## equilateral triangle ABC

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• 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
Message 1 of 2 , Mar 12, 2004
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
Message 2 of 2 , Apr 2, 2004
--- 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.
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