Reflections of a Line through the Orthocenter in the Sides of an Acute Triangle

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A line is drawn through the orthocenter (the intersection point of the altitudes) of an acute-angled triangle. Prove that the symmetric images , , of with respect to the sides , , have a point in common, which lies on the circumcircle of .

Contributed by: Jaime Rangel-Mondragon (July 2013)
Open content licensed under CC BY-NC-SA


Snapshots


Details

This Demonstration provides four locators to change the shape of triangle and to rotate the (blue) line that goes through the point . The result is true even if is not inside the triangle; that is, even when is not an acute triangle. Rotate around . The point in common mentioned in the statement of the problem is the intersection of the brown lines. This is problem 41 taken from the ninth International Mathematical Olympiad (IMO) held at Celtinje, Yugoslavia, July 2-13, 1967.

Reference

[1] D. Djukić, V. Janković, I. Matić, and N. Petrović, The IMO Compendium, 2nd ed., New York: Springer, 2011.



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