Six Concyclic PointsThe applet below illustrates a generalization of Problem 1 from the 2008 International Mathematics Olympiad:
|Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander BogomolnySolutionOne solution is based on the fact that P is the radical center of the three circles whilst the cevians AP, BP, CP are the radical axes of the circles taken in pairs.
Points A1, A2, B1, B2 are concyclic because the powers of point C with respect to the two circles (Oa) and (Ob) are equal implying
By the converse of the Intersecting Chords Theorem, A1, A2, B1, B2 lie on a circle which we denote (Xc). We shall show that C1 and C2 also lie on (Xc). The power of point A with respect to (Xc) is
which is the power of A with respect to (Oc). It follows that A lies on the radical axis of (Xc) and (Oc). By symmetry, the same holds for B. In other words, both A and B lie on the radical axis of the two circles. The radical axis then of (Xc) and (Oc) is exactly AB implying that the points of intersection of (Xc) and (Oc) that define their radical axis lie on AB. But, since (Oc) meets AB in C1 and C2, so does (Xc). This shows that (Xc) passes through C1 and C2. Remark: The statement just proved leads to a simple 6 to 9 Point Circle proof of the existence of the 9-point circle.) Radical Axis and Radical Center
|Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
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