Difference between revisions of "1991 OIM Problems/Problem 6"
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Case 1: <math>\angle NHM = 90^{\circ}</math> | Case 1: <math>\angle NHM = 90^{\circ}</math> | ||
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| + | If you measure the angle <math>\angle NHM</math> on teh given points and it happens to be a right angle, then | ||
* 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 
, 
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: 
If you measure the angle 
on teh given points and it happens to be a right angle, then
- 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.
