The Fermat Point of a triangle ABC is the point F such that AF+BF+CF is minimised.
Pierre Fermat (1601-1665) challenged Evangelista Torricelli (1608-1647) to find this point.
If any of the angles of the triangle is 120º or more, F coincides with the vertex of that angle. Otherwise, F is the point inside ABC such that the angles AFB, BFC, CFA are all 120º.
To locate F (assuming all angles are less than 120º), construct the equilateral triangles BCX, CAY, ABZ all pointing outwards so that they do not overlap with ABC. Then the lines AX, BY, CZ all intersect at F. Also, the circumcircles of these three triangles all pass through F. AX, BY and CZ are of equal length, this being AF+BF+CF.