC'$, where $M$ is the intersection point of $BC$ and $B'C'$. But these triangles have the same altitude upon bases $BM$ and $CM$, while $CM>BM$: it follows that the area of $MCC'$ is greater than the area of $MBB'$ and thus the area of $ABCD$ is greater than the area of $AB'C'D$, as it was to be proved.

EDIT.

The argument given above only proves that the "equal angles" configuration is a **local** maximum. To prove it is a global maximum one could use Bretschneider's formula for the area of a quadrilateral, which in our case gives:
$$
area = {1over4}sqrt{(a+d)^3(3a-d)-16a^3dcos^2{alpha+gammaover2}},
$$
where $a=AB=BC=CD$, $d=AD$, $alpha=angle DAB$ and $gamma=angle BCD$.

This formula obviously gives the maximum area when $cos{alpha+gammaover2}=0$, that is when $alpha+gamma=pi$. But that condition implies the quadrilateral is cyclic and is then equivalent to $angle ABC=angle BCD$.

Correct answer by Intelligenti pauca on December 9, 2020

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