Problem
GEO-B3-M06-P013 Fermat via Trig Ceva
In triangle \(ABC\), all angles are less than \(120^\circ\). Prove that there exists a point \(T\) such that \(\angle ATB=\angle BTC=\angle CTA=120^\circ\), reducing the problem to trig Ceva for suitable cevians.
C. Hint 1. Construct from each vertex the ray that should pass through \(T\).
D. Hint 2. The angles at vertices are expressed as \(60^\circ\) plus triangle angles.
Define rays from \(A,B,C\) so that if they meet at \(T\), the angles around \(T\) become \(120^\circ\). In the standard construction, these are the lines toward the outer vertices of external equilateral triangles.
For these three rays, the angles in trig Ceva are of the form \(B+60^\circ\), \(C+60^\circ\), and cyclically. Substitution gives a telescoping product: \[ \frac{\sin(B+60^\circ)}{\sin(C+60^\circ)} \frac{\sin(C+60^\circ)}{\sin(A+60^\circ)} \frac{\sin(A+60^\circ)}{\sin(B+60^\circ)}=1. \] Hence the rays are concurrent. Their point of concurrence is \(T\), and the construction gives the \(120^\circ\) angles.
This can be given as a trigonometric version of the Fermat construction.