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

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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 <math>LC</math>, then <math>2CN</math>=<math>LC</math>, the given information is the same as \frac{LN}{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>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>, then </math>A<math>,</math>C<math>,</math>R<math> are collineal.
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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$, as requiered.
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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.
  
 
==See also==
 
==See also==

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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