What are the $\inf$ and $\sup$ of the area of quadrilateral given its sides length?

Question

I asked this question on MSE here.

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Given a quadrilateral with side lengths $a,b,c$ and $d$ (listed in order around the perimeter), t's known that the area, is always less than or equal to $\frac{(a+b+c+d)^2}{16}$ i.e This establishes an upper bound for any quadrilateral with these side lengths.

A question that came to my mind is if $\frac{(a+b+c+d)^2}{16}$ is an upper bound Then the least upper bound exits so what is the infimum and supremum of the area of quadrilateral ($0$ is an oblivious lower bound so the infimum exists too)?

If $a=b, \ c=d$ "a kite" or if $a=c, \ d=b$ "a parallelogram" then it is easy to prove the infimum is $0$.

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The infimum and supremum depends on $a,b,c,d$ in this examples the side lengths are $19,13,7,4$ and the minimum area is around $9.5$ so it is not $0$.

Claim 1: If $a<b<c<d$ and the area of the quadrilateral is $S$ then $\inf S $ is the area of the triangle $a,b,d-c$ which is

$$\sqrt{\frac{a+b+d-c}{2}\cdot\frac{a-b+d-c}{2}\cdot\frac{-a+b+d-c}{2}\cdot\frac{a+b-d+c}{2}}$$ This also works when $a=b=c=d, \ \ \ a=b, \ c=d \ \ \ ,a=c, \ d=b $.

I couldn't rigorously proof that claim but I am pretty sure it is correct.

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In the previous example when side lengths are $19,13,7,4$ the naximum area is around $73.5$ This time it seems that this is a a cyclic quadrilateral.

Claim 2: The supremum of the area of a quadrilateral with side lengh $a,b,c,d$ is when its vertices lie on a circle.

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How to determine the infimum and the supremum of the area given $a,b,c,d$?

Another question is: Can these results be generalized for any $n$-gon? What are the infimum and supremum of the area for a polygon with n sides given its side lengths?

Answer 1

The area, call it $K$, is indeed maximized when the quadrilateral is cyclic, in which case $K$ is given by Brahmagupta's remarkable generalization of Heron's formula: $K = \sqrt{(s-a)(s-b)(s-c)(s-d)}$ where as in Heron $s = (a+b+c+d)/2$ is the semiperimeter. This is a known consequence of Bretschneider's formula $$ K^2 = (s-a)(s-b)(s-c)(s-d) - abcd \cos^2 \theta $$ where $2\theta$ is the sum of two opposite angles, so $\cos^2 \theta = 0$ iff opposite angles add up to $\pi$, which is a classical condition for a quadrilateral to be cyclic. (There are two pairs of opposite angles, but the two choices yield values of $\theta$ that sum to $\pi$ and thus have the same $\cos^2 \theta$.)

For proofs and references see for example the Wikipedia pages for the Brahmagupta and Bretschneider formulas.

Answer 2

Another proof that the area is maximized when the quadrilateral is cyclic is the following:

First we consider a cyclic quadrilateral with the given sides lengths (the existence of such a quadrilateral requires an argument). We now “glue” an arc of the circumcircle on each side.

If you consider another quadrilateral with the same side lengths and the previous arcs glued on each side, the figure has a smaller total area than the circle with the same perimeter (that is, the original circumcircle) by the classical isoperimetric inequality, hence the quadrilateral has smaller area than the original cyclic quadrilateral.

The same proof works for any number of sides.

Answer 3

I argue that the expression for infimum is more complicated and cannot be given by a single formula unless certain inequalities involving $a,b,c,d$ are assumed.

Let us work in the setting that OP considered for infimum: $a<b<c<d$. Denote the angle between edges of lengths $a,b$ by $\alpha$ and the one between edges of lengths $c,d$ by $\beta$. They are related only by the following constraint (which comes from considering the diagonal length $e$ in the picture):

