## Re: [EMHL] An Isocubic

Expand Messages
• ... Let s improve the configuration: Let ABC be a triangle, P a point and A*B*C* the circumcevian triangle of P. Denote: A b := AB* / BC A c := AC* / BC Oab
Message 1 of 4 , Aug 6, 2004
[APH]:

>Let ABC be a triangle, and P a point.
>
>Denote:
>
>(Reflection of BP in BC) /\ AC := Ab
>(Reflection of CP in BC) /\ AB := Ac
>
>Oab := The circumcenter of ABAb
>Oac := The circumcenter of ACAc
>
>A' := BOab /\ COac. A" := AA' /\ BC
>
>Similarly B',B"; C',C"
>
>Which is the locus of P such that:
>
>(1) ABC, A'B'C' are perspective ?
>
>(2) A", B", C" are collinear?

Let's improve the configuration:

Let ABC be a triangle, P a point and A*B*C* the circumcevian
triangle of P.

Denote:

A'b := AB* /\ BC
A'c := AC* /\ BC

Oab := The circumcenter of ABA'b
Oac := The circumcenter of ACA'c

A' := BOab /\ COac. A" := AA' /\ BC

Similarly B',B"; C',C"

Which is the locus of P such that:

(1) ABC, A'B'C' are perspective ?

(2) A", B", C" are collinear?

The points ABAbA'b are concyclic, so
the Oab is the same point in both configurations.

Now in the new configuration, denote:

Hab := The Orthocenter of ABA'b
Hac := The Orthocenter of ACA'c

A1 := BHab /\ CHac. A2 := AA1 /\ BC

Similarly B1,B2; C1,C2

Which is the locus of P such that:

(1') ABC, A1B1C1 are perspective ?

(2') A2, B2, C2 are collinear?

We have Locus(1) = Locus(1'), and Locus(2) = Locus(2')

Antreas

--
Your message has been successfully submitted and would be delivered to recipients shortly.