Six Concyclic Points

The applet below illustrates a generalization of Problem 1 from the 2008 International Mathematics Olympiad:

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.

Points A₁, A₂, B₁, B₂ are concyclic because the powers of point C with respect to the two circles (O_a) and (O_b) are equal implying

By the converse of the Intersecting Chords Theorem, A₁, A₂, B₁, B₂ lie on a circle which we denote (X_c). We shall show that C₁ and C₂ also lie on (X_c).

The power of point A with respect to (X_c) is

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 C₁ and C₂, so does (X_c). This shows that (X_c) 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

• A Property of the Line IO: A Proof From The Book
• Cherchez le quadrilatère cyclique II
• Circles On Cevians
• Circles And Parallels
• Circles through the Orthocenter
• Coaxal Circles Theorem
• Isosceles on the Sides of a Triangle
• Properties of the Circle of Similitude
• Six Concyclic Points
• Radical Axis and Center, an Application
• Radical axis of two circles
• Radical Axis of Circles Inscribed in a Circular Segment
• Radical Center
• Radical center of three circles
• Steiner's porism
• Stereographic Projection and Inversion
• Stereographic Projection and Radical Axes
• Tangent as a Radical Axis
• Two Circles on a Side of a Triangle
• Pinning Butterfly on Radical Axes