Romanian Mathematical Olympiad

We assume that $O$ lies, for example, on the diagonal $AC$. Let $OE \perp AB$, $E \in AB$ and $OF \perp AD$, $F \in AD$. Then we have successively $OE = OF$, $\triangle SOE \equiv \triangle SOF$ (C.C.), $SE = SF$. Then $M \in SE$ and $OM \perp SF$, $Q \in SF$ and $OQ \perp SF$, $\triangle SOM \equiv \triangle SOQ$, $SM = SQ$, $\frac{SM}{SE} = \frac{SQ}{SF}$, $QM \parallel EF$, so $QM \parallel BD$. Analogously we get $NP \parallel BD$, so $QM \parallel NP$, which means that points $M, N, P, Q$ are coplanar.

Conversely, assume that $M, N, P, Q$ are coplanar in a plane $\alpha$. Take $OG \perp CD$, $G \in CD$ and $OH \perp BC$, $H \in BC$. Then the points $E, O, G$ are collinear, so the lines $SE, SO, SG$ are coplanar. It follows from this that the lines $SO$ and $MP$ are coplanar, and $SO \cap MP$ is the same as $SO \cap \alpha$. Similarly we'll get $NQ \cap SO$ is the same as $SO \cap \alpha$, so the lines $MP$ and $NQ$ intersect $SO$ at the same point $R$.

We calculate the ratio in which point $R$ divides $OS$, in terms of $OS, OE$ and $OG$.
Take $MT \perp OS$, $PU \perp OS$, $U, T \in OS$. From the cyclic quadrilateral $MOPS$ we get $\triangle MSR \sim \triangle OPR$, so $\frac{MS}{OP} = \frac{MR}{OR} = \frac{SR}{PR}$, hence $\frac{MS^2}{OP^2} = \frac{SR}{OR} \cdot \frac{MR}{PR} = \frac{SR}{OR} \cdot \frac{MT}{PU}$. It follows $\frac{SR}{OR} = \frac{PU}{MT} \cdot \frac{MS^2}{OP^2} = \frac{PS \cdot PO \cdot MS^2}{MO \cdot MS \cdot PS \cdot PG} = \frac{PO}{PG} \cdot \frac{MS}{MO}$. But $\frac{PO}{PG} = \tan \angle SGO = \frac{SO}{OG}$ and $\frac{MS}{MO} = \cot \angle MSO = \frac{SO}{OE}$, so $\frac{SR}{OR} = \frac{SO^2}{OE \cdot OG}$.

Since line $NQ$ intersects $OS$ also in $R$, we obtain $OE \cdot OG = OF \cdot OH$. On the other hand we have $OE + OG = l = OF + OH$, where $l = AB = AD$. It follows from this that $OE(l - OE) = OF(l - OF)$, then $(OE - OF)(l - OE - OF) = 0$, so $(OE - OF)(OH - OE) = 0$. Thus $OE = OF$ or $OE = OH$, which implies $O \in AC$ or $O \in BD$.