Pedal Collinearities
Hubert Shutrick came up with the following statement:
 |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny Pedal CollinearitiesBy 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.
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. |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
| 40619984 |
