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Gergonne and Medial Triangles Are Orthologic: What is this about?
A Mathematical Droodle


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Explanation

Copyright © 1996-2008 Alexander Bogomolny

 

 

 

 

 

 

 

 

 

 

 

 

 

The applet suggests the following statement and one of its proofs:

In ABC, let Ma, Mb, Mc be the midpoints of sides BC, AC, and AB. Let A', B', C' be the points where the incircle touches those sides. (A'B'C' is known as Gergonne or contact triangle of ABC.) Then the perpendiculars from the midpoints Ma, Mb, Mc onto the opposite sides of A'B'C' are concurrent.


This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


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According to R. Honsberger, the problem has been submitted by Bulgaria to an IMO, but was left unused. It was discussed and solved by J. T. Groenman, The Netherlands, in Crux Mathematicorum, 1987, 75. Here's Groenman's solution:

The center of the incircle of a triangle lies at the intersection of its angle bisectors. The incenter is thus naturally equidistant from the sides of the triangle. It follows that the tangents AB', AC' from vertex A of ABC to its incircle are equal and AB'C' is isosceles. The bisector of angle A is perpendicular to B'C', hence parallel to the perpendicular from Ma. The latter, being parallel to the angle bisector in ABC, serves as the angle bisector of angle MbMaMc in the medial triangle MaMbMc, since the two triangles are homothetic. The same holds for the other two perpendiculars. The three intersect as being angle bisectors of MaMbMc.

The fact just proven means that the Gergonne and medial triangles are orthologic, which, as we know from Maxwell's theorem, is a symmetric relationship. We conclude that the perpendiculars from the vertices of Gergonne triangle A'B'C' onto the opposite sides of MaMbMc also concur. But this fact is obvious. Indeed the sides of the medial triangle are parallel to those of ABC, so that the perpendiculars at the vertices of A'B'C' where the incircle of ABC touches its sides meet at the incenter of ABC, naturally. Going backwards, we get another proof of the Bulgarian problem.

References

  1. R. Honsberger, From Erdös To Kiev, MAA, 1996, pp. 99-102

Maxwell's Theorem (of orthologic triangles)

  1. Gergonne and Medial Triangles Are Orthologic
  2. Pedal Triangle and Isogonal Conjugacy
  3. Orthologic Triangles in a Quadrilateral

Copyright © 1996-2008 Alexander Bogomolny

28727341Page copy protected against web site content infringement by Copyscape


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