## Re: [EMHL] The pedal triangles of Ia, Ib, Ic

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• Thank you Bernard You are right. The three intersections are very close. For the triangle ABC with sin(A/2) = 3/5 sin(B/2) = 5/13 I found with MathCad for the
Message 1 of 35 , Feb 28, 2002
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Thank you Bernard

You are right.
The three intersections are very close.
For the triangle ABC
with sin(A/2) = 3/5 sin(B/2) = 5/13
I found with MathCad for the three lines OI
a rational determinant with
nominator and denominator integer numbers
with about 20 digits and the value of the

Best regards
Nik

--- In Hyacinthos@y..., Bernard Gibert <b.gibert@f...> wrote:
> Dear Antreas, Paul and Nik,
>
> >>> No surprise if the OI lines [after the OG and OK lines]
> >>> of the pedal triangles of Ia,Ib,Ic
> >>> are concurrent on some point on the Euler line of ABC as well!
> >>>
> >>> ***
> >>> A quick sketch suggest that they are.
>
> Cabri says they are NOT !
> although the three intersections are very close.
>
> >>>It would be interesting to know
> >>> the
> >>> point of concurrency.
> >
> [PY]
> > Now I have no confidence that the three lines are concurrent.
> > May be Edward can help check this.
> >
>
> Best regards
>
> Bernard
• Dear Antreas ... I apologize, you are right.. I am mistaked in my notations. Gilles
Message 35 of 35 , Mar 3, 2002
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Dear Antreas

> Dear Gilles,
>
> [APH]
> Let IaaIabIac be the pedal triangle of Ia.
>
> A' := IaIaa /\ IabIac
>
> Similarly B', C'
>
> The triangles ABC, A'B'C' are perspective.
>
> [...]
>
>
> ==> the point of concurrence is the centroid G.
>
> [...]
>
> 1. A" := the orth. projection of Iaa on IabIac.
>
> Similarly B", C"
>
> Then the triangles ABC, A"B"C" are perspective.
>
> [GB]:
> The triangle MaMbMc with sides IabIac, IbaIbc, IcaIcb is homothetic to
> IaIbIc. A' is the feet on IabIac of the altitude, as we prove with
> A'B'C' is the orthic triangle of MaMbMc, and ABC is the orthic triangle
> of IaIbIc. So ABC and A'B'C' are homothetic, therefore they are
> perspective. Perspector is the Mittelpunkt X_9 of ABC.
>
> *********
>
> I guess your A'B'C' is mine A"B"C".

I apologize, you are right.. I am mistaked in my notations.

Gilles

>
>
> Let's see what the perspector of ABC, A"B"C" is:
>
>
> A
> ..................
> / \
> / \
> B----A1-Iaa--------C
> / \
> A3 A2
> Iac \
> / A" Iab
> / \
> Ia
>
> [...]
>
> ==> AA" is the a-cevian of Mittenpunkt (cot(A/2) ::)
>
>
> Antreas
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