- 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

On Sat, Mar 9, 2013 at 9:54 PM, C�sar Lozada <cesar_e_lozada@...>wrote:

> **

>

>

> 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.

>

> Cesar Lozada

>

>

[Non-text portions of this message have been removed] - -----Mensaje original-----
>> You might also want to try these:

Yes: At X(3296)

>>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

No.

>BC,CA,AB in the excircles).

>>Are JaJbJc and KaKbKc perspective?

Cesar Lozada - 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]
> Ja, Jb, Jc = the centers of inversion of BC, CA, AB in the Incircle

Equivalently:

>

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

>

> Are the triangles ABC, LaLbLc perspective?

>

> Excircles version?

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