Dear Antreas,

In Hyacinthos message #9756, you wrote:

>> The parallel through P to BC intersects AB,

>> AC at Ab, Ac, resp.

>>

>> The parallel through P to CA intersects BC,

>> BA at Bc, Ba, resp.

>>

>> The parallel through P to AB intersects CA,

>> CB at Ca, Cb, resp.

For any point P with homogeneous trilinears

P ( x : y : z ), we have

BcCb = a^2 x / (ax + by + cz),

and similarly we can find CaAc and AbBa.

>> 1. For which P we have that BcCb = CaAc = AbBa ?

This is for P = X(75), the isotomic conjugate

of the incenter of triangle ABC. In this case,

BcCb = CaAc = AbBa = 1 / (1/a + 1/b + 1/c).

>> 2. Which is the locus of P such that

>>

>> k * (BcCb + CaAc + AbBa) = (BC + CA + AB) ?

>>

>> where k = given real

This locus is the line with the trilinears

( ka^2 - a(a+b+c) : kb^2 - b(a+b+c) : kc^2 - c(a+b+c) ).

Especially, for k = 1, we get the line with

the trilinears

( a(b+c) : b(c+a) : c(a+b) ).

And for k = infinity, i. e. for

BcCb + CaAc + AbBa = 0, we get the line with

the trilinears

( a^2 : b^2 : c^2 ),

i. e. the isogonal tripolar of X(75).

Nice problem, Antreas!

Friendly,

Darij Grinberg