Pedal Collinearities

 

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Hubert Shutrick came up with the following statement:

  Let a straight line through the circumcenter of ΔABC meet the sides BC, AC, AB at P, Q, R, respectively. Then the pedal circles of the three points are coaxal: they pass through two common points and their centers are collinear.

Proof

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Copyright © 1996-2018 Alexander Bogomolny

Pedal Collinearities

By Griffiths' theorem, the pedal circles of the points on the same line through the circumcenter of a triangle concur at a point on the 9-point circle of that triangle. As Prof. Shutrick surmised, for the three points at which the line crosses the sides of the triangle, the three pedal circles have an additional point in common so that they are coaxal and, in particular, have collinear centers.

 

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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Looking for a proof, Dr. Shutrick discovered that the transversal line does need to pass through the circumcenter. The applet shows that the statement indeed holds in a more general case. The line can be translated or rotated (if dragged near the border of the applet region.)

As a matter of fact, he also found that the statement generalizes even further. The base observation was that the center of Griffiths; circle, say the one through A lies on the midline of ΔABC parallel to BC. A midline connects two midpoints. Three midpoints are feet of three cevians through the center of the triangle. So a general case deals with one transversal (a Menelaus' line I would say) and a triple of concurrent cevians (a Ceva's triple they may be called. Dr. Shutrick finally proved that some intersections in this configuration are indeed collinear.

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Copyright © 1996-2018 Alexander Bogomolny

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