Browse Groups

• Dear Alexey, Darij and other yacithists it is very easy, starting from ABC to construct a Brocardian hexagon ABCDEF, I mean with two points P, Q such as
Message 1 of 2 , Jun 8, 2004
View Source
Dear Alexey, Darij and other yacithists
it is very easy, starting from ABC to construct a Brocardian hexagon
ABCDEF, I mean with two points P, Q such as
<PAB=<PBC=<PCD=<PDE=<PEF=<PFA=<QBA=<QCB=<QDC=<QED=<QFE=<QAF
See the figure Brocardian Hexagon in Hyacinthos files.
L is the isogonal conjugate wrt ABC of the homothetic of the
midpoint of AC in (B,-2); then DEF is the circumcevian triangle of L.
W1,W2,W3 are the projections of O upon AL, BL, CL.
P lies on the circles ABW2,BCW3,CDW1, DEW2, EFW3, FAW1 and on the
circle with diameter OL.
Q lies on the circles ABW1, BCW2, CDW3, DEW1, EFW2, FAW and on the
circle with diameter OL.
Q is the reflection of P wrt OL
Note that the quadrilaterals ABCE, BCDF, CDEA, DEFB, EFAC, FABD are
harmonic.
Let's try a conjecture : the points DEF above are the only points in
the plane for which the hexagon ABCDEF is harmonic.
Friendly. Jean-Pierre
• Dear Alexey, Darij and other Hyacinthists I wrote ... hexagon ... L. ... are ... Note that the triples of lines (AE,BF,CD), (AD,BC,EF), (AB,CF,ED) concur on
Message 1 of 2 , Jun 8, 2004
View Source
Dear Alexey, Darij and other Hyacinthists
I wrote
> it is very easy, starting from ABC to construct a Brocardian
hexagon
> ABCDEF, I mean with two points P, Q such as
> <PAB=<PBC=<PCD=<PDE=<PEF=<PFA=<QBA=<QCB=<QDC=<QED=<QFE=<QAF
> See the figure Brocardian Hexagon in Hyacinthos files.
> L is the isogonal conjugate wrt ABC of the homothetic of the
> midpoint of AC in (B,-2); then DEF is the circumcevian triangle of
L.
> W1,W2,W3 are the projections of O upon AL, BL, CL.
> P lies on the circles ABW2,BCW3,CDW1, DEW2, EFW3, FAW1 and on the
> circle with diameter OL.
> Q lies on the circles ABW1, BCW2, CDW3, DEW1, EFW2, FAW and on the
> circle with diameter OL.
> Q is the reflection of P wrt OL
> Note that the quadrilaterals ABCE, BCDF, CDEA, DEFB, EFAC, FABD
are
> harmonic.

Note that the triples of lines (AE,BF,CD), (AD,BC,EF), (AB,CF,ED)
concur on the radical axis of the circumcircle and the circle with
diameter OL.
Any inversion with pole a Poncelet point of these two circles will
map ABCDEF to a regular hexagon.
Friendly. Jean-Pierre
Your message has been successfully submitted and would be delivered to recipients shortly.
• Changes have not been saved
Press OK to abandon changes or Cancel to continue editing
• Your browser is not supported
Kindly note that Groups does not support 7.0 or earlier versions of Internet Explorer. We recommend upgrading to the latest Internet Explorer, Google Chrome, or Firefox. If you are using IE 9 or later, make sure you turn off Compatibility View.