--- In

Hyacinthos@yahoogroups.com, "orl_ml" <orlando.doehring@h...>

wrote:

> Given an equilateral triangle ABC and a point M in the plane (ABC).

> Let A', B', C' be respectively the symmetric through M of A, B, C.

>

> a) Prove that there exists a unique point P equidistant from A and

> B', from Band C' and from C and A'.

>

> b) Let D be the midpoint of the side AB. When M varies (M does not

> coincide with D), prove that the circumcircle of triagle MNP (N is

> the intersection of the line DM and AP) pass through a fixed point.

For (a): We can see that triangle (B'C'A') is obtained from (ABC) by

a rotation of -60 degrees followed by a translation, and because if

we compose a translation and a rotation we get a rotation, it means

that triangle (B'C'A') is obtained from (ABC) by a rotation of -60

deg, and the center of that rotation is the point we're looking for.

Moreover, PAB', PBC" and PCA' are equilateral.

For (b): Are you saying that M moves on a line? I drew some dynamic

cketches and found that if M moves on a line l then that point moves

on a line l' perpendicular to l. If M moves on a circle, then the

circumcircle of MNP also moves on a circle.