HYACINTHOS 4053

Dear Paris Pamfilos,

[PP]: On the sides of an oriented triangle define A' on BC, B' on CA, C' on AB
s.t. BA' = CB' = AC' = k const. The 3 circles (A'B'C), (B'C'A), (C'A'B) pass
through a point P, which, as k varies, describes a circle passing through
one Brocard point V. Doing the same with the other orientation of the
triangle, you get another point Q, describing another circle through the
other Brocard point V'. What are these circles? Are they known? Are their
centers and radii easily expressible through the triangle's data?


*** In homogeneous barycentric coordinates, the intersection of the three
circles
(A'B'C), (B'C'A), (C'A'B) is the point

( a(c^2a - (c^2+ca+a^2)k + (a+b+c)k^2)
: b(a^2b - (a^2+ab+b^2)k + (a+b+c)k^2)
: c(b^2c - (b^2+bc+c^2)k + (a+b+c)k^2)).

As k varies, this traverses the circle

(a+b+c)(a^2yz+b^2zx+c^2xy) - (a+b+c)(bc^2x+ca^2y+ab^2z) = 0.

This circle clearly passes through the Brocard point
V = (1/b^2 : 1/c^2 : 1/a^2) as you mentioned. It also passes through the
incenter
I = (a : b : c) [for k = infinity].

The center of the circle is the point

(a(a^3-2ab^2+b^3-a^2c+abc-b^2c)
:b(b^3-2bc^2+c^3-b^2a+abc-c^2a)
:c(c^3-2ca^2+a^3-c^2b+abc-a^2b)).

The circle in the other orientation can be similarly written down.
It is clear that this second circle also passes through the incenter, and has
equation

(a+b+c)(a^2yz+b^2zx+c^2xy) - (a+b+c)(b^2cx+c^2ay+a^2bz) = 0.

Apart from the incenter, the two circles should have another intersection
which is also a triangle center. In fact, the radical axis of the two
circles is the line

bc(b-c)x + ca(c-a)y + ab(a-b)z = 0

which contains the centroid G and the symmedian point K = (a^2 : b^2 : c^2).

The second common point of the circles has coordinates

(a(a^4-a^3(b+c)-a^2bc+2abc(b+c)-bc(b^2+c^2): ... : ...).

This triangle center does not appear in ETC.

The value of k that gives this point as a point on the FIRST circle is

k = (a^3b+b^3c+c^3a-abc(a+b+c))/((a+b+c)(a^2+b^2+c^2-ab-bc-ca)).

Best regards
Sincerely
Paul Yiu