Shortlist JBMO, 2008

Solution:
Let us consider a parallelogram $ABCD$, with $AB = a$, $BC = b$, $AC = x$, $BD = y$, $\widehat{AOD} = \theta$.
For the area of $ABCD$ we know $(ABCD) = ab \sin A$.
But it is also true that $(ABCD) = 4(AOD) = 4 \cdot \frac{OA \cdot OD}{2} \sin \theta = 2 OA \cdot OD \sin \theta = 2 \cdot \frac{x}{2} \cdot \frac{y}{2} \sin \theta = \frac{xy}{2} \sin \theta$.
So $ab \sin A = \frac{xy}{2} \sin \theta$ and since $ab = \frac{xy}{2}$ by hypothesis, we get
$$\sin A = \sin \theta$$
Thus
$$\theta = \widehat{A} \text{ or } \theta = 180^\circ - \widehat{A} = \widehat{B}$$
If $\theta = A$ then (see Figure below) $A_2 + B_1 = A_1 + A_2$, so $B_1 = A_1$ which implies that $AD$ is tangent to the circumcircle of triangle $OAB$. So
$$DA^2 = DO \cdot DB \Rightarrow b^2 = \frac{y}{2} \cdot y \Rightarrow b = \frac{y}{\sqrt{2}}$$

Then by $ab = \frac{xy}{2}$ we get $a = \frac{x}{\sqrt{2}}$.
If $\theta = B$ we similarly get $a = \frac{x}{\sqrt{2}}, b = \frac{y}{\sqrt{2}}$.

Let us consider a parallelogram $ABCD$, with $AB = a$, $BC = b$, $AC = x$, $BD = y$, $\widehat{BOC} = \theta$, and let us produce the line $AD$ towards $D$ and consider $M \in (AD)$ so that $AD = DM$. Then $BCMD$ is a parallelogram, so $CM = BD = y$.
Observe also that $(ABCD) = 2(ACD) = (ACM)$ which is written equivalently as
$$CB \cdot CD \cdot \sin C = \frac{AC \cdot CM \cdot \sin \theta}{2} \text{ i.e. } ab \sin C = \frac{xy \sin \theta}{2}$$
Because of the given relation $ab = \frac{xy}{2}$ the last relation becomes $\sin C = \sin \theta$, i.e.
$$\theta = \widehat{C} \text{ or } \theta = 180^\circ - \widehat{C} = \widehat{B}$$
If $\theta = \widehat{C}$, then the triangles $ACM$ and $BCD$ are similar because their angles at $C$ are equal, as well as their angles at $B, M$ (remember $BCMD$ is a parallelogram).

Then
$$\frac{b}{y} = \frac{a}{x} = \frac{y}{2b} \Rightarrow \left(b = \frac{y}{2}, a = \frac{x}{2}\right)$$
If $\theta = \widehat{B}$, then similarly we prove that the triangles $ACM$ and $ACD$ are similar, which then implies
$$\frac{a}{y} = \frac{b}{x} = \frac{x}{2b} \Rightarrow \left(a = \frac{y}{2}, b = \frac{x}{2}\right)$$

The Parallelogram Law states that, in any parallelogram, the sum of the squares of its diagonals is equal to the sum of the squares of its sides.
In our case, this translates to $x^2 + y^2 = 2(a^2 + b^2)$. First adding $2xy = 4ab$, then subtracting the same equality, yields $(x + y)^2 = 2(a + b)^2$ and $(x - y)^2 = 2(a - b)^2$. It follows that $x + y = a \sqrt{2} + b \sqrt{2}$ and either $x - y = a \sqrt{2} - b \sqrt{2}$, or $x - y = b \sqrt{2} - a \sqrt{2}$. In the first case one obtains $x = a \sqrt{2}, y = b \sqrt{2}$, in the latter case, $x = b \sqrt{2}, y = a \sqrt{2}$.
For the proof of the Parallelogram Law, simply apply the Law of cosines in triangles $ABC$ and $ABD$ and use the fact that $\text{( ABC) = - ( BAD)}$. Adding the two relations gives the desired condition.