Difference between revisions of "2006 Romanian NMO Problems/Grade 7/Problem 1"

Problem

Let $ABC$ be a triangle and the points $M$ and $N$ on the sides $AB$ respectively $BC$, such that $2 \cdot \frac{CN}{BC} = \frac{AM}{AB}$. Let $P$ be a point on the line $AC$. Prove that the lines $MN$ and $NP$ are perpendicular if and only if $PN$ is the interior angle bisector of $\angle MPC$.

Solution

Let $L$ be a point on $BC$ such that $N$ is the midpoint of LC, then $2CN$= $LC$, the given information is the same as $\frac{LC}{BC} = \frac{AM}{AB}$, applicating Thales theorem it follows that $ML$ is parallel to $AC$.

Let $R$ be the point on $MN$ such that $MN$= $NR$, in view of $MN$= $NR$ and $LN$= $NC$ it follows that $RLMC$ is a parallelogram, implying that $CR$ is parallel to $ML$, but we know that $ML$ is parallel to $AC$, then $A$, $C$, $R$ are collineal. $MN$ is perpendicular to $PN$ if and only if $NP$ is the perpendicular bisector of $MC$ if and only if $PN$ is the angle bisector of $\angle MPR$ if and only if $PN$ is the angle bisector of $\angle MPC$, as requiered.