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Difference between revisions of "2002 OIM Problems/Problem 3"
 
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== Problem ==
 
== Problem ==
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Pablo was copying the following problem:
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A point <math>P</math> is interior to the equilateral triangle <math>ABC</math> and satisfies that <math>\angle APC = 120^{\circ}</math>.  Let <math>M</math> be the intersection of <math>CP</math> with <math>AB</math> and <math>N</math> be the intersection of <math>AP</math> with <math>BC</math>. Find the locus of the circumcenter of the triangle <math>MBN</math> by varying <math>P</math>.
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''<cmath>\text{Consider all sequences of 2004 real numbers}(x_0,x_1,x_2,\cdots , x_{2003})\text{, such that}</cmath>
 
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<cmath>x_0=1\text{,}</cmath>
 
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<cmath>0\le x_1 \le 2x_0 \text{,}</cmath>
 
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<cmath>0\le x_2 \le 2x_1 \text{,}</cmath>
 
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<cmath>\vdots </cmath>
 
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<cmath>0\le x_{2003} \le 2x_{2004} \text{.}</cmath>
 
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<cmath>\text{Among all these sequences, find the one for which the following expression takes its largest value:}</cmath>
 
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<cmath>S = ...</cmath>''
 
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When Pablo was going to copy the expression for <math>S</math>, they erased the blackboard. The only thing he could remember was that <math>S</math> was of the form
 
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<cmath>S=\pm x_1\pm x_2\pm \cdots\pm x_{2002}+x_{2003}</cmath>
 
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where the last term, <math>x_{2003}</math>, had a coefficient +1, and the previous ones had a coefficient +1 or -1. Show that Paul, despite not having the complete statement, can find with certainty the solution to the problem.
 
  
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
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== See also ==
 
== See also ==
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https://www.oma.org.ar/enunciados/ibe18.htm
 

Latest revision as of 03:41, 14 December 2023

Problem

A point $P$ is interior to the equilateral triangle $ABC$ and satisfies that $\angle APC = 120^{\circ}$. Let $M$ be the intersection of $CP$ with $AB$ and $N$ be the intersection of $AP$ with $BC$. Find the locus of the circumcenter of the triangle $MBN$ by varying $P$.

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

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See also

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