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

(when a problem doesn't have a solution, put a {{solution}} tag under it if there isn't already.)
Line 2: Line 2:
 
Let <math>ABC</math> be a triangle and the points <math>M</math> and <math>N</math> on the sides <math>AB</math> respectively <math>BC</math>, such that <math>2 \cdot \frac{CN}{BC} = \frac{AM}{AB}</math>. Let <math>P</math> be a point on the line <math>AC</math>. Prove that the lines <math>MN</math> and <math>NP</math> are perpendicular if and only if <math>PN</math> is the interior angle bisector of <math>\angle MPC</math>.
 
Let <math>ABC</math> be a triangle and the points <math>M</math> and <math>N</math> on the sides <math>AB</math> respectively <math>BC</math>, such that <math>2 \cdot \frac{CN}{BC} = \frac{AM}{AB}</math>. Let <math>P</math> be a point on the line <math>AC</math>. Prove that the lines <math>MN</math> and <math>NP</math> are perpendicular if and only if <math>PN</math> is the interior angle bisector of <math>\angle MPC</math>.
 
==Solution==
 
==Solution==
 +
{{solution}}
 +
 
==See also==
 
==See also==
 
*[[2006 Romanian NMO Problems]]
 
*[[2006 Romanian NMO Problems]]
  
 
[[Category:Olympiad Geometry Problems]]
 
[[Category:Olympiad Geometry Problems]]

Revision as of 07:33, 27 August 2008

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

This problem needs a solution. If you have a solution for it, please help us out by adding it.

See also

Invalid username
Login to AoPS