## Re: [EMHL] A problem of I. Yaglom

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• 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
Message 1 of 2 , Nov 19, 2002
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

>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
>
• Dear Ricardo ... I believe this question originates from EMHL (Lemoine) himself, I was given the reference: E.M.H. Lemoine, Question 864, Nouvelles Annales de
Message 2 of 2 , Nov 20, 2002
Dear Ricardo

[RC]:
> 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)

I believe this question originates from EMHL (Lemoine) himself, I was
given the reference:

E.M.H. Lemoine, Question 864, Nouvelles Annales de Mathematiques 2-7
191 (1868).

A very easy solution can be found by applying the given in section 4 sub
4 of

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 equilateral
triangles 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!

Kind regards,
Sincerely,
Floor.
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