Dear Alex,
[AM]: A week ago I've found one property of 184-point in Clark List.
That point generates three equil triangles, similar to the base triangle ...
***
Fred Lang, Hyacinthos message 1191, has found another interesting property
of X(184) that leads to a very easy construction of this point:
Given a reference triangle ABC, let the perpendicular bisectors of BC, CA,
AC intersect the other pairs of sides at B1, C1; C2, A2; A3, B3
repsectively. The perpendicular bisectors of the segments B1C1, C2A2 and
A34B3 bound a triangle homothetic to ABC at X(184).
Having constructed this point, it is now not difficult to construct three
congruent triangles similar to the reference triangle ABC.
I wonder if the similarity ratio ZBaCa/ABC has a simple expression.
Best regard
Sincerely
Paul Yiu
[AM]: A week ago I've found one property of 184-point in Clark List.
That point generates three equil triangles, similar to the base triangle ...
***
Fred Lang, Hyacinthos message 1191, has found another interesting property
of X(184) that leads to a very easy construction of this point:
Given a reference triangle ABC, let the perpendicular bisectors of BC, CA,
AC intersect the other pairs of sides at B1, C1; C2, A2; A3, B3
repsectively. The perpendicular bisectors of the segments B1C1, C2A2 and
A34B3 bound a triangle homothetic to ABC at X(184).
Having constructed this point, it is now not difficult to construct three
congruent triangles similar to the reference triangle ABC.
I wonder if the similarity ratio ZBaCa/ABC has a simple expression.
Best regard
Sincerely
Paul Yiu
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