### Two Triples of Concurrent Circles

What is this about?

A Mathematical Droodle

What if applet does not run? |

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Copyright © 1996-2018 Alexander BogomolnyThe applet is supposed to illustrate an additional property of the six circles configuration observed by Bui Quang Tuan.

_{1}A

_{2}A

_{3}, start with inscribing a circle C

_{1}into ∠A

_{3}A

_{1}A

_{2}and note point T

_{12}of tangency on the side A

_{1}A

_{2}. Next inscribe circle C

_{2}into ∠A

_{1}A

_{2}A

_{3}so that it is tangent to A

_{1}A

_{2}at T

_{12}and note point T

_{23}of tangency with side A

_{2}A

_{3}. Continue inscribing circles C

_{3}, C

_{4}, C

_{5}, C

_{6}, and so on, into angles A

_{3}, A

_{1}, A

_{2}, A

_{3}and so on, tangent to the previous circle. Then

_{7}= C

_{1}.

_{1}, C

_{3}, C

_{5}concur in a point, they concur at the Gergonne point of

_{1}A

_{2}A

_{3}

_{2}, C

_{4}, C

_{6}.

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

A_{1}A2 is a common tangent of circles C_{1} and C2 and, as such, it is also their radical axis. Similarly, A_{1}A3 is the radical axis of C_{1} and C6, implying that A_{1} is the radical center of three circles, C_{1}, C2 and C_{6}. A2A3 (or T_{56}T_{23}) is a common tangent of C2 and C_{6} so that the midpoint M_{1} of T_{56}T_{23} lies on radical axis of C2 and C_{6}. In other words, A_{1}M_{1} is the radical axis of C2 and C_{6}.

On the other hand, as we have seen, the six points of tangency T_{12}, T_{23}, ..., T_{61} are concyclic with their circumcircle centered at the incenter I of _{1}A_{2}A_{3}._{1} is perpendicular to A_{2}A_{3}, making M_{1} the point of tangency of the incircle of _{1}A_{2}A_{3}_{2}A_{3}. Thus, A_{1}M_{1} is one of the Gergonne cevians in _{1}A_{2}A_{3}.

Define M_{2} and M_{3} similarly to M_{1}. Then, on one hand, A_{2}M_{2} and A_{3}M_{3} are radical axes of pairs C_{4}, C_{6} and C_{2}, C_{4}. Three lines A_{1}M_{1}, A_{2}M_{2}, A_{3}M_{3} concur at the Gergonne point of _{1}A_{2}A_{3}_{4}, C_{6} 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 _{1}A_{2}A_{3}

Similarly, if circles C1, C_{3}, C_{5} are concurrent, their common point is again the Gergonne point of _{1}A_{2}A_{3}

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

The published solution has been supplied by H. Eves:

_{1}, P

_{2}, P

_{3}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

_{i}P

_{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

_{i}P

_{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.

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