HYACINTHOS 3849

Dear Lev,

[LE]: Let A1B1C1 be the N-cevian triangle (N=Nagel point), A'B'C' be the
intouch
triangle. Let line a be perpendicular to B1C1 through A'. Similarly we
define lines b and c. Lines a, b and c are concurrent. What is the point of
concurrency?

***
Let P be a point with traces X, Y, Z on the side lines of ABC, and
Let Q its isotomic conjugate, with traces X', Y', Z'.
The perpendiculars from X to Y'Z', from Y to Z'X', and from Z to X'Y' are
concurrent if and only if P = (u:v:w) [homogeneous barycentric coordinates]
lies

(i) the side lines of the dilated triangle of ABC, or
(ii) the Lucas cubic, the isotomic cubic with pivot the isotomic conjugate
of the orthocenter, i.e.

cyclic sum (b^2+c^2-a^2)u(v^2-w^2) = 0.

For P = Gergonne point, this intersection is the point

(a(bb+cc-aa)(a^3(b+c)-a^2(b-c)^2-a(b+c)(b-c)^2+(b^2-c^2)^2) : ... : ...),

which apparently is not in the current edition of ETC.

On the other hand, if P = Nagel point, this intersection is the point

(a(b+c)(bb+cc-aa) : ... : ...)

and is the point X(72), the intersection of IK and HN.
[I = incenter, K = symmedian point, H = orthocenter, and N = Nagel point].

Best regards
Sincerely
Paul Yiu