Maximizing Quadrilateral Area

We claim that the maximum area of a quadrilateral with given side length is achieved when the quadrilateral is cyclic. To prove this, we note that Bretschneider's Formula states that the area of a quadrilateral is $\sqrt{(s-a)(s-b)(s-c)(s-d) - abcd\,\cos^2\left(\frac{A+C}{2}\right)}$ , where $A$ and $C$ are angles that are opposite one another. Clearly, the area of this quadrilateral is maximized if $\frac{A+C}{2}=\frac{\pi}{2}\implies A+C=\pi$ . This condition is a necessary and sufficient property of cyclic quadrilaterals, so the area is maximized when these side lengths form a cyclic quadrilateral.

Now, we will prove that if given the initial side lengths, we can form a cyclic quadrilateral. To prove this, order the sides such that $AB=41$ , $BC=22$ , $CD=18$ , and $DA= 39$ . While we keep $AB$ fixed, we can move point $C$ such that $BCD$ is a straight line, forming a non-degenerate triangle $ABD$ (the side lengths are $41$ , $40$ , and $39$ ), and we note that $\angle BAD+\angle BCD>\pi$ . Alternatively, we can move point $C$ such that $CDA$ is a straight line, forming a non-degenerate triangle $ABC$ (the side lengths are $41$ , $22$ , and $57$ ), and we note that $\angle BAD+\angle BCD<\pi$ . Thus, by geometric continuity, there exists a point where $\angle BAD+\angle BCD=\pi$ , so the maximum area above can be obtained, and that area is

$\sqrt{(s-a)(s-b)(s-c)(s-d)}$ $=\sqrt{(60-18)(60-22)(60-39)(60-41)}= \boxed{798}$ .

As a note, this is true for all quadrilaterals $ABCD$ , with side lengths $a$ , $b$ , $c$ , and $d$ given that $a,b,c,d<s$ .

First, note that the order of the sides does not in fact matter. Let the known lengths of the sides be a, b, c and d. There are essentially three configurations, namely having the side opposite a be b, c or d. There are two possibilities for the remaining sides, but a reflection of the plane maps them to each other. We can take the three cases to be abcd, acbd and abdc. Given a quadrilateral with the side order abcd, constructing one with the side order acbd and the same area is done by taking the triangle ABC spanned by the sides b and c (AB=b, BC=c, AC = a diagonal) and mirroring it so that the points A and C change places. Similarly, abcd can be made into abdc by mirroring the triangle spanned by the sides c and d.

We will use the side order 18, 41, 22, 39. Divide the quadrilateral into one triangles with two sides 18,41 and one triangle with two sides 22,39. By the sine theorem, the area A of a triangle with two known sides a and b and their angle $\gamma$ is $\frac{1}{2}ab\sin\gamma$ . This area is maximized when $sin\gamma = 1$ , so $\gamma = 90^\circ$ . The maximum area for the 18-41 triangle is thus $\frac{18\times41}{2} = 369$ , and for the 22-39 triangle is $\frac{22\times39}{2} = 429$ . The diagonal of the 18-41 triangle is, by Pythagoras's theorem, $\sqrt{18\times18+41\times41} = \sqrt{2005}$ . The diagonal of the 22-39 triangle is $\sqrt{22\times22+39\times39} = \sqrt{2005}$ . Since the two diagonals have the same length, we can let them coincide and be the diagonal of the quadrilateral. Its maximal area is thus 369+429=798.

Given any 4 line segments a,b,c,d satisfying a+b+c>d, it is possible to form a cyclic quadrilateral. A rigorous proof can be found here : http://www.17centurymaths.com/contents/briggsagain/ALfrevCh24.pdf Now I move on to proof that the cyclic quadrilateral has the biggest area. By Bretschneider's formula: K=sqrt((s-a)(s-b)(s-c)(s-d)-abcd*cos^2((alpha+gamma)/2)) where alpha,gamma are opposite angles,a,b,c,d are the side lengths and s is the semiperimeter. cos^2((alpha+gamma)/2) has minimum value of 0, when (alpha+gamma)/2=90, alpha+gamma=180 is true iff ABCD is a cyclic quadrilateral. Thus, the maximum possible value is sqrt((s-a)(s-b)(s-c)(s-d))=798, satisfied iff ABCD is a cyclic quadrilateral.

