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Reflection In Orthocenter And Projection From Altitude Foot
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Dear All My Friends,
Given triangle ABC with orthocenter H, altitudes AHa, BHb, CHc.
A', B', C' are reflections of A, B, C in H respectively.
Ab = orthogonal projection of Ha on AC
Ac = orthogonal projection of Ha on AB
A'' = intersection of B'Ab and C'Ac
Similarly define B'', C''.
The result:
Two triangles ABC and A''B''C'' are perspective at perspector P with barycentrics coordinate:
(a^4  b^4  c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
Search value: 3.57840086062088, may be not in current ETC?
Thank you and best regards,
Bui Quang Tuan

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Dear Quang,
your perspector is X(2883)
It is the isogonal conjugate of H wrt the medial triangle of ABC
The searchvalue in the ETC is different from you value
I did not check which one is correct
Kind regards
Eric
> Dear All My Friends,
with barycentrics coordinate:
> Given triangle ABC with orthocenter H, altitudes AHa, BHb, CHc.
> A', B', C' are reflections of A, B, C in H respectively.
> Ab = orthogonal projection of Ha on AC
> Ac = orthogonal projection of Ha on AB
> A'' = intersection of B'Ab and C'Ac
> Similarly define B'', C''.
> The result:
> Two triangles ABC and A''B''C'' are perspective at perspector P
>
> (a^4  b^4  c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
>
> Search value: 3.57840086062088, may be not in current ETC?
>
> Thank you and best regards,
> Bui Quang Tuan 0 Attachment
Dear Peter and Eric,
Thank you very much for your messages!
After checking I think the perspector could not be X(2883).
Please kindly check your construction again, Dear Eric!
Thank you and best regards,
Bui Quang Tuan
Eric Danneels <ERIC.DANNEELS@...> wrote:
Dear Quang,
your perspector is X(2883)
It is the isogonal conjugate of H wrt the medial triangle of ABC
The searchvalue in the ETC is different from you value
I did not check which one is correct
Kind regards
Eric
> Dear All My Friends,
with barycentrics coordinate:
> Given triangle ABC with orthocenter H, altitudes AHa, BHb, CHc.
> A', B', C' are reflections of A, B, C in H respectively.
> Ab = orthogonal projection of Ha on AC
> Ac = orthogonal projection of Ha on AB
> A'' = intersection of B'Ab and C'Ac
> Similarly define B'', C''.
> The result:
> Two triangles ABC and A''B''C'' are perspective at perspector P
>

> (a^4  b^4  c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
>
> Search value: 3.57840086062088, may be not in current ETC?
>
> Thank you and best regards,
> Bui Quang Tuan
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Dear Eric and Tuan, I get the same value as Tuan
3.578400860620876159553999905...
 In Hyacinthos@yahoogroups.com, "Eric Danneels"
<ERIC.DANNEELS@...> wrote:>
> Dear Quang,
>
> your perspector is X(2883)
> It is the isogonal conjugate of H wrt the medial triangle of ABC
> The searchvalue in the ETC is different from you value
> I did not check which one is correct
>
> Kind regards
>
> Eric
>
>
> > Dear All My Friends,
> > Given triangle ABC with orthocenter H, altitudes AHa, BHb, CHc.
> > A', B', C' are reflections of A, B, C in H respectively.
> > Ab = orthogonal projection of Ha on AC
> > Ac = orthogonal projection of Ha on AB
> > A'' = intersection of B'Ab and C'Ac
> > Similarly define B'', C''.
> > The result:
> > Two triangles ABC and A''B''C'' are perspective at perspector P
> with barycentrics coordinate:
> >
> > (a^4  b^4  c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
> >
> > Search value: 3.57840086062088, may be not in current ETC?
> >
> > Thank you and best regards,
> > Bui Quang Tuan
> 0 Attachment
Dear Quang and Garcia,
Quang's searchvalue is correct
I made a wrong construction with
Ab = orthogonal projection of A' (instead of Ha) on AC
Ac = orthogonal projection of A' (instead of Ha) on AB....
Sorry
Kind regards
Eric
 In Hyacinthos@yahoogroups.com, "garciacapitan"
<garciacapitan@...> wrote:>
CHc.
> Dear Eric and Tuan, I get the same value as Tuan
>
> 3.578400860620876159553999905...
>
>
>
>  In Hyacinthos@yahoogroups.com, "Eric Danneels"
> <ERIC.DANNEELS@> wrote:
> >
> > Dear Quang,
> >
> > your perspector is X(2883)
> > It is the isogonal conjugate of H wrt the medial triangle of ABC
> > The searchvalue in the ETC is different from you value
> > I did not check which one is correct
> >
> > Kind regards
> >
> > Eric
> >
> >
> > > Dear All My Friends,
> > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
> > > A', B', C' are reflections of A, B, C in H respectively.
