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{LN}{BC} = \frac{AM}{AB} $, applicating Thales theorem it follows that$ ML $is parallel to$ AC$.

Let$ (Error compiling LaTeX. Unknown error_msg)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.

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2006 Romanian NMO Problems/Grade 7/Problem 1

Problem

Solution

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