Using Bretschneider's formula, we have that the area of the quadrilateral is

$\sqrt{(60-22)(60-18)(60-39)(60-41)-22\cdot 18 \cdot 39\cdot 41 \cos^2(\frac{A+C}{2})}$ . Since $22\cdot 18 \cdot 39\cdot 41 \cos^2(\frac{A+C}{2}$ is nonnegative, it must be greater than or equal to 0. The expression is maximized when $22\cdot 18 \cdot 39\cdot 41 \cos^2(\frac{A+C}{2} = 0$ , so $\sqrt{(60-22)(60-18)(60-39)(60-41)-22\cdot 18 \cdot 39\cdot 41 \cos^2(\frac{A+C}{2})}$ $=\sqrt{(60-22)(60-18)(60-39)(60-41)} = 798.$ It is easy to see that this is attainable (let ABCD be cyclic).

The area of a quadrilateral is maximized when the quadrilateral is a cyclic quadrilateral. This can be proved using calculus. Since we know the side lengths, we can use Brahmagupta's formula. Note: we could also use Bretschneider's formula, but since the quadrilateral is cyclic, $\cos 90^\circ=0$ so the last term cancels out, giving us Brahmagupta's. Thus, the area of the quadrilateral is $\sqrt{(60-18)(60-22)(60-39)(60-41)}=798$

The semi-perimeter of the quadrilateral, $s=\frac{18+22+39+41}{2}=60$ . Using Bretschneider's formula, the area of the quadrilateral is $\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd(\cos^2{(\frac{A+C}{2})})}$ $\leq \sqrt{(60-18)(60-22)(60-39)(60-41)} = 798$ where $A,C$ are opposite angles in the quadrilateral because $abcd(\cos^2{(\frac{A+C}{2})}) \geq 0$ with equality when $A+C=180^\circ$ resulting in a cyclic quadrilateral.

Bretschneider's Formula is: square root of (s-a)(s-b)(s-c)(s-d) - abcd . cos2(x/2)

a, b, c, and d are the sides of the cuadrilateral, and x is the sum of any pair of opposite angles of the cuadrilateral. s is the semiperimeter which is equal to (a + b + c + d)/2. In order for the area to be the largest possible, it is necessary that (s-a)(s-b)(s-c)(s-d) - abcd . cos2(x/2) be the largest possible.

Note that abcd . cos2(x/2) is negative since it is being substracted. To the greatest posible area it is necessary for that part to be zero. Note that cos(90) = 0. In other words, the sum of opposite angles,x, should equal 180. This immediately means that the cuadrilateral is cyclic. This means the area of the cuadrilateral is the maxmum when it is cyclic. So we assume it is cyclic and then apply the formula. It doesnt matter what exact numbers are a,b,c, and d.

s= (41 + 39 + 22 + 18) /2 = 120/2 = 60

AREA = square root of (60 - 18)(60 - 22)(60 - 39)(60 - 41)

        = square root of (42)(38)(21)(19)

        = square root of 636804

        = 798

Note: you dont actually have to multiply (42)(38)(21)(19), just decompose it in prime factors and then root it.

Bretschneider's formula states that the area of a general convex quadrilateral with sides $a$ , $b$ , $c$ , and $d$ is: $\sqrt{(s-a)(s-b)(s-c)(s-d)-abcd*\cos^2{\frac{\alpha+\beta}{2}}}$ where $s$ is the semiperimeter and $\alpha$ and $\beta$ are opposite angles.

To maximize this expression, we must minimize $abcd*\cos^2{\frac{\alpha+\beta}{2}}$ . This can be done by having $\alpha$ and $\beta$ be supplementary. Then, $abcd*\cos^2{\frac{\alpha+\beta}{2}} = 0$ . Now, to find the area, we compute $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ to get our answer of $798$ .

Note: Quadrilaterals with supplementary angles are known as cyclic quadrilaterals. The formula $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ that we have found is known as Brahmagupta's formula, which finds the area of cyclic quadrilaterals. It follows that given four sides of a quadrilateral, the cyclic quadrilateral with those sidelengths will have the maximum area.

By Bretschnedier's Formula , we get area of the quadrilateral equals to

$\sqrt {(60 - 18)(60 - 22)(60 - 39)(60 - 41) -(18)(22)(39)(41)[cos (\frac{A + C}{2})^{2}]}$ , where $A, C$ are opposite angles.

