Six Concyclic Points
The applet below illustrates a generalization of Problem 1 from the 2008 International Mathematics Olympiad:
Let P be a point in the plane of ΔABC with points A', B', C' on the cevians AP, BP, CP, respectively. Assume that the circumcircles (O_{a}), (O_{b}), (O_{c}), of the triples |
What if applet does not run? |
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Copyright © 1996-2018 Alexander BogomolnySolution
One 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.
What if applet does not run? |
Points A_{1}, A_{2}, B_{1}, B_{2} are concyclic because the powers of point C with respect to the two circles (O_{a}) and (O_{b}) are equal implying
CA_{1} × CA_{2} = CP × CC' = CB_{1} × CB_{2}. |
By the converse of the Intersecting Chords Theorem, A_{1}, A_{2}, B_{1}, B_{2} lie on a circle which we denote (X_{c}). We shall show that C_{1} and C_{2} also lie on (X_{c}).
The power of point A with respect to (X_{c}) is
AB_{1} × AB_{2} | = AP × AA' | |
= AC_{1} × AC_{2} |
which is the power of A with respect to (O_{c}). It follows that A lies on the radical axis of (X_{c}) and (O_{c}). 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 (X_{c}) and (O_{c}) is exactly AB implying that the points of intersection of (X_{c}) and (O_{c}) that define their radical axis lie on AB. But, since (O_{c}) meets AB in C_{1} and C_{2}, so does (X_{c}). This shows that (X_{c}) passes through C_{1} and C_{2}.
Remark: The statement just proved leads to a simple 6 to 9 Point Circle proof of the existence of the 9-point circle.)
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Copyright © 1996-2018 Alexander Bogomolny