Re: [EMHL] More about X(1000) [Isogonal conjugate of midpoint of X(1),X(57)]

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• You might also want to try these: Let Ja,Jb, Jc be the centers of inversions of BC,CA,AB in the incircle, resp. Are the triangles ABC, JaJbJc perspective? And
Message 1 of 5 , Mar 9, 2013
You might also want to try these:

Let Ja,Jb, Jc be the centers of inversions
of BC,CA,AB in the incircle, resp.

Are the triangles ABC, JaJbJc perspective?

And let Ka, Kb, Kc your centers (ie the centers of the inversions of
BC,CA,AB in the
excircles).

Are JaJbJc and KaKbKc perspective?

APH

> **
>
>
> Dear teachers:
>
> Let A� the center of the inversive figure of line BC on A-excircle. Build
> B�
> and C� ciclycally.
>
> Lines (AA�), (BB�) and (CC�) meet in X(1000)
>
> Regards.
>
>
>

[Non-text portions of this message have been removed]
• ... Yes: At X(3296) ... No. Cesar Lozada
Message 2 of 5 , Mar 9, 2013
-----Mensaje original-----
>> You might also want to try these:

>>Let Ja,Jb, Jc be the centers of inversions of BC,CA,AB in the incircle,
>>resp.

>>Are the triangles ABC, JaJbJc perspective?

Yes: At X(3296)

>And let Ka, Kb, Kc your centers (ie the centers of the inversions of
>BC,CA,AB in the excircles).

>>Are JaJbJc and KaKbKc perspective?

No.

• Ja, Jb, Jc = the centers of inversion of BC, CA, AB in the Incircle La, Lb, Lc = the Incenters of JaBC, JbCA, JbAB Are the triangles ABC, La,Lb,Lc perspective?
Message 3 of 5 , Mar 9, 2013
Ja, Jb, Jc = the centers of inversion of BC, CA, AB in the Incircle

La, Lb, Lc = the Incenters of JaBC, JbCA, JbAB

Are the triangles ABC, La,Lb,Lc perspective?

Excircles version?

aph
• [APH] ... Equivalently: Let A B C be the pedal triangle of I (intouch triangle) and Ja,Jb,Jc the midpoints of IA , IB , IC and La, Lb, Lc the incenters of
Message 4 of 5 , Mar 9, 2013
[APH]
> Ja, Jb, Jc = the centers of inversion of BC, CA, AB in the Incircle
>
> La, Lb, Lc = the Incenters of JaBC, JbCA, JbAB
>
> Are the triangles ABC, LaLbLc perspective?
>
> Excircles version?

Equivalently:

Let A'B'C' be the pedal triangle of I (intouch triangle)
and Ja,Jb,Jc the midpoints of IA', IB', IC'
and La, Lb, Lc the incenters of JaBC, JbCA, JbAB, resp.

Are the triangles ABC, LaLbLc perspective?

We have cot(JaBA') = BA'/(r/2), BA' = rcot(B/2) ==>

cot(JaBA') = 2cot(B/2)
Similarly cot(JaCA') = 2cot(C/2)

cotB - 2cot(B/2) / cotC - 2cot(C/2) *

cotC - 2cot(C/2) / cotA - 2cot(A/2) *

cotA - 2cot(A/2) / cotB - 2cot(B/2) = 1

==> the triangles ABC, JaJbJc are perspective

Perspector in barycentrics:

1 / (cotA - 2cot(A/2)) :: =.......

Now, cot(JaBA') = [cot(LaBA')^2 - 1] / 2cot(LaBA')

==> cot(LaBA'): = f(B)

Similarly we define f(C)

And we have

cotB - f(B) / cotC - f(C) *

cotC - f(C) / cotA - f(A) *

cotA - f(A) / cotB - f(B) = 1

==> the triangles are perspective.

Perspector in barycentrics:

1/ (cotA - f(A)) ::

APH
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