Note that the maximum value is obtained when the quadrilateral is cyclic because $-(18)(22)(39)(41)[cos (\frac{A + C}{2})^{2}] \leq 0$ , with equality if $A + C = 180$ degrees, so the quadrilateral is cyclic.

So, the maximum area is $\sqrt{(42)(21)(38)(19)} = 798$

Solution 1: Consider any quadrilateral with side lengths 18, 22, 39 and 41 and maximal area. Let the vertices be $A, B, C$ and $D$ and let $[ABCD]$ denote the area of the quadrilateral $ABCD$ . If $D$ is within triangle $ABC$ , we may reflect $D$ across $AC$ to $D'$ and quadrilateral $ABCD'$ would satisfy the length requirement and have a larger area. Hence, we may assume that this figure is convex.

Draw the diagonals of the quadrilateral. Consider the reflection of triangle $ABC$ on the perpendicular bisector of $BC$ where $A$ is reflected to $A'$ . We have that $A'C = AB$ and $A'B = AC$ and $[A'BCD] = [ABCD]$ . Thus, we may reflect any of the triangles $ABC, BCD, CDA$ and $DAB$ about the perpendicular bisector of the corresponding diagonal, and still obtain a quadrilateral with equal area that fulfills the length requirement.

We have $[ABC] = \frac{1}{2} (AB) (BC) \sin \angle ABC$ . The maximum area occurs when $\angle ABC = 90^\circ$ as $AB$ and $BC$ are fixed. Thus $AB$ is perpendicular to $BC$ . Similarly, the maximum area for the triangle $CDA$ occurs when $CD$ is perpendicular to $DA$ . We notice that $18^2 + 41 ^2 = 2005 = 39^2 + 22^2$ . Thus we can build a quadrilateral from 2 right triangles, $ABC$ and $CDA$ , with $AB = 18,$ $BC = 41,$ $CD = 39,$ $DA = 39$ and hypotenuse $AC = \sqrt{2005}$ . Hence $[ABCD] = [ABC] + [CDA] = \frac {1}{2} (18)(41) + \frac {1}{2} (39)(22) = 798$ .

Solution 2: The generalized form of Brahmagupta's formula (also known as Bretschneider's formula), gives the area of a quadrilateral with side lengths $a, b, c$ and $d$ as $\sqrt{ (s-a)(s-b)(s-c)(s-d) -abcd\cos \theta}$ , where $s = \frac {a+b+c+d}{2}$ is the semiperimenter and $\theta$ is half the sum of 2 opposite angles (doesn't matter which pair). Since $a, b, c$ and $d$ are fixed, the area is maximized when $\theta = 90^\circ$ (which implies that the quadrilateral is cyclic), and has value $\sqrt{ (60-18)(60-22)(60-39)(60-41)}=798$ .

Note: The takeaway from Solution 2 is that for a quadrilateral with fixed side lengths, its area is maximal when it is cyclic ( $\theta = 90^\circ$ ).

For a quadrilateral where the side lengths are given, the maximum area can be computed using Brahmagupta's formula which is just a generalization of Heron's formula for area of triangle.

For a given quadrilateral with side lengths a, b, c, and d:

Area = sqrt{(s-a)(s-b)(s-c)(s-d)}

where

s = (a+b+c+d)/2

Substituting the given side lengths,

s = (18+39+41+22)/2 = 60

Area = sqrt{(60-18)(60-39)(60-41)(60-22)} = 798

Since we know that cyclic quadrilateral give maximum area, with the given sides let us examine how we get same diagonal for a pair of sides. Only direct possibility is there are two righted angled triangles with a pair of sides that give same hypotenuse. Obviously if there are such pairs, only possibility is (41,18) and (39,22).
$41^2+18^2=2005, ~~ and ~~39^2+22^2=2005 !!$
So the area is 1/2{ 41 * 18 + 39 * 22} =798.

We know that the maximum possible area is attained when we have a cyclic quadrilateral. We notice that 18^2+41^2=324+1681=2005 and 22^2+39^2=484+1521=2005. Thus we have a cyclic quadrilateral, with AB=18, BC=41, CD=22, DA=39, and the diagonal AC is the diameter of the circumcircle. Then the area of the entire quadrilateral is the sum of the areas of the 2 right triangles, which is (18 41+39 22)/2=798.