$$a^2+b^2-2ab\cos\alpha=c^2+d^2-2cd\cos\beta.$$

Thus the goal is to obtain the optima of the area function $f(\alpha,\beta)=\frac{1}{2}ab\sin\alpha+\frac{1}{2}cd\sin\beta$ subject to the constraint $g(\alpha,\beta)=(a^2+b^2-2ab\cos\alpha)-(c^2+d^2-2cd\cos\beta)=0$. Using Lagrange multipliers, let us consider the points where $\nabla f=\left(\frac{1}{2}ab\cos\alpha,\frac{1}{2}cd\cos\beta\right)$ is a multiple of $\nabla g=(2ab\sin\alpha,-2cd\sin\beta)$. This amounts to $\tan\alpha=-\tan\beta$, i.e. $\alpha+\beta=\pi$. This is the case of maximum area mentioned in other answers. Therefore, to derive the infimum of $f(\alpha,\beta)$ given $g(\alpha,\beta)=0$, we should consider boundary values for $\beta$. The assumptions $d>a$ and $c>b$ imply $\beta\leq\pi$. Thus $\beta$ belongs to $[0,\pi]$. The boundary values for $\beta$ are thus uniquely determined by those of $\cos\beta$. Based on the constraint, the latter yield $\cos\alpha$ and hence two opposite choices for $\sin\alpha$. Substituting the negative one (which amounts to $\alpha>\pi$ and hence a concave quadrilateral) in the area formula yields a candidate for the minimum. In view of this discussion, let's analyze the extreme values that $\cos\beta$ takes.

Solving the equation above for $\cos\beta$, the fraction $\frac{c^2+d^2-a^2-b^2+2ab\cos\alpha}{2cd}$ must be between $-1$ and $1$. Conversely, any such fraction in $[-1,1]$ determines $\beta$ uniquely. The range for $\cos\beta=\frac{c^2+d^2-a^2-b^2+2ab\cos\alpha}{2cd}$ is

$$\left[\max\left(\frac{c^2+d^2-a^2-b^2-2ab}{2cd},-1\right), \min\left(\frac{c^2+d^2-a^2-b^2+2ab}{2cd},1\right)\right].$$

Notice that: $\frac{c^2+d^2-a^2-b^2-2ab}{2cd}\geq -1\Leftrightarrow \require{enclose}\enclose{horizontalstrike}{d\geq a+b+c} \, c+d\geq a+b$ which always hold; and $\frac{c^2+d^2-a^2-b^2+2ab}{2cd}\geq 1\Leftrightarrow a+d\geq b+c$. So to simplify further, three two different regimes should be considered:

• $\require{enclose}\enclose{horizontalstrike}{d\geq a+b+c}$ (in which case the quadrilateral is concave unlike the picture above),
• $b+c-a<d\leq a+b+c$ ($d\leq a+b+c$ must hold for a quadrilateral with side lengths $a,b,c,d$ to exist),

This pertains to the limit case that OP mentions: $\beta$ attains all values in $[0,\pi]$, and $\beta=0$ yields the minimum possible area as the area of a triangle with sides $a,b,d-c$. Thus minimum area is $$\sqrt{\frac{a+b+d-c}{2}\cdot\frac{a-b+d-c}{2}\cdot\frac{-a+b+d-c}{2}\cdot\frac{a+b-d+c}{2}}$$.

• $d\leq b+c-a$,

In this case, $\cos\beta$ attains all values in $\left[\frac{c^2+d^2-a^2-b^2-2ab}{2cd},\frac{c^2+d^2-a^2-b^2+2ab}{2cd}\right]$. In view of the constraint $a^2+b^2-2ab\cos\alpha=c^2+d^2-2cd\cos\beta$, the endpoints of the interval correspond to $\cos\alpha=\pm 1$. In these cases we are dealing with triangles of side lengths $c,d,a+b$ or $c,d,b-a$. Their areas are given by
$$\sqrt{\frac{a+b+c+d}{2}\cdot\frac{-a-b+c+d}{2}\cdot\frac{a+b-c+d}{2}\cdot\frac{a+b+c-d}{2}},$$ $$\sqrt{\frac{-a+b+c+d}{2}\cdot\frac{a-b+c+d}{2}\cdot\frac{-a+b-c+d}{2}\cdot\frac{-a+b+c-d}{2}}.$$ It is not immediately clear to me if one of them is always larger than the other.

Conclusion) When $a<b<c<d$, the minimum of area is achieved at one of degenerate cases where the angle $\beta$ between sides of lengths $c,d$ is $0$, or when the angle $\alpha$ between sides of lengths $a,b$ is $0$ or $\pi$. To see which of them happens and which of the resulting triangles yields the least possible area, further assumptions are required, e.g. if $a+d>b+c$ holds or not.

Note on the update) I corrected an error in the earlier version.

Answer 4

The basic fact is that one can express the required extreme values of the areas in the generic case (our use of this term is explained below). The simple formulae involved can be found below. To do this we introduce a series of functions of four positive variables, the side lengths. These are $$f_1=-(a^4+b^4+c^4+d^4)+2(a^2b^2+2a^2c^2+2a^2d^2+2b^2c^2+2b^2d^2+2c^2d^2)+8abcd.$$ $f_2$ is the same, with a minus at the $abcd$ term.

$$f_3=4a^4-(a^4+b^4+c^4+d^4-2(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)).$$

We then have $f_4=a^2+b^2-c^2-d^2$ , $f_5=a^2-b^2-c^2+d^2$, $f_6=ab+cd$, $f_7=bc+ad$, $f_8=ab-cd$, $f_9=ad-bc$.

