1991 OIM Problems/Problem 6
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 
on the given points and it happens to be a right angle, then constructing the triangle is easy because point is also point 
of the triangle 
. One can notice if this angle is a right angle or not if you can draw a perpendicular from point to line 
and it passes through . If this happens to be the case, then since 
and 
then one can simply draw a circle with the compass at points and with radiuses measuring 
and 
respectively. Then extend the lines 
and 
to the intersection on their respective circles at 
and 
respectively. Then draw triangle .
Case 2: 
Let the red circle in the image above be the circumcircle of triangle . Let 
be a diameter of the circle. This means that 
and 
are both equal to 
because right angle triangles inscribed in circles with the hypothenuse on the diameter. Therefore 
is parallel to 
and 
is parallel to 
. Thus quadrilateral 
is a parallelogram with in the center and 
. So, one can draw point 
using and . Since 
, then 
is the diameter of a circle that also passes through . This means that one can find point from the intersection of this circle and the perpendicular to 
that passes through and we can now start our construction as follows:
- 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.
