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 (Oa), (Ob), (Oc), of the triples {P, B', C'}, {P, C', A'}, {P, A', B'}, intersect sides BC, AC, AB in {A1, A2}, {B1, B2}, {C1, C2}, respectively. Then the six points A1, A2, B1, B2, C1, C2 are concyclic.

 

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.


What if applet does not run?

Solution

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Copyright © 1996-2017 Alexander Bogomolny

Solution

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.

 

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.


What if applet does not run?

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

  CA1 × CA2 = CP × CC' = CB1 × CB2.

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

 AB1 × AB2= AP × AA'
  = AC1 × AC2

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

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Copyright © 1996-2017 Alexander Bogomolny

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