HYACINTHOS 3190

Dear Antreas,

[APH]:

> Let ABC be a triangle, and PaPbPc the pedal triangle of point P.
> The perp. bisectors of PPc, PPb meet at A', and intersect BC at Ab, Ac, resp.
>
> A
> /\
> / \
> / \
> / A' \
> Pc Pb
> / P \
> / B' C' \
> B-Ab-------Ac--C
>
> Similarly we define the points B', C' and Bc,Ba on CA and Ca, Cb on AB.
> The triangle A'B'C' is the Midway triangle of P.

We can simply take Ab to be A'B' /\ BC and Ac to be A'C'/\BC (etc.).

[APH]:

> For which point P we have that: AbAc = BcBa = CaCb ?

This is the point in barycentrics (-3bc+ac+ab : bc-3ac+ab : -3ab+ab+ac).
I have not checked whether it is in ETC, but you probably have, Antreas?

[APH]:

> Generalization:
> Let P be a point and A', B', C' points on PA, PB, PC, resp., such that
> PA'/PA = PB'/PB = PC'/PC = t
> [A'B'C' = t-way triangle of P]
>
> Let Ab = A'B' /\ BC, Ac = A'C' /\ BC
> Similarly we define the points Bc,Ba; Ca,Cb.
>
> Which is the locus of P such that AbAc = BcBa = CaCb, as t varies?

I get the point ( (-2-t)bc+ab+ac : ... : ... ) but here t= AA'/PA'.

This shows that the locus is the line through the centroid and the
isotomic conjugate of the incenter.

Kind regards,
Sincerely,
Floor van Lamoen