## 22068Re: Locus

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• Mar 25, 2014
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[APH]:

Let ABC be a triangle, P = I and A'B'C' the antipedal
triangle of P.

Denote:

Ab = (perpendicular from A' to BC) /\ PB
Ac = (perpendicular from A' to BC) /\ PC

Similarly:

Bc = (perpendicular from B' to CA) /\ PC
Ba = (perpendicular from B' to CA) /\ PA

Ca = (perpendicular from C' to AB) /\ PA
Cb = (perpendicular from C' to AB) /\ PB

The lines BcCb, CaAc, AbBa are concurrent.

Point?

In general, the locus of P such that the lines BcCb, CaAc, AbBa are
concurrent is Darboux cubic.

[Angel ]:
For every point P the lines BcCb, CaAc and  AbBa  are concurrent.
If P = (x: y: z), then the point of concurencia of BcCb, CaAc and  AbBa  is Q=(x (SB y^2 + SC z^2 + SA (y + z)^2):...:...)

Some pairs {P, Q}:
{1, 8}, {2, 599}, {3, 3}, {4, 3}, {6, 599}, {8, 2098}, {13, 298},
{14, 299}, {15, 298}, {16, 299}, {36, 4511}, {56, 2098},  {80, 4511}, ...

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Dear Angel

Whenever the locus is the entire plane, the concurrence  can be
proved synthetically.

In our case we have:

1. BaCa, CbAb, BcAc are concurrent at P.

2. AbAc, BcBa, CaCb are concurrent, since ABC, A'B'C'
are orthologic by conctruction, so they concur at the other
than P orthologic center.

Now by the Double Perspectivity Theorem we get that
BcCb, CaAc, AbBa are concurrent.

APH

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