P
> > > Ab = orthogonal projection of Ha on AC
> > > Ac = orthogonal projection of Ha on AB
> > > A'' = intersection of B'Ab and C'Ac
> > > Similarly define B'', C''.
> > > The result:
> > > Two triangles ABC and A''B''C'' are perspective at perspector
> > with barycentrics coordinate:
> > >
> > > (a^4  b^4  c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
> > >
> > > Search value: 3.57840086062088, may be not in current ETC?
> > >
> > > Thank you and best regards,
> > > Bui Quang Tuan
> >
> 0 Attachment
Dear Quang,
I would like to present the following generalization:
Let PaPbPc be the cevian triangle of P wrt triangle ABC
A', B', C' are reflections of A, B, C in P respectively.
Ab = intersection of AC with parallel to PB through Pa
Ac = intersection of AB with parallel to PC through Pa
A'' = intersection of B'Ab and C'Ac
Similarly define B'', C''.
Triangles ABC and A''B''C'' are perspective
Kind regards
Eric
 In Hyacinthos@yahoogroups.com, "garciacapitan"
<garciacapitan@...> wrote:>
CHc.
> Dear Eric and Tuan, I get the same value as Tuan
>
> 3.578400860620876159553999905...
>
>
>
>  In Hyacinthos@yahoogroups.com, "Eric Danneels"
> <ERIC.DANNEELS@> wrote:
> >
> > Dear Quang,
> >
> > your perspector is X(2883)
> > It is the isogonal conjugate of H wrt the medial triangle of ABC
> > The searchvalue in the ETC is different from you value
> > I did not check which one is correct
> >
> > Kind regards
> >
> > Eric
> >
> >
> > > Dear All My Friends,
> > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
> > > A', B', C' are reflections of A, B, C in H respectively.
P
> > > Ab = orthogonal projection of Ha on AC
> > > Ac = orthogonal projection of Ha on AB
> > > A'' = intersection of B'Ab and C'Ac
> > > Similarly define B'', C''.
> > > The result:
> > > Two triangles ABC and A''B''C'' are perspective at perspector
> > with barycentrics coordinate:
> > >
> > > (a^4  b^4  c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
> > >
> > > Search value: 3.57840086062088, may be not in current ETC?
> > >
> > > Thank you and best regards,
> > > Bui Quang Tuan
> >
> 0 Attachment
Dear Francisco and Eric,
Thank you very much for your messages! I am very happy to receive your messages.
I would like to thank Peter J.C. Moses for remark that the barycentrics can be more simple as:
(S^2 + 2*SA^2)*(2*SB*SC  S^2)/SA^2 : :
and this perspector is intersection of lines {{2,1105},{4,6}}
One small remark:
We construct A* = intersection of B'Ac and C'Ab similary construct B*, C* then two triangles ABC and A*B*C* are perspective at isogonal conjugate of X(1593)
Thank you Dear All again!
Have you very nice weekend!
Bui Quang Tuan
Eric Danneels <ERIC.DANNEELS@...> wrote: Dear Quang and Garcia,
Quang's searchvalue is correct
I made a wrong construction with
Ab = orthogonal projection of A' (instead of Ha) on AC
Ac = orthogonal projection of A' (instead of Ha) on AB....
Sorry
Kind regards
Eric
 In Hyacinthos@yahoogroups.com, "garciacapitan"
wrote:>
CHc.
> Dear Eric and Tuan, I get the same value as Tuan
>
> 3.578400860620876159553999905...
>
>
>
>  In Hyacinthos@yahoogroups.com, "Eric Danneels"
> wrote:
> >
> > Dear Quang,
> >
> > your perspector is X(2883)
> > It is the isogonal conjugate of H wrt the medial triangle of ABC
> > The searchvalue in the ETC is different from you value
> > I did not check which one is correct
> >
> > Kind regards
> >
> > Eric
> >
> >
> > > Dear All My Friends,
> > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
> > > A', B', C' are reflections of A, B, C in H respectively.
P
> > > Ab = orthogonal projection of Ha on AC
> > > Ac = orthogonal projection of Ha on AB
> > > A'' = intersection of B'Ab and C'Ac
> > > Similarly define B'', C''.
> > > The result:
> > > Two triangles ABC and A''B''C'' are perspective at perspector
> > with barycentrics coordinate:

> > >
> > > (a^4  b^4  c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
> > >
> > > Search value: 3.57840086062088, may be not in current ETC?
