Re: [EMHL] A problem of I. Yaglom
- 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.
>Dear Hyacinthos:that are the outer vertcies of equilateral triangles
>Construct a triangle, given the tree points in the plane
constructed outward on the sides of the desired triangle.
>from the Russian by Allen Shields, A. Shenitzer (p 12)
>Yaglom, I (1973): Geometric transformations / translated
>Ricardo Barroso (Universidad de Sevilla)
- Dear Ricardo
> 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.I believe this question originates from EMHL (Lemoine) himself, I was
> Yaglom, I (1973): Geometric transformations / translated from the Russian by Allen Shields, A. Shenitzer (p 12)
given the reference:
E.M.H. Lemoine, Question 864, Nouvelles Annales de Mathematiques 2-7
A very easy solution can be found by applying the given in section 4 sub
F.M. van Lamoen and P. Yiu, The Kiepert pencil of Kiepert hyperbolas,
Forum Geom., 1 (2001) 125-132.
>From this source we conclude that if we first build equilateraltriangles outwardly on the sides of ABC, resulting in a new triangle
A'B'C' of the new vertices, and then build equilateral triangles
inwardly on the sides of A'B'C', resulting in new vertices forming a
triangle A"B"C", then ABC is the medial triangle of A"B"C".
This yields a very easy construction!