### 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 https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Explanation 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, PPi lying opposite APi. Then, since the six Aij are on a circle C, it is clear that I is the center of C, and therefore that all the segments PiPjk, (i≠j≠k≠i), are equal. Hence Pi and Ai lie on the radical axis of Cj and Ck. Thus the common point of the three circles Ci must be the point of concurrence of the three lines AiPi, 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.  