> > >
> > > Thank you and best regards,
> > > Bui Quang Tuan
> >
>
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Dear Eric & Tuan,
A' = {p  q  r, 2 q, 2 r}
Ab = {p q, 0, r (p + q + r)}
Ac = {p r, q (p + q + r), 0}
A" = {p q r (3 p + q + r), q (p + q  r) (p q + q^2 + q r + 2 r p), r (p  q + r) (2 p q + q r + r p + r^2)}
ABC & A"B"C" are perspective at p (p  q  r) (p^2 + p q + 2 q r + r p) ::
When P = I, the perspector is X(958),
When P = G, the perspector is X(2),
When P = O, the perspector is X(578),
When P = H, the perspector is the point Tuan found, (S^2 + 2 SA^2) (2 SB SC  S^2) / SA^2 ::
Ab, Ac ... live on an conic centered on p (q + r) (p^2 + q^2 + r^2 + 2 p q + q r + 2 r p) ::
When P = G, the conic is an ellipse (Sqrt[7] / 4 size of Steiner Circum) centered on G.
When P = H, the conic is an ellipse centered on X(389)
When P = X(20), the conic is centered on X(2883).
Best regards,
Peter.
 Original Message 
From: Eric Danneels
To: Hyacinthos@yahoogroups.com
Sent: Friday, February 29, 2008 9:07 PM
Subject: [EMHL] Re: Reflection In Orthocenter And Projection From Altitude Foot
Dear Quang,
I would like to present the following generalization:
Let PaPbPc be the cevian triangle of P wrt triangle ABC
A', B', C' are reflections of A, B, C in P respectively.
Ab = intersection of AC with parallel to PB through Pa
Ac = intersection of AB with parallel to PC through Pa
A'' = intersection of B'Ab and C'Ac
Similarly define B'', C''.
Triangles ABC and A''B''C'' are perspective
Kind regards
Eric
 In Hyacinthos@yahoogroups.com, "garciacapitan"
<garciacapitan@...> wrote:
>
> Dear Eric and Tuan, I get the same value as Tuan
>
> 3.578400860620876159553999905...
>
>
>
>  In Hyacinthos@yahoogroups.com, "Eric Danneels"
> <ERIC.DANNEELS@> wrote:
> >
> > Dear Quang,
> >
> > your perspector is X(2883)
> > It is the isogonal conjugate of H wrt the medial triangle of ABC
> > The searchvalue in the ETC is different from you value
> > I did not check which one is correct
> >
> > Kind regards
> >
> > Eric
> >
> >
> > > Dear All My Friends,
> > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
CHc.
> > > A', B', C' are reflections of A, B, C in H respectively.
> > > Ab = orthogonal projection of Ha on AC
> > > Ac = orthogonal projection of Ha on AB
> > > A'' = intersection of B'Ab and C'Ac
> > > Similarly define B'', C''.
> > > The result:
> > > Two triangles ABC and A''B''C'' are perspective at perspector
P
> > with barycentrics coordinate:
> > >
> > > (a^4  b^4  c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
> > >
> > > Search value: 3.57840086062088, may be not in current ETC?
> > >
> > > Thank you and best regards,
> > > Bui Quang Tuan
> >
>
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Dear Eric,
Thank you for interesting generalization. If I am not wrong, if barycentrics of P = (p : q : r) then results are:
Two triangles ABC and A''B''C'' are perspective at:
p*(p  q  r)*(p*(p + q + r) + 2*q*r) : :
We construct A* = intersection of B'Ac and C'Ab similary construct B*, C* then two triangles ABC and A*B*C* are perspective at:
1/(p*(p + q + r) + 2*q*r) : :
Barycentric product of these two point is a point:
p*(p  q  r) : :
Best regards,
Bui Quang Tuan
Eric Danneels <ERIC.DANNEELS@...> wrote: Dear Quang,
I would like to present the following generalization:
Let PaPbPc be the cevian triangle of P wrt triangle ABC
A', B', C' are reflections of A, B, C in P respectively.
Ab = intersection of AC with parallel to PB through Pa
Ac = intersection of AB with parallel to PC through Pa
A'' = intersection of B'Ab and C'Ac
Similarly define B'', C''.
Triangles ABC and A''B''C'' are perspective
Kind regards
Eric

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Dear Eric and All My Friends,
We can also take one more point and get one cubic locus. Detail as following:
Barycentrics of P = (p : q : r), X = (x : y : z). XaXbXc is cevian triangle of X wrt ABC
A', B', C' are reflections of A, B, C in P respectively
Ab = intersection of AC with parallel to XB through Xa
Ac = intersection of AB with parallel to XC through Xa
A'' = intersection of B'Ab and C'Ac
A* = intersection of B'Ac and C'Ab
The locus of X such that ABC and A''B''C'' are perspective is union of some lines and one cubic C(P).
