What angles of a plane scalene quadrilateral maximize its area? By 'scalene' I mean the four lengths are unequal.

It is known that if a quadrilateral has opposite sides equal and parallel as a rectangle, it has maximum area. Also if one pair is parallel and the other pair of equal length the trapezium has maximum area.

i believe cyclic quadrilateral maximizes the area. if you want to minimize you can get almost zero area.

here is a proof. let the vertices of quadrilateral be labelled by $A, B, C$ and $D.$ also set $AB = a, BC = b, CD = c$ and $DA = d$ by cosine rule you have $$e^2 = a^2 + d^2 - 2ad\cos A, \ e^2 = b^2 + c^2 - 2bc \cos C$$ and the area of the quadrilateral be $\Delta = {1 \over 2} ad\sin A + {1 \over 2} bc \sin C.$

now comes the manipulation: note that $$4abcd\cos A \cos C = (a^2+d^2-e^2)(b^2+c^2-e^2),$$ we make use of it later.

$$16\Delta^2 = 4a^2d^2 -(a^2+d^2-e^2)^2 + 4b^2c^2 -(b^2+c^2-e^2)^2 + 8abcd \sin A \sin C \\ = 4a^2d^2 + 4b^2c^2-(a^2+d^2-e^2)^2 -(b^2+c^2-e^2)^2 +2(a^2+b^2-e^2)(b^2+c^2 -e^2) -8abcd\cos A\cos C + 8abcd \sin A \sin C \\= 4a^2d^2+4b^2c^2-(a^2+d^2-b^2-c^2)^2-8abcd\cos(A+C) \\ \le 4a^2d^2+4b^2c^2-(a^2+d^2-b^2-c^2)^2+8abcd \\=4(ad+bc)^2-(a^2+d^2-b^2+c^2)^2 \\=(2ad+2bc-a^2-d^2+b^2+c^2)(2ad+2bc+a^2+d^2-b^2-c^2) \\=(a+b+c+d)(b+c+a-d)(a+b+d-c)(a+c+d-b)$$

the last expression is symmetric in $a,b,c$ and $d.$ this is the expression i alluded to in my comments. this expression reduces to heron's formula for the area of the triangle if you let $D$ coincide with $C.$

the equality is obtained when $A+C = 180^\circ$ which makes $ABCD$ a cyclic quad. the min value could be put in nicer form, but it will have to wait.

• I also remember reading it before, but there should a nice proof. – Narasimham Dec 17 '14 at 14:43
• @Narasimham, i will see if i can come up with a proof. i know that it it involves ptolemy's theorem or a generalization of heron's theorem attributed to bramagupta(?). – abel Dec 17 '14 at 14:44
• Could you include some sketches ? – Narasimham Feb 16 at 6:49