# A problem that was not interesting for Erdős

As a young man (I was about 15 years old) I had the opportunity to meet Paul Erdős in Budapest who visited young people at an event. At a certain point of the meeting I could go the blackboard and show the following problem to Erdős:
Given the triangle ABC. Let us erect squares on its sides externally. Choose a point on an external side of one of the squares, say J (see the figure below), join it with C and consider the intersection point K with the external side of another square. Now repeat this idea and create point L by finding the intersection of KA and HI. Finally, do this again by intersecting LB and DE, getting the final point M.
Question: How to choose J to get back to this initial point, namely, to have M=J?

Erdős was listening to the solution I quickly disclosed, but he was not interested in the problem. I was a bit disappointed, but I had to admit that mathematicians can have different interests...
That time, around 1990, this type of problem was difficult to study. Noone could solve it in my class, even in the school, but the father of one of my classmates, György Gyurkó, having doctorate in mathematics, found an easy and very elegant solution (see below).
Surprisingly, such questions can be quickly solved by GeoGebra today, actually, just for particular cases, but it is much more than we had in the nineties! The applet above includes the GeoGebra command

**LocusEquation(M==J,J)**and the equation 0=-1 is shown that there is no solution. (If you cannot see that, just drag one of the vertices of the triangle to recompute the equation.)## Gyurkó's solution

Let us assume that JKLM is a triangle, that is, M=J. Consider the angles of the triangle JKL, and let us focus, for example, on the one on side FG. This angle (at K) is at least 45 degrees, because if K coincides with F or G, the triangle CKA is isosceles and have a right angle at C or A. On the other hand, the angle at K is at most 53,13... degrees, and this will occur only when K is the midpoint of FG.
This is an immediate consequence of the inscribed angle theorem. By using this idea, but from a different point of view, the angle CKA can be 60 degrees only if K lies on a circle that is a circumcircle of the equilateral triangle erected on side CA, externally. Clearly, that circle has no intersection point with the side FG, and, what is more, taking a point K' outside of this circle, the angle CK'A must be less than 60 degrees.
By combining this result for all vertices of the triangle JKL, we learn that the sum of the internal angles is less than 3 times 60 degrees. This contradiction finishes the proof.

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