Let $P$ and $Q$ be the foci of the ellipse $\mathcal{E}$, and let $\mathcal{R} = ABCD$ be a circumscribed rhombus. First, we prove that $\mathcal{E}$ and $\mathcal{R}$ are aligned. Points $P$ and $Q$ are isogonal conjugates in $\mathcal{R}$, which implies $\angle APB + \angle CPD = 180^{\circ}$. It is well-known and easy to show that this implies $P$ and $Q$ lie on one diagonal of the rhombus $x$ and are reflections over the other diagonal $y$. Thus the centers of $\mathcal{E}$ and $\mathcal{R}$ coincide at a point $O$, and the ellipse has axes $x$ and $y$. Now apply an affine transform to turn $\mathcal{E}$ into a circle $\omega$ while preserving $x$ and $y$; this maps the rhombus $\mathcal{R}$ to another rhombus $\mathcal{R}'$ circumscribed about $\omega$, with diagonals $x$ and $y$. It is easy to see that the area of $\mathcal{R}'$ is minimized exactly when it is a square, solving (1). In particular, it follows that the vertices of $\mathcal{R}'$ lie on the image of $\mathcal{E}$ under a homothety of scale $\sqrt{2}$ with center $O$. Now we solve (2). First we construct $O$. Fix a point at infinity $\infty$; draw two chords $\overline{AB}$ and $\overline{CD}$ of $\mathcal{E}$ which contain $\infty$. By Brokard, the points $\overline{AC} \cap \overline{BD}$ and $\overline{AD} \cap \overline{BC}$ lie on the polar of $\infty$, so this line (which contains $O$) may be constructed. Construct another polar, and then their intersection point is $O$. Construct a circle with center $O$ which intersects $\mathcal{E}$ at four points; they make a rectangle $WXYZ$. In particular, $\overline{WX}$ and $\overline{WZ}$ are parallel to $x$ and $y$, so the lines $x$ and $y$ may be constructed. The problem is almost solved: let $x$ and $y$ intersect $\mathcal{E}$ at points $A_0$, $B_0$, $C_0$, $D_0$ such that $A_0B_0C_0D_0$ is a rhombus; then the rhombus $ABCD$ obtained by applying a homothety to $A_0B_0C_0D_0$ with scale $\sqrt{2}$ is the desired rhombus $\mathcal{R}$.

I wonder if there are any "elementary" methods for $(2)$ (i.e. ones that do not use Brokard) because I heard that CMO doesn't allow people to use such "advanced methods" without proof? Part (1) and the characterization of the rhombus used in (2) can be proved using Cartesian coordinates.

stroller wrote: I wonder if there are any "elementary" methods for $(2)$ (i.e. ones that do not use Brokard) because I heard that CMO doesn't allow people to use such "advanced methods" without proof? Part (1) and the characterization of the rhombus used in (2) can be proved using Cartesian coordinates. If you're worried about Brokard, we don't need to cite it. The main claim is the following. Claim. Let $\mathcal{E}$ be an ellipse with center $O$ and parallel chords $\overline{AB}$ and $\overline{CD}$. Then $O$, $\overline{AC} \cap \overline{BD}$, $\overline{AD} \cap \overline{BC}$ are collinear. Proof. Affine transform the ellipse to a circle, after which it's obvious.

Sorry I forgot to mention this, but I think affine/projective transformations aren't regarded as elementary by them either.

I think Pascal and Brianchon is enough.

stroller wrote: Sorry I forgot to mention this, but I think affine/projective transformations aren't regarded as elementary by them either. Sorry if I'm being dumb, but the affine transformation used by CantonMathGuy can be viewed as a dilation of the ellipse so it is a circle, which seems pretty elementary....

just to construct it take the parallelogram formed by the vertices and the covertices then consider $MNPQ$ the parallelogram formed by the midpoints of the sides if $O$ is the center of the ellipse let $OM,ON,OP,OQ$ cut the ellipse at $D,E,F,G$ .DEFG is exactly the required rhombus tangecy points.

yayups wrote: stroller wrote: Sorry I forgot to mention this, but I think affine/projective transformations aren't regarded as elementary by them either. Sorry if I'm being dumb, but the affine transformation used by CantonMathGuy can be viewed as a dilation of the ellipse so it is a circle, which seems pretty elementary.... It is elementary, but... if you say that on The National College Entrance Examinations in China (and depending on the province you are in, the qualifying round(s) ) you won't get any points.

