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

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==Problem==
 
==Problem==
 
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>.
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==Solution==
 
==Solution==
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Let <math>L</math> be a point on <math>BC</math> such that <math>N</math> is the midpoint of LC, then <math>2CN</math>=<math>LC</math>, the given information is the same as <math>\frac{LC}{BC} = \frac{AM}{AB}</math>, applicating Thales theorem it follows that <math>ML</math> is parallel to <math>AC</math>.
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Let <math>R</math> be the point on <math>MN</math> such that <math>MN</math>=<math>NR</math>, in view of <math>MN</math>=<math>NR</math> and <math>LN</math>=<math>NC</math> it follows that <math>RLMC</math> is a parallelogram, implying that <math>CR</math> is parallel to <math>ML</math>, but we know that <math>ML</math> is parallel to <math>AC</math>, hence <math>A</math>,<math>C</math>,<math>R</math> are collineal.
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<math>MN</math> is perpendicular to <math>PN</math> if and only if <math>NP</math> is the perpendicular bisector of <math>MC</math> if and only if <math>PN</math> is the angle bisector of <math>\angle MPR</math> if and only if <math>PN</math> is the angle bisector of <math>\angle MPC</math>, as requiered.
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==See also==
 
==See also==
 
*[[2006 Romanian NMO Problems]]
 
*[[2006 Romanian NMO Problems]]
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[[Category:Olympiad Geometry Problems]]

Latest revision as of 20:17, 11 October 2013

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 $AC$, hence $A$,$C$,$R$ 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

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