Elliptical Exercise

For a quadrilateral to be maximal, the point $A$ must be as far as possible from segment $BD$ as to maximize the area of triangle $ABD$ . Therefore, we must have the tangent at $A$ parallel to $BD$ . Same applies to the other three points and that means that $C$ is symmetrical to $A$ in respect to the origin $O(0;0)$ and same with $B$ and $D$ .

Let $A=(x;y)$ , $C=(-x;-y)$ , $B=(-u;v)$ and $D=(u;-v)$ and suppose to simplify they're all non-negative as in the picture. Given $x$ , there's only one possible value for the three other variables if $OB$ is parallel to the tangent at $A$ . We'll then have to find the right $x$ so that $OA$ is parallel to the tangent at $B$ .

$A$ and $B$ are on $y=f_+(x)=\frac{1}{3}\sqrt{9-x^2}$ , with $f_+'(x)=\frac{-x}{3\sqrt{9-x^2}}$ .

$B(-u;v)$ is the solution of the system $\left\{ \begin{array}{l}v=\frac{1}{3}\sqrt{9-(-u)^2}\\ \frac{v}{-u}=\frac{-x}{3\sqrt{9-x^2}}\end{array}\right.$ . Substituting $v$ in the second equation gives $\frac{\sqrt{9-u^2}}{-3u}=\frac{-x}{3\sqrt{9-x^2}}\,\Longleftrightarrow\,xu=\sqrt{(9-u^2)(9-x^2)}\,\underset{x,u\geq0}{\Longleftrightarrow}\,x^2u^2=(9-u^2)(9-x^2)\,\Longleftrightarrow\,x^2+u^2=9$ .

The symmetry of $x^2+u^2=9$ implies that if we were to find for which point $A'(x';y')$ we would have the tangent at $B$ parallel to $OA'$ , we would find $x'=x$ . That is, for any $x$ , the tangent at $B$ is already parallel to $OA$ !

If we suppose that for any $x$ , the quadrilateral is optimal and so they would all have the same area, we just take $x=3$ which puts the four points on the axes and the area is easy to find.

If we're not so sure of that, let's find the area of $ABCD$ in respect to $x$ . $\overrightarrow{AB}=\begin{pmatrix}-u-x\\v-y\end{pmatrix}$ and $\overrightarrow{AD}=\begin{pmatrix}u-x\\-v-y\end{pmatrix}$ .

Area of parallelogram built on $\begin{pmatrix}a\\b\end{pmatrix}$ and $\begin{pmatrix}c\\d\end{pmatrix}$ is $|a·d-b·c|$ . Here, we get Area $=\Big|(-u-x)(-v-y)-(v-y)(u-x)\Big|=2(xv+uy)$ .

$u=\sqrt{9-x^2}$ and $v=\frac{1}{3}\sqrt{9-u^2}=\frac{1}{3}x$ . So, Area $=2(x·\frac{1}{3}x+\sqrt{9-x^2}·\frac{1}{3}\sqrt{9-x^2})=2(\frac{1}{3}x^2+\frac{1}{3}(9-x^2))=2·\frac{9}{3}=6$ . The area is always the same and is 6.

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Answer to bonus question: there is an infinity of optimal quadrilaterals; just pick a point $A$ wherever you want.

Let's first solve the much simpler problem of inscribing the largest quadrilateral in a unit circle. Using Laurent's argument (from his first paragraph) we see that those are the squares, with area 2.

Now we stretch out the circle in $x$ -direction by a factor of 3 to generate our ellipse. All areas are scaled by 3 in this transformation, so that the largest quadrilaterals now have an area of $\boxed{6}$ .

There are infinitely many of those quadrilaterals with maximal area. Two "special" ones are the rectangle with its sides parallel to the axes and the rhombus with its vertices at the points $(\pm 3,0)$ and $(0,\pm 1)$ .