## Locus

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• Let ABC be a triangle, P a point and A B C the 1. cevian 2. the pedal triangle of P. Denote: Ab = (Parallel to AB through B ) / BC Ac = (Parallel to AC
Message 1 of 240 , Mar 20, 2013
Let ABC be a triangle, P a point and A'B'C' the 1. cevian 2. the pedal
triangle of P.

Denote:

Ab = (Parallel to AB through B') /\ BC
Ac = (Parallel to AC through C') /\ BC
Aa = (Parallel to AB through B') /\ (Parallel to AC through C')

Similarly:

Bc = (Parallel to BC through C') /\ CA
Ba = (Parallel to BA through A') /\ CA
Bb = (Parallel to BC through C') /\ (Parallel to BA through A')

Ca = (Parallel to CA through A') /\ AB
Cb = (Parallel to CB through B') /\ AB
Cc = (Parallel to CA through A') /\ (Parallel to CB through B')

A* = AbBb /\ AcCc, B* = BcCc /\ BaAa, C* = CaAa /\ CbBb

Which is the locus of P such that:

1. ABC, A*B*C*

2. A'B'C', A*B*C*

are perspective?

Variations:

A* = AbCc /\ AcBb, B* = BcAa /\ BaCc, C* = CaBb /\ CbAa

Antreas
• [APH]: Let ABC be a triangle. A line L passing through A intersects the circle with diameter AB again at Ab and the circle with diameter AC again at Ac.
Message 240 of 240 , Feb 16

[APH]:

Let ABC be a triangle.

A line L passing through A intersects the circle with diameter AB again at Ab and the circle with diameter AC again at Ac.

[Equivalently: Let Ab, Ac be the orthogonal projections of B, C on L, resp.]
Let A* be the intersection of BAc and CAb.

Which is the locus of A* as L moves around A?

Parametric trilinear equation:

1/u(t) = a*((b^2+c^2-a^2)^2-4*b^2*c^2*c os(2*t)^2)/(2*S)

1/v(t) = 2*(cos(2*t)*c-b)*S - c*sin(2*t)*(a^2+3*b^2-2*cos(2* t)*b*c-c^2)

1/w(t) = 2*(cos(2*t)*b-c)*S + b*sin(2*t)*(a^2+3*c^2-2*cos(2* t)*b*c-b^2)

Regards,