Problem
GEO-B3-M05-P009 Constructing the Fermat Point
All angles of \(ABC\) are less than \(120^\circ\). External equilateral triangles \(BCX\) and \(CAY\) are constructed. Prove that lines \(AX\) and \(BY\) meet at a point \(T\) such that \(\angle ATB=\angle BTC=\angle CTA=120^\circ\).
C. Hint 1. Rotate the picture by \(60^\circ\) about \(C\).
D. Hint 2. The rotation sends one equilateral construction into the direction of the other.
The \(60^\circ\) rotation about \(C\) sends \(B\) to \(X\), since \(BCX\) is equilateral. It also sends the line associated with \(A\) into the line associated with the construction on \(CA\). Thus \(T=AX\cap BY\) is arranged so that the directions \(TB\) and \(TC\), and then \(TC\) and \(TA\), differ by \(120^\circ\).
More explicitly, from \(CB=CX\), \(CA=CY\), and the \(60^\circ\) angles, similar triangles around \(T\) appear. They give \(\angle BTC=120^\circ\), and cyclically the other two angles. Hence \(T\) is the Fermat point.
Ask students to explain why all angles below \(120^\circ\) ensure the point is inside.