Difference between revisions of "1991 OIM Problems/Problem 6"
| Line 1: | Line 1: | ||
== Problem == | == Problem == | ||
| − | Given 3 non-aligned points <math>M</math>, <math>N</math> and <math> | + | 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 | ||
| Line 6: | Line 6: | ||
== Solution == | == Solution == | ||
| − | + | 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 19:03, 22 December 2023
Problem
Given 3 non-aligned points 
, 
and 
, we know that and are midpoints of two sides of a triangle and that 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: 
- 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.
