If regular triangles are drawn outwards on the sides of an acute-angled triangle ABC to vertices A'B'C' the lines AA', BB', CC', are concurrent and form a star; that is the angles between them are all 60°. This is the Fermat centre of the triangle. It is the only point within the triangle at which the lines from the vertices form a star.
The Fermat centre is also the point of concurrence of the circumcircles of the three regular triangles. This result remains true for obtuse angled triangles with angle less than 120°. When the angle equals 120° the Fermat centre coincides with the obtuse vertex. For a triangle with angle greater than 120° the Fermat centre is external. In the limiting case of three collinear points Fermat points can still be identified, one above and one below the line.
In the case of a regular triangle besides the centre of symmetry all the points of the circumcircle are star points.
The lines AA', BB', CC' are all equal in length to PA + PB + PC.
If ABC represent holes in a flat smooth tabletop and PA, PB, PC, represent three smooth strings, knotted at P, to which equal weights are suspended below the table, then the equilibrium position of P is at the Fermat point.
The centres of the three regular triangles themselves form a regular triangle, called Napoleon's triangle. Its sides are perpendicular bisectors of PA, PB, PC.