## Parallels through P

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• Let ABC be a triangle, and P a point. 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,
Message 1 of 2 , Apr 29 8:30 AM
Let ABC be a triangle, and P a point.

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.

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

2. Which is the locus of P such that

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

where k = given real

APH
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• Dear Antreas, ... 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.
Message 2 of 2 , Apr 29 10:37 AM
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
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