Dear Ricardo,

If ABC is the required triangle and A'B'C' is the

given triangle then AA' = BB' = CC'

and the three lines AA', BB', CC' pass through

the first Fermat point F of triangle ABC.

We construct on the sides of triangle A'B'C'

equilateral triangles and let A", B", C" be the vertices

of these triangles outside of A'B'C'.

Since B'FC' = 120 and B'A"C' = 60

the points F, B', A", C' are concyclic. (1)

The rotation about C' by angle 60 moves

B to A

B' to A".

Hence A"A = BB' = AA' and

angle C'A"A = angle C'B'B

= angle C'B'F

= angle C'A"F because of (1)

This means that the line A"A passes through F,

the points A", A, A' are collinear

and the required point A is the mid point of

the segment A'A". Similarly we construct the

points B, C of triangle ABC.

Best regards

Nikos Dergiades

>Dear Hyacinthos:

>

>

>Construct a triangle, given the tree points in the plane

that are the outer vertcies of equilateral triangles

constructed outward on the sides of the desired triangle.

>

>

>Yaglom, I (1973): Geometric transformations / translated

from the Russian by Allen Shields, A. Shenitzer (p 12)

>

>

>Best regards

>

>Ricardo Barroso (Universidad de Sevilla)