Difference between revisions of "1991 OIM Problems/Problem 6"
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Latest revision as of 08:41, 23 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 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:
Given the points ,
, and
, we first draw a perpendicular from point
to
with straight edge and compass using the traditional known method of finding tow points from
on
equidistant to
and from those two points drawing the perpendicular bisector to them. That way we can draw the black perpendicular line.
Then, with the compass draw the red circle with center at and radius
. From
we extend the line
and where it intersects the red circle that's our point
. We then draw blue line
and find it's bisection point using traditional bisection method with compass and straight edge. From this bisection point and the compass using that as its center we can draw the blue circle. This blue circle intersects the black perpendicular line at
. 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
.
But the given points do not give us one unique triangle. It gives us two because the blue circle also intersects the black perpendicular line at another point. We will call this point . Then using the same methods as described in the procedure for triangle
we find triangle
which also has the given points
and
And this provides all of the cases and all of the triangles we can draw from those given points.
- Note. I actually competed at this event in Argentina when I was in High School representing Puerto Rico. If I remember correctly I solved the case 1 described here as it was the easiest and then wasted a lot of time trying to figure out a way unsuccessfully. I also noted that two triangles could be build instead of just one. I think they awarded me 3 points out of 10.
~Tomas Diaz. ~orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.