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We then have $f_4=a^2+b^2-c^2-d^2$ , $f_5=a^2-b^2-c^2+d^2$.

Then $f_6=ab+cd$, $f_7=bc+ad$, $f_8=ab-cd$, $f_9=ad-bc$.

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 $ (d \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).

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

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).

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

There are constraints on the variables which are necessary to ensure that suitable quadrilaterals with these side lengths exist. The formulation is not quite complete—it is not quite clear whether the situation allows 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. Then in the generic case, there are two quadrilaterals where the sup and inf are obtained and these are cyclic. They can be realised as those quadrilaterals with vertices $(0,0)$, $(a,0)$, $(,)$ and $(,)$ which is the maximum. The miminum is attained at the vertices as above but with $f_1$ replaced by $f_2$, $f_6$ by $f_8$ and $f_7$ by $f_9$. (I am writing this on a pad, so trying to reduce the use of mathematical symbols).

The computations are rather intricate (I used Mathematica).

In order to be concrete, the coordinates of the vertices of the quadrilaterals which attain the sup and in are firstly $(0,0)$, $(a,0)$, $(\frac{f_3 }{4a f_4},\frac{b \sqrt {f_1}}{f_6}) $ and $ (d \frac{f_5}{2 f_7},\frac{d\sqrt{f_1}}{f_3})$.

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 mention here. where the denominators do not vanish.

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

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 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 $ (d \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 $(1,0)$.

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.

The condition which $F$ must fulfill is the positivity of an explicit sextic polynomial withcoefficients 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) but the result is quite simple:the squares of the two areas are $$\frac 1 {16}(-a^4+2 a^2b^2-b^4+2a^2c^2 +2b^2c^2+2 c^2d^2-c^4\pm 8abcd+2a^2d^2+2b^2 d^2-d^4).$$

Of course, it can happen that one of these expressions is negative. This means is that there is no quadrilateral with the assigned side lengths.

The plus sign gives the maximum.

In order to be concrete, the coordinates of the vertices of the quadrilaterals which attain the sup and in are firstly $(0,0)$, $(a,0)$, $(\frac{f_8 }{f_4},\frac{b f_1}{f_6}) $ and $ (\frac{f_6}{f_3},\frac{df_1}{f_3})$.

where the numerated $f$ in order the functions $f_1$ the first area function, $f_2$ the second one, $f_3=2(bc+ad)$, $f_4=4a(a^2+b^2-c^2-d^2)$, $f-5=d(a^2-b^2-c^2+d^2)$,$f_6=2(ab+cd)$, $f_7=2( ab-cd)$, $f_8=2(ad-bc)$ and $f_9=3a^4-b^4-(c^2-d^2)^2+2b^2(c^2+d^2)-2a^2(b^2+c^2d^2)$.

The rather ungainly formatting is due to the fact that I am using a pad. As mentioned above, this is for the generic case where the denominators do not vanish. In this form, they are, of course, of no use to man or beast but the computations which lead to the formula produce a coded version which allows one to compute the vertices for any $a,b,c,d$ input at the press of a button.

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.

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

We then have $f_4=a^2+b^2-c^2-d^2$ , $f_5=a^2-b^2-c^2+d^2$.

Then $f_6=ab+cd$, $f_7=bc+ad$, $f_8=ab-cd$, $f_9=ad-bc$.

There are constraints on the variables which are necessary to ensure that suitable quadrilaterals with these side lengths exist. The formulation is not quite complete—it is not quite clear whether the situation allows 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. Then in the generic case, there are two quadrilaterals where the sup and inf are obtained and these are cyclic. They can be realised as those quadrilaterals with vertices $(0,0)$, $(a,0)$, $(,)$ and $(,)$ which is the maximum. The miminum is attained at the vertices as above but with $f_1$ replaced by $f_2$, $f_6$ by $f_8$ and $f_7$ by $f_9$. (I am writing this on a pad, so trying to reduce the use of mathematical symbols).

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).

In order to be concrete, the coordinates of the vertices of the quadrilaterals which attain the sup and in are firstly $(0,0)$, $(a,0)$, $(\frac{f_3 }{4a f_4},\frac{b \sqrt {f_1}}{f_6}) $ and $ (d \frac{f_5}{2 f_7},\frac{d\sqrt{f_1}}{f_3})$.

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 mention here. where the denominators do not vanish.