## RE: [EMHL] line, circumcenters

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• Correction: _________________________________ Dear Antreas, [APH]: Let ABC be a triangle, L a line, A , B , C the orth. projections of A,B,C on L, resp.
Message 1 of 4 , Feb 15, 2013
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Correction:
_________________________________
Dear Antreas,

[APH]: Let ABC be a triangle, L a line, A', B', C' the orth. projections
of A,B,C on L, resp. Denote:

Ab, Ac = the circumcenters of AB'C, ABC', resp.
Bc, Ba = the circumcenters of BC'A, BCA', resp
Ca, Cb = the circumcenters of CA'B, CAB', resp.

The lines AbAc, BcBa, CaCb are concurrent
and the radical axes [common chords] of (AB'C, ABC'), (BC'A, BCA'),
(CA'B, CAB') are concurrent.

Points for special cases L = OH,OK,OI ??

*** There are only three circles involved. Your Ba=Ca etc.
If L = ux+vy+wz = 0, then the radical center is the point

((S_A(v-w)^2 + S_B w(w-u) - S_C v(u-v))/(u(b^2S_Bv + c^2S_Cw - S_{BC}u )) : ... : ...).

For the Euler line, this is ( (a^2(S_{AA}+S_{BC})-4S_{ABC})/(S_{BB}S_{CC}(S_B-S_C)^2) : ... : ...)

For the Brocard axis: (a^4(b^4+c^4-2^2(b^2+c^2-^2)/(b^2-c^2)^2 : ... : ...)

*** (a^4(b^4+c^4-2a^2(b^2+c^2-a^2))/(b^2-c^2)^2 : ... : ...)

For the OI line: (a^2(2a^3-2a^2(b+c)-a(b^2-4bc+c^2)+(b-c)^2(b+c))/((b-c)^2(b+c-a)^2) : ... : ...)

None of these is in the current edition of ETC.

Best regards
Sincerely
Paul
• Dear Paul So the the 6 circles are only 3..... now the 12 are only 9..... ABB , AB C AC B, AC C BCC , BC A BA C, BA A CAA , CA B CB A, CB B The 9 circumcenters
Message 2 of 4 , Feb 15, 2013
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Dear Paul

So the the 6 circles are only 3..... now the 12 are only 9.....

ABB', AB'C
AC'B, AC'C

BCC', BC'A
BA'C, BA'A

CAA', CA'B
CB'A, CB'B

The 9 circumcenters are grouped in three lines passing through O
and parallel to altitudes.
Probably they have more properties....

APH

--- In Hyacinthos@yahoogroups.com, Paul Yiu <yiu@...> wrote:
>
> Correction:
> _________________________________
> Dear Antreas,
>
> [APH]: Let ABC be a triangle, L a line, A', B', C' the orth. projections
> of A,B,C on L, resp. Denote:
>
> Ab, Ac = the circumcenters of AB'C, ABC', resp.
> Bc, Ba = the circumcenters of BC'A, BCA', resp
> Ca, Cb = the circumcenters of CA'B, CAB', resp.
>
> The lines AbAc, BcBa, CaCb are concurrent
> and the radical axes [common chords] of (AB'C, ABC'), (BC'A, BCA'),
> (CA'B, CAB') are concurrent.
>
> Points for special cases L = OH,OK,OI ??
>
> *** There are only three circles involved. Your Ba=Ca etc.
> If L = ux+vy+wz = 0, then the radical center is the point
>
> ((S_A(v-w)^2 + S_B w(w-u) - S_C v(u-v))/(u(b^2S_Bv + c^2S_Cw - S_{BC}u )) : ... : ...).
>
> For the Euler line, this is ( (a^2(S_{AA}+S_{BC})-4S_{ABC})/(S_{BB}S_{CC}(S_B-S_C)^2) : ... : ...)
>
> For the Brocard axis: (a^4(b^4+c^4-2^2(b^2+c^2-^2)/(b^2-c^2)^2 : ... : ...)
>
> *** (a^4(b^4+c^4-2a^2(b^2+c^2-a^2))/(b^2-c^2)^2 : ... : ...)
>
> For the OI line: (a^2(2a^3-2a^2(b+c)-a(b^2-4bc+c^2)+(b-c)^2(b+c))/((b-c)^2(b+c-a)^2) : ... : ...)
>
> None of these is in the current edition of ETC.
>
> Best regards
> Sincerely
> Paul
>
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