75th 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 $OE=OF$, $△SOE\equiv △SOF$ (C.C.), $SE=SF$. Then $M\in SE$ and $OM\perp SF$, $Q\in SF$ and $OQ\perp SF$, $△SOM\equiv △SOQ$, $SM=SQ$, $\frac{SM}{SE}=\frac{SQ}{SF}$, $QM\parallel EF$, 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 $△MSR\sim △OPR$, so $\frac{MS}{OP}=\frac{MR}{OR}=\frac{SR}{PR}$, hence $\frac{M{S}^2}{O{P}^2}=\frac{SR}{OR}\cdot \frac{MR}{PR}=\frac{SR}{OR}\cdot \frac{MT}{PU}$. It follows $\frac{SR}{OR}=\frac{PU}{MT}\cdot \frac{M{S}^2}{O{P}^2}=\frac{PS\cdot PO\cdot M{S}^2}{MO\cdot MS\cdot PS\cdot PG}=\frac{PO}{PG}\cdot \frac{MS}{MO}$. But $\frac{PO}{PG}=\tan \angle SGO=\frac{SO}{OG}$ and $\frac{MS}{MO}=\cot \angle MSO=\frac{SO}{OE}$, so $\frac{SR}{OR}=\frac{S{O}^2}{OE\cdot OG}$.

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 $O\in BD$.