## 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
Message 1 of 12 , Feb 28, 2008
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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
Message 2 of 12 , Feb 28, 2008
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Dear Quang,

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
• 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
Message 3 of 12 , Feb 28, 2008
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Dear Peter and Eric,
Thank you very much for your messages!
After checking I think the perspector could not be X(2883).
Thank you and best regards,
Bui Quang Tuan

Eric Danneels <ERIC.DANNEELS@...> wrote:
Dear Quang,

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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[Non-text portions of this message have been removed]
• Dear Eric and Tuan, I get the same value as Tuan -3.578400860620876159553999905...
Message 4 of 12 , Feb 29, 2008
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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,
>
> 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
>
• 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
Message 5 of 12 , Feb 29, 2008
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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:
>
> 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
> >
>
• 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
Message 6 of 12 , Feb 29, 2008
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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
> >
>
• 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
Message 7 of 12 , Feb 29, 2008
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Dear Francisco and Eric,
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:
>
> 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,
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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[Non-text portions of this message have been removed]
• 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
Message 8 of 12 , Mar 1, 2008
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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
> >
>

[Non-text portions of this message have been removed]
• 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
Message 9 of 12 , Mar 1, 2008
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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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[Non-text portions of this message have been removed]
• 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).
Message 10 of 12 , Mar 1, 2008
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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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[Non-text portions of this message have been removed]
• Dear Eric In barycentrics coordinates P = [m:n:p] the perspector is [2np(n+p) +m(n²+p²-m²):2mp(m+p)+n(p²+m²-n²):2mn(m+n)+p(m²+n²-p²)] King regards
Message 11 of 12 , Mar 3, 2008
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Dear Eric
In barycentrics coordinates P = [m:n:p]
the perspector is
[2np(n+p) +m(n²+p²-m²):2mp(m+p)+n(p²+m²-n²):2mn(m+n)+p(m²+n²-p²)]

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
> > >
> >
>
• Dear Tuan, [snip] ... this is the pivotal cubic with pivot taaP and isopivot P (tX = isotomic conj. of X, aX = anticomplement of X) thus, C(X3) = pK(X1073,
Message 12 of 12 , Mar 4, 2008
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Dear Tuan,

[snip]

> [QTB] 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

this is the pivotal cubic with pivot taaP and isopivot 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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