Problem
GEO-B3-M05-P021 Minimal Property of the Fermat Point
Let all angles of \(ABC\) be less than \(120^\circ\), and let \(T\) be the Fermat point. Prove that for any point \(X\) inside the triangle, \(XA+XB+XC\ge TA+TB+TC\).
C. Hint 1. Rotate one of the segments \(XB\) by \(60^\circ\).
D. Hint 2. Reduce the sum of three lengths to the length of a broken line.
Construct an external equilateral triangle \(BCU\). For an arbitrary point \(X\), rotate segment \(XB\) by \(60^\circ\) so that \(B\) maps to \(U\); let \(X'\) be the image of \(X\). Then \(XB=UX'\), and \(XX'=XB\) with \(\angle BXX'=60^\circ\).
The sum \(XA+XB+XC\) becomes the length of a broken line connecting fixed points through \(X\) and \(X'\). By the triangle inequality, this broken line is at least the corresponding straight segment. Equality occurs when the broken line is straight, which is equivalent to the \(120^\circ\) angles at \(T\). Therefore the minimum is attained at the Fermat point.
An olympiad problem on straightening a broken line by rotation.