0
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I also remember reading it before, but there should a nice proof. $\endgroup$ – Narasimham Dec 17 '14 at 14:43
  • $\begingroup$ @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(?). $\endgroup$ – abel Dec 17 '14 at 14:44
  • $\begingroup$ Could you include some sketches ? $\endgroup$ – Narasimham Feb 16 at 6:49

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.