Belarus — Selection and Training Session

Let $O = AC \cap BD$; and, without loss of generality, $BO \geq DO$, $CO \geq AO$. If $AB \parallel DC$ then $1 \leq \frac{BO}{DO} = \frac{AO}{CO} \leq 1$, whence $BO = OD$ and $ABCD$ is a parallelogram. Then by the problem condition $AC \leq 1$, $BD \leq 1$, hence $S(ABCD) \leq \frac{1}{2} AC \cdot BD \leq \frac{1}{2} < \frac{2}{3}$.

Let now $AB \nparallel DC$, and let $M$ be the intersection point of the rays $BA$ and $CD$; denote $BC = b$, $\angle ABC = \beta$, $\angle DCB = \gamma$. Then $S = S(ABCD) < S(BMC)$, while
$$S(BMC) = \frac{1}{2} MC \cdot MB \sin(\beta + \gamma) = \frac{1}{2} \cdot \frac{b \sin \beta}{\sin(\beta + \gamma)} \cdot \frac{b \sin \gamma}{\sin(\beta + \gamma)} \cdot \sin(\beta + \gamma) = \\ = \frac{1}{2} \cdot \frac{b^2}{\operatorname{ctg} \beta + \operatorname{ctg} \gamma},$$
hence
$$S \cdot (\operatorname{ctg} \beta + \operatorname{ctg} \gamma) < \frac{b^2}{2}. \qquad (1)$$
If $A_1 \in [AB]$ and $D_1 \in [CD]$ are the points such that $S(BA_1C) = \frac{1}{2}S = S(BD_1C)$ then by the problem condition $CA_1 \leq 1$, $BD_1 \leq 1$. Thus
$$CA_1^2 = b^2 + \frac{S^2}{b^2 \sin^2 \beta} - 2S \cdot \frac{\cos \beta}{\sin \beta} = \left(b - \frac{S}{b} \operatorname{ctg} \beta\right)^2 + \frac{S^2}{b^2} \leq 1,$$
and, similarly, $\left(b - \frac{S}{b} \operatorname{ctg} \gamma\right)^2 + \frac{S^2}{b^2} \leq 1$. Then, in view of (1),
$$2 \geq \frac{2S^2}{b^2} + \frac{1}{2}\left(2b - \frac{S}{b}\left(\operatorname{ctg} \beta + \operatorname{ctg} \gamma\right)\right)^2 > \frac{2S^2}{b^2} + \frac{1}{2}\left(2b - \frac{b}{2}\right)^2 = \\ = \frac{2S^2}{b^2} + \frac{9b^2}{8} \geq \frac{1}{2}\sqrt{\frac{2S^2}{b} \cdot \frac{9b^2}{8}} = 3S.$$
It follows that $S < \frac{2}{3}$.