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 {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.

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Solution

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.

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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

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 (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

	AB₁ × AB₂	= AP × AA'
		= AC₁ × AC₂

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

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