Let MNP any equilateral triangle circumscribed to ABC, that is with A,B,C on the sides of MNP (or extensions).

Locus of the centroid G of MNP

Let M'N'P' any equilateral triangle inscribed in ABC, that is M',N',P' on the sides of ABC (or extensions)

Locus of the centroid G' of M'N'P'

Let M"N"P" the centers of three equal circles inscribed in ABC, (that is tangent "inside" the lines ABC, the circles themselves may be outside)

Locus of center M" of one of these circles (resp. N",P")

Locus of the centroid G" of M"N"P"

(inscribed triangle, equal inscribed circles)

In the applet, draggable point M defines the directions of the equilateral triangles.

We can "see" (without any proof) that :

The locus of G is a circle, going through the Fermat-Toricelli point F of ABC.

The locus of G' is a straight line

The locus of M" is an hyperbola

The locus of G" is an hyperbola.

Of course, only portions of these locii are really used, depending on the conditions
(on sides or extensions) and the circles being really inside, or outside.

For locus of M" and G", the applet "jumps" when the MNP triangle collapses into the Fermat-Toricelli point.
The circle to consider is the pink circle with center M" **and** the "associated" blue circle.

This associated circle comes in the construction of M" from M' when choosing the other side of line AB.
(see the general construction)

They swap "suddenly" when M goes through F, that is M' goes "through" infinity.
The applet displays then a spurious transversal line.

The other branch of the hyperbola appears when the assiociated blue circles are all three "outside" tangent to ABC.

GM goes through the midpoint Ia of arc BC in the circle locus of M.

Similarily, GN goes through midpoint Ib of arc AC and GP through midpoint Ic of arc AB.

The locus of G is therefore the circle from where we see IaIb under a 120°/60° angle.

We then deduce that IaIbIc is an equilateral triangle and the locus of G is the circumcircle to that triangle.

For Fermat point G of ABC to be on this circle is obvious, when triangle MNP collapses into M = N = P = G = F.

This property is general for any circumscribed triangle MNP of constant shape, and any noticeable point in MNP.

Draggable points A, B, C define triangle ABC.

The shape of MNP is defined through the draggable intersection point F common to the circles locus of M,N,P,
hence defining the angles of this triangle.

M is draggable on its circle to rotate triangle MNP.

An arbitrary triangle mnp is used to define the barycentric coordinates of G in MNP by dragging point g.

To simplify the writing, we call here just MNP the inscribed triangle, former M'N'P'.

Miquel Theorem :

Let M,N,P three points on the sides of ABC.
The circumcircles to ANP, BMP and CMN have a common point S |

We then deduce the following corollary :

All similar triangles inscribed in ABC have the same Miquel point S,
which is the similitude center of all these triangles. |

Here, all the equilateral triangles MNP inscribed in ABC have the same Miquel point S,
and S is the similitude center of any two such triangles.

From the MNP point of view, point S is the only point from where we see the sides of MNP
under angles π-A, π-B, π-C.

Hence there is a constant similitude of center S transforming M into N.
It transforms line BC into AC.

In the same way, there is a constant similitude of center S which transforms M into P.

Any point Q with constant barycentric coordinates in MNP is then obtained from M
by some constant similitude of center S.

The locus of this point Q is then the transformed of locus of M, that is line BC, by that similitude,
this locus is therefore a straight line.

In the applet, a such point Q is defined by the yellow points with constant ratios qM/qN and QP/Qq.

Specifically the centroid G of MNP, with coordinates (1,1,1) in MNP, describes a straight line.

This line is at easiest defined by any two positions of MNP.