### Circles And Parallels: What is this about?

A Mathematical Droodle

What if applet does not run? |

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Copyright © 1996-2018 Alexander BogomolnyThe applet hints at the following problem [Honsberger, pp. 102-103]:

From an arbitrary point P on side AB of ΔABC draw parallels PM and PN to the sides AC and BC, respectively. The circumcircles of triangles APN and BPM intersect in P and Q. Prove that as P varies on AB, all the lines PQ are concurrent. |

What if applet does not run? |

A couple of remarks are in order. First, the problem obviously requires some adjustment for the case where the circles at hand are tangent to each other. If they are, their common tangent still satisfies the conditions of the problem. Second, Honsberger also indicates that the problem has been plucked from a collection by V. V. Prasolov. Although I do have the book, I was unable to locate the problem there.

I shall first give a proof that appears in [Honsberger] and then suggest a somewhat different approach.

Construct ∠C at A and B to give an isosceles ΔABX outwardly on AB; then

Now, the parallel lines make ∠C equal to both ∠ANP and ∠BMP. Thus AX and BX are equal **tangents** to the circles, implying X is always a point on their radical axis which, in the case of these intersecting circles, is their common chord PQ. Thus, as P varies on AB, PQ always goes through the fixed point X.

My variant starts with the observation that the radical axis of two intersecting circles passes through the two points of intersection. The radical axis of two tangent circles is their common tangent. Thus we wish to establish the immutability of a point on the radical axis of two circles. It would have been nice had there been a third circle such that the fixed point in the problem were the radical center of the triad of circles. What circle that is related to ΔABC can we try? The circumcircle is a natural first choice and a successful one at that.

Indeed, triangles ABC and ANP are homothetic at A and so are their respective elements, the circumcircles in particular. Therefore, the radical axis of the circumcircles of triangles ABC and ANP is their common tangent at A. Similarly, the radical axis of the circumcircles of triangles ABC and BMP is their common tangent at B. We now have three circles, three radical axis and therefore a radical center where the three lines meet. Note, that the tangents to the circumcircle of ΔABC at A and B do not depend on the position of P. These two tangents define a point, call it X. Add to this two circles, one tangent to the circumcircle at A, the other at B. This step confirms on the two tangents the honor of being radical axes of two pairs of circles. The third radical axis - the one of the added pair - is bound to pass through the same point X.

### References

- R. Honsberger,
*Mathematical Delights*, MAA, 2004 - V. V. Prasolov,
*Problems in Planimetry*, v 1, Nauka, Moscow, 1986, in Russian

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Copyright © 1996-2018 Alexander Bogomolny