2006 Romanian NMO Problems/Grade 7/Problem 1

Revision as of 19:18, 11 October 2013 by Rodrigo1702 (talk | contribs) (Problem)

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. ! Missing $ inserted.)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.

See also

Invalid username
Login to AoPS