Romanian Mathematical Olympiad - Final Round

Since $P \neq M$ and $P \neq N$, if $X, Y, P$ are collinear, we have $X \notin \{A, B\}$ and $Y \notin \{C, D\}$. Indeed, if $X = A$, then $XY \subset (ACD)$ and thus $P \in (ACD)$, false. The other situations are analogous.

2023 ROMANIAN MATHEMATICAL OLYMPIAD – FINAL ROUND

Existence: Since $P \in (MN)$, the point $P$ lies in the interior of the triangle $ANC$. Denote $\{Q\} = AP \cap CN$.
From $(CN) \subset \text{Int}(BCD)$, it follows that $Q \in \text{Int}(BCD)$, thus the line $BQ$ intersects the open segment $CD$. Denote $\{Y\} = BQ \cap (CD)$. Because $P \in (AQ) \subset (ABY)$, we deduce that $P$ lies in the interior of the triangle $ABY$, thus the line $YP$ intersects the open segment $AB$. Denote $\{X\} = PY \cap (AB)$. The pair $(X, Y)$ satisfies the statement.

Uniqueness: Assume that a pair of points $(X', Y')$ exists, with $X' \in (AB)$, $Y' \in (CD)$, $(X, Y) \neq (X', Y')$, such that the points $X'$, $P$ and $Y'$ are collinear. We consider $Y' \neq Y$ (the situation $X' \neq X$ is analogous). If $X' = X$, then $Y' \in XP \cap CD$, therefore $Y' = Y$, false. Consequently $X' \neq X$, so the distinct straight lines $XY$ and $X'Y'$ intersect at $P$. If $\alpha = (XY, X'Y')$, we obtain $XX' = AB \subset \alpha$ and $YY' = CD \subset \alpha$, thus the points $A, B, C$ and $D$ are coplanar, false. Consequently, the pair $(X, Y)$ is unique.