There are constraints on the variables which are necessary to ensure that suitable quadrilaterals with these side lengths exist. The formulation in the original question seems to be not quite complete—it is not clear whether the situation allows the solution to be non convex (which, in the cyclic case, means self-crossing). We are assuming that this is the case. Then the positivity of $f_2$ and hence of $f_1$ ensure the existence. In the generic case, there are two quadrilaterals where the sup and inf are obtained and these are cyclic. An explicit description in terms of the above functions can be found below (General remark: I am writing this on a pad, so trying to reduce the use of mathematical symbols).

The decisive fact is that for any value $F$ which satisfies conditions to be described below there are precisely two quadrilaterals $ABCD$ which have area $F$ and side lengths $a,b,c,d$. To be precise, I am using directed areas (i.e. $A\wedge B+B\wedge C+C\wedge D+D\wedge A$) and so the area can be negative or zero in non-trivial ways (example —the quadrilateral with vertices $(0,0),(1,0),(0,1),(1,1)$). The side lengths are $|AB|=a$ and so on cyclically. The required max and min are the largest and smallest values for $|F|$ which satisfy the given condition.

The condition which $F$ must fulfill is the positivity of an explicit sextic polynomial with coefficients functions of the side lengths. Hence the required optimal values are roots of this polynomials. The polynomial is, in fact, a cubic in $F^2$ and so can be solved by radicals.

The computations are rather intricate (I used Mathematica). The quadrilateral which attains the sup can be realised as the one with vertices $(0,0)$, $(a,0)$, $(\frac{f_3 }{4a f_4},\frac{b \sqrt {f_1}}{f_6}) $ and $ (\frac{f_5}{2 f_7},\frac{d\sqrt{f_1}}{f_3})$. The minimum is analogous with changes in the subscripts (which I can add if requested).

As mentioned above, this is for the generic case, by which I mean where the denominators do not vanish. The remaining cases are treated by other computations which I shall not describe here. To see what can happen in the non generic case consider the simplest one with all side lengths equal (to $1$). The maximum is clearly a square but the minimum is achieved with $A=(0,0)$, $B=D=(1,0)$ but $C$ any point on the unit circle centred at $(1,0)$.

Caveat. Of the three related topics—triangles, tetrahedra and quadrilaterals—from elementary geometry, the latter displays some subtleties not present in the other two. This is due to non-rigidity. For example, one can lose uniqueness. This is often in a relatively weak sense—instead of one solution, there are two (typically one convex, the other non-convex—example, Brahmagupta). This has been taken care of here. The second potential problem, which is relevant here, is that there are singular cases which have to be dealt with separately This can arise when terms in the denominator vanish. One important class—kites (in particular, rhomba) and parallelograms— can be subsumed under the pythagorean quadrilaterals (the sums of the squares of the lengths of two opposite pairs of adjacent sides coincide). I have also computed this case but will spare the reader the details.

Explanatory remarks about the relation to Brahmagupta’s formula for the area of a cyclic quadrilateral: There are, in fact, two such formulae, one for the convex case, one for the non-convex one (this distinction is sometimes not explicitly formulated in the literature). They correspond exactly to the above two expressions for the sup and inf. Since the validity of either of these expressions is equivalent to the quadrilateral being cyclic, this shows that (despite comments here) both the sup and inf are attained and, indeed, at cyclic quadrilaterals.

The following is off topic but I would like to add it since I feel that it makes what is going on here more transparent. One can prove the Brahmagupta formulae with a few calculations (which can be done by hand—I checked them with Mathematica to be sure). The idea is very simple— wlog one can assume that $A_1=(\cos (\theta_1),\sin(\theta_1))$, etc., and then compute the corresponding side lengths and the (square of) the area. Showing that the corresponding differences vanish then reduces to the kind of manipulation of trigonometric identities that we all learned in high school. The difference between the convex and non-convex cases arises through the occurrence of square roots of expressions of the form $1-\cos(\theta_2-\theta_1)$ and so on. This is a sine expression, with choice of sign depending on the orientation of $A_1$ and $A_2$ as one traverses the circle. In the convex case one can assume that this is always $+$, otherwise that there is one $-$. This explains the dichotomy between the convex and the non-convex case.