Two Circles in a Square: What Is It About?
A Mathematical Droodle

Explanation

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Copyright © 1996-2012 Alexander Bogomolny

The applet suggests the following statement [Greitzer, p. 57, Honsberger, p. 118]:

Assume points M and N are selected on the sides AD and BC of the square ABCD. Let K be an arbitrary point on MN. Besides K, the circumcircles of triangles AMK and CNK intersect at a point P. Prove that P always lies on the diagonal AC.

Proof

Assume that M is strictly inside segment AD, while N is strictly inside segment BC. Assume the quadrilateral AMKP is convex. (Slight modifications may be necessary if those conditions do not hold.)

First, ∠APK + ∠AMK = 180°. Also ∠KNC = ∠KPC and ∠AMK = ∠KNC. Therefore, ∠APK + ∠KPC = 180°, which exactly means that P lies on the diagonal AC.

Remark 1

Nathan Bowler has observed that the proposition holds not only for squares, but also for rectangles and, in fact, parallelograms.

Remark 2

As an afterthought of handling a different problem it became clear that the diagram depicted by the applet offers more properties than has been suggested by either of the references. These will be investigated along with the extended problem.

References

1. S. Greitzer, Arbelos, v 5, MAA, 1991
2. R. Honsberger, Mathematical Chestnuts From Around the World, MAA, 2001

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Copyright © 1996-2012 Alexander Bogomolny