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>.
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 ''LC/BC=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, then A,C,R are collineal.
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==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>.
  
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 MPR if and only if PN is the angle bisector of MPC, as requiered.
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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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</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.
  
 
==See also==
 
==See also==

Revision as of 19:18, 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{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

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