For part (a), we will show that any circumscribed rhombus is axis-aligned. We rely on the following lemma. Lemma. If an affine transform sends a rhombus to another rhombus, then its direction is parallel to some axis of the rhombus. Proof. Consider the following picture: [asy][asy] size(3.5cm); defaultpen(fontsize(10pt)); pair A,B,C,D,W,X,Y,Z; A=(-46.5,15.5); B=(13.5,40.5); C=-A; D=-B; W=(-46.5,40.5); X=(46.5,40.5); Y=-W; Z=-X; draw(A--B--C--D--A); draw(W--X--Y--Z--W); dot(A); dot(B); dot(C); dot(D); dot(W); dot(X); dot(Y); dot(Z); label("$y_1$",A--W,dir(180)); label("$x_1$",W--B,dir(90)); label("$x_2$",B--X,dir(90)); label("$y_2$",X--C,dir(0)); [/asy][/asy] Assume that a horizontal affine transformation $x\mapsto\lambda x$, $|\lambda|\ne1$, sends it to another rhombus. We have from the original rhombus that $x_1^2+y_1^2=x_2^2+y_2^2$. Furthermore since the image is another rhombus, we have $\lambda^2x_1^2+y_1^2=\lambda^2x_2^2+y_2^2$, or \[\lambda^2\left(x_1^2-x_2^2\right)=y_2^2-y_1^2=x_1^2-x_2^2\implies x_1=x_2,\]as desired. $\blacksquare$ Thus consider the affine transformation sending the ellipse to a circle. The image of the rhombus is a parallelogram with an inscribed circle and is therefore a rhombus. By the lemma, the rhombus must be axis-aligned. Considering the image of this transformation, we just need to show that the rhombus tangent to a circle at all four sides with minimum area whose diagonals lie on fixed lines is unique; this is just a square, so (a) is done. For part (b), we first construct the center and axes of this ellipse. Consider any two parallel chords; by affine transformation, the line between their midpoints passes through the center of the ellipse. Applying this twice, the center is constructable. Intersecting a circle whose center coincides with that of the ellipse, the four points of intersection form a rectangle whose sides are parallel to the axes of the ellipse, so we can construct the axes of the ellipse. Let the axes intersect the ellipse at four points. By the affine transformation, the image of the rhombus formed by the four points under dilation at the center by scale factor $\sqrt2$ is the desired rhombus, so we are done.

Here is a different, projective-style argument for the key lemma in (1), which may appeal more to certain readers. Something tells me the China MO won't approve though... Claim. If a rhombus $\mathcal R$ is circumscribed about an ellipse $\mathcal E$, then $\mathcal R$ and $\mathcal E$ have the same center and are aligned. Actually, this is a corollary of the following claim. Lemma. If a kite $\mathcal K$ with diagonal of symmetry $d$ is circumscribed about an ellipse $\mathcal E$, then $\mathcal E$ is also symmetric about $d$. (In particular, $d$ coincides with one of the axes of $\mathcal E$.) Proof. Let $\mathcal E'$ be the reflection of $\mathcal E$ over $d$, and let $\ell$ be a line perpendicular to $d$ which is tangent to $\mathcal E$. Then $\ell$ is also tangent to $\mathcal E'$. Now $\mathcal E$ and $\mathcal E'$ have five different common tangents (the four sidelines of $\mathcal K$, along with $\ell$), which forces $\mathcal E = \mathcal E'$. $\square$

Take a homography sending the ellipse to a circle. This is equivalent to simply strecthing the ellipse. Note that parralel lines stay parralel. We can see this through either noting that two parralel lines stay concurrent at the point at infinity or that if we simply plot the ellipse in a Cartesian plane, after stretching the slopes stay the same. As polar of a point at infinity with respect to a circle is a diameter, by Brokard, if we have a trapezoid inscribed in a circle, the center of the circle will be collinear with the intersection of diagonals of the trapezoid and the intersection of the two non-parralel opposite sides. Same thing applies to an ellipse due to a homography. To draw a pair of parralel lines, first draw a line with straightedge, and using the compass, make two adjacent segments with the same length. Then just do the midpoint trick on one side, and connect those two, to get a line parralel to the first. Do this on the ellipse twice to find the center of the ellipse. Now, we need to find the smallest area of a rhombus where the size of the incircle is the same. Let the radius of the circle be $1.$ Connect opposite vertices of rhombus to split the diagram into $4$ right triangles that are rotations or reflections of eachother. The height of each of these where the base is the hypotenuse is $1.$ By $AM-GM,$ the minimum length of the hypotenuse is $2.$ Thus, we want a rhombus where the vertices' distances to the center are $\frac{\sqrt{2}}{2}$ times the length of either axis. We can draw these out by simply drawing a circle at the center of the ellipse, which then gives an inscribed rectangle, which gives the axes by constructing the perpendicular bisectors. Finally, to get a length that is $\sqrt{2}$ times the length of half an axis, simply get the length of half the axis with compass, then on a line drawn with straightedge, mark a starting point, and then two more points spaced out by half an axis. Then draw the perpendicular bisector of the first and third point, mark with compass a side with length half the axis. Finally, get the hypotenuse of this triangle. Do this again for the other axis, and we're done.$\blacksquare$

(1) Replace rhombus with parallelogram, then affine (vertical stretch) the ellipse to a circle and the minimum area is clearly when the parallelogram becomes a square. Since the square's diagonals are perpendicular, unless they are aligned with the affining direction, since otherwise undoing the affine pushes two upward-facing diagonal rays down, increasing their angle above $90^\circ$. Thus only one square, the one with axis aligned diagonals, was originally a rhombus as well. (2) Intersect an arbitrary circle with the ellipse, the angle bisectors of any pair of lines through the four points are aligned to the ellipse, by DIT on the line at infinity and by symmetry. Construct the center of the conic by taking two pairs of parallel lines and intersecting the lines through each pair's midpoints. Thus we may construct the four vertices of the ellipse, now homothety them by $\sqrt2$ at the center to get the vertices of the rhombus (by affine).