Two Triples of Concurrent Circles

Two Triples of Concurrent Circles
What is this about?
A Mathematical Droodle

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The applet is supposed to illustrate an additional property of the six circles configuration observed by Bui Quang Tuan.

Given Δ A₁A₂A₃, start with inscribing a circle C₁ into ∠A₃A₁A₂ and note point T₁₂ of tangency on the side A₁A₂. Next inscribe circle C₂ into ∠A₁A₂A₃ so that it is tangent to A₁A₂ at T₁₂ and note point T₂₃ of tangency with side A₂A₃. Continue inscribing circles C₃, C₄, C₅, C₆, and so on, into angles A₃, A₁, A₂, A₃ and so on, tangent to the previous circle. Then C₇ = C₁. If circles C₁, C₃, C₅ concur in a point, they concur at the Gergonne point of Δ A₁A₂A₃ and the same holds for the triple of circles C₂, C₄, C₆.

The statement and the proof are due to Bui Quang Tuan.

A₁A2 is a common tangent of circles C₁ and C2 and, as such, it is also their radical axis. Similarly, A₁A3 is the radical axis of C₁ and C6, implying that A₁ is the radical center of three circles, C₁, C2 and C₆. A2A3 (or T₅₆T₂₃) is a common tangent of C2 and C₆ so that the midpoint M₁ of T₅₆T₂₃ lies on radical axis of C2 and C₆. In other words, A₁M₁ is the radical axis of C2 and C₆.

On the other hand, as we have seen, the six points of tangency T₁₂, T₂₃, ..., T₆₁ are concyclic with their circumcircle centered at the incenter I of Δ A₁A₂A₃. It follows that IM₁ is perpendicular to A₂A₃, making M₁ the point of tangency of the incircle of Δ A₁A₂A₃ with A₂A₃. Thus, A₁M₁ is one of the Gergonne cevians in Δ A₁A₂A₃.

Define M₂ and M₃ similarly to M₁. Then, on one hand, A₂M₂ and A₃M₃ are radical axes of pairs C₄, C₆ and C₂, C₄. Three lines A₁M₁, A₂M₂, A₃M₃ concur at the Gergonne point of Δ A₁A₂A₃ on one hand and at the radical center of circles C2, C₄, C₆ on the other. This assertion holds regardless whether the three circles concur or not, but, when they do, the point of concurrency being their radical center is necessarilly the Gergonne point of Δ A₁A₂A₃.

Similarly, if circles C1, C₃, C₅ are concurrent, their common point is again the Gergonne point of Δ A₁A₂A₃.

The above construction provides a solution to the problem E457 (The American Mathematical Monthly, Vol. 48, No. 9 (Nov., 1941)) proposed by V. Thebault:

Construct three circles which have one common point and which are such that each touches two sides of a given triangle, the six points of contact being concyclic.

The published solution has been supplied by H. Eves:

Let I be the incenter of the given triangle, and let P₁, P₂, P₃ be the feet of the perpendiculars from I on the sides of the triangle, PP_i lying opposite AP_i. Then, since the six A_ij are on a circle C, it is clear that I is the center of C, and therefore that all the segments P_iP_jk, (i≠j≠k≠i), are equal. Hence P_i and A_i lie on the radical axis of C_j and C_k. Thus the common point of the three circles C_i must be the point of concurrence of the three lines A_iP_i, namely the Gergonne point of the triangle. The construction is now evident.

As a side note there is another way of getting six concycling points on the side lines of a triangle.

Radical Axis and Radical Center

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