Difference between revisions of "1991 OIM Problems/Problem 6"

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== Problem ==
 
== Problem ==
Given 3 non-aligned points <math>M</math>, <math>N</math> and <math>P</math>, we know that <math>M</math> and <math>N</math> are midpoints of two sides of a triangle and that <math>P</math> is the point of intersection of the heights of said triangle. Build the triangle.
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Given 3 non-aligned points <math>M</math>, <math>N</math> and <math>H</math>, we know that <math>M</math> and <math>N</math> are midpoints of two sides of a triangle and that <math>H</math> is the point of intersection of the heights of said triangle. Build the triangle.
  
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
 
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
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== Solution ==
 
== Solution ==
  
 
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Case 1: <math>\angle NHM = 90^{\circ}</math>
  
 
* Note.  I actually competed at this event in Argentina when I was in High School representing Puerto Rico.  I think I may have been able to build some cases of this.  I don't remember much of it.
 
* Note.  I actually competed at this event in Argentina when I was in High School representing Puerto Rico.  I think I may have been able to build some cases of this.  I don't remember much of it.

Revision as of 18:03, 22 December 2023

Problem

Given 3 non-aligned points $M$, $N$ and $H$, we know that $M$ and $N$ are midpoints of two sides of a triangle and that $H$ is the point of intersection of the heights of said triangle. Build the triangle.

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

Solution

Case 1: $\angle NHM = 90^{\circ}$

  • Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. I think I may have been able to build some cases of this. I don't remember much of it.

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

https://www.oma.org.ar/enunciados/ibe6.htm