The locus of X such that ABC and A*B*C* are perspective is union of some lines and the same cubic C(P).
Barycentric equation of C(P):
CyclicSum[y*z*p*((r + p  3*q)*y  (q + p  3*r)*z)] = 0
This cubic always passes A, B, C, X(2), P
 If P = X(3), C(P) passes:
X(2), X(3), X(64), X(69), X(253), X(1073), X(3146)
 If P = X(5), C(P) passes:
X(2), X(5), X(20), X(69), X(264), X(343)
 If P = X(10) , C(P) passes:
X(2), X(7), X(10), X(65), X(75), X(145), X(226), X(3175)
These cubics are not listed in ETC and in Bernard cubics sites.
Any good idea about these cubics?
Thank you and best regards,
Bui Quang Tuan
Eric Danneels <ERIC.DANNEELS@...> wrote: Dear Quang,
I would like to present the following generalization:
Let PaPbPc be the cevian triangle of P wrt triangle ABC
A', B', C' are reflections of A, B, C in P respectively.
Ab = intersection of AC with parallel to PB through Pa
Ac = intersection of AB with parallel to PC through Pa
A'' = intersection of B'Ab and C'Ac
Similarly define B'', C''.
Triangles ABC and A''B''C'' are perspective
Kind regards
Eric

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Dear Eric
In barycentrics coordinates P = [m:n:p]
the perspector is
[2np(n+p) +m(n²+p²m²):2mp(m+p)+n(p²+m²n²):2mn(m+n)+p(m²+n²p²)]
King regards
Michel
 In Hyacinthos@yahoogroups.com, "Eric Danneels" <ERIC.DANNEELS@...>
wrote:>
> Dear Quang,
>
> I would like to present the following generalization:
>
> Let PaPbPc be the cevian triangle of P wrt triangle ABC
> A', B', C' are reflections of A, B, C in P respectively.
> Ab = intersection of AC with parallel to PB through Pa
> Ac = intersection of AB with parallel to PC through Pa
> A'' = intersection of B'Ab and C'Ac
> Similarly define B'', C''.
> Triangles ABC and A''B''C'' are perspective
>
> Kind regards
>
> Eric
>
>
>  In Hyacinthos@yahoogroups.com, "garciacapitan"
> <garciacapitan@> wrote:
> >
> > Dear Eric and Tuan, I get the same value as Tuan
> >
> > 3.578400860620876159553999905...
> >
> >
> >
> >  In Hyacinthos@yahoogroups.com, "Eric Danneels"
> > <ERIC.DANNEELS@> wrote:
> > >
> > > Dear Quang,
> > >
> > > your perspector is X(2883)
> > > It is the isogonal conjugate of H wrt the medial triangle of ABC
> > > The searchvalue in the ETC is different from you value
> > > I did not check which one is correct
> > >
> > > Kind regards
> > >
> > > Eric
> > >
> > >
> > > > Dear All My Friends,
> > > > Given triangle ABC with orthocenter H, altitudes AHa, BHb,
> CHc.
> > > > A', B', C' are reflections of A, B, C in H respectively.
> > > > Ab = orthogonal projection of Ha on AC
> > > > Ac = orthogonal projection of Ha on AB
> > > > A'' = intersection of B'Ab and C'Ac
> > > > Similarly define B'', C''.
> > > > The result:
> > > > Two triangles ABC and A''B''C'' are perspective at perspector
> P
> > > with barycentrics coordinate:
> > > >
> > > > (a^4  b^4  c^4 + 2*SA^2)*(b^2*c^2 + SA^2)/SA^2 : :
> > > >
> > > > Search value: 3.57840086062088, may be not in current ETC?
> > > >
> > > > Thank you and best regards,
> > > > Bui Quang Tuan
> > >
> >
> 0 Attachment
Dear Tuan,
[snip]
> [QTB] Barycentric equation of C(P):
this is the pivotal cubic with pivot taaP and isopivot P
>
> CyclicSum[y*z*p*((r + p  3*q)*y  (q + p  3*r)*z)] = 0
>
> This cubic always passes A, B, C, X(2), P
(tX = isotomic conj. of X, aX = anticomplement of X)
thus,
C(X3) = pK(X1073, X253)
C(X5) = pK(X343, X69)
C(X10) = pK(X226, X7)
Best regards
Bernard
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