Lat ABC be a triangle and (Ab), (Ac) the circles passing through H and

touching BC at B, C, respectively.

A

/\

/ \

/ \

/ \

/ \ Ab

Ac / \

/ H \

/ \

/ A' \

/ \

B-----------A"-------C

A' = BAb /\ CAc

A" = orth. Proj. of A' on BC

Similarly we define B',C'; B",C".

The triangles A'B'C', A"B"C" are in perspective with ABC.

Perspectors in Barycentrics:

1. A'B'C', ABC:

(sinAsec^2Acos(B-C) ::)

2. A"B"C", ABC:

(sec^2A ::)

GENERALIZATION:

Lat ABC be a triangle, P a point and (Ab), (Ac) the circles passing through P

and touching BC at B, C, respectively.

A

/\

/ \

/ \

/ \

/ \ Ab

Ac / \

/ P \

/ \

/ A' \

/ \

B-----------A"-------C

A' = BAb /\ CAc

A" = orth. Proj. of A' on BC

Similarly we define B',C'; B",C".

1. Which is the locus of P such that A'B'C', ABC are perspective?

2. Which is the locus of P such that A"B"C", ABC are perspective?

Answers:

1. The locus is a sextic.

2. The locus is the entire plane ie for every P the triangles A"B"C", ABC

are perspective.

Perspector:

(L^(-2) : M^(-2) : N^(-2)), where L,M,N are the tripolar distances of P

(ie AP, BP, CP, resp.) in Barycentrics

(sinA / (y^2 + z^2 + 2yzcosA) ::) in Normals

P Perspector in Normals

---------------------------------------

I (tan(A/2) ::)

G (1/a(m_a)^2 ::) = (1/a(-a^2 + 2(b^2 + c^2)) ::)

where m_a = A-median

O (1/a ::) = G

H (cscAsec^2A ::)

K (a/(m_a)^2 ::) = (a/(-a^2 + 2(b^2 + c^2)) ::)

etc

Hope that my calculations were correct!

Antreas