HYACINTHOS 4204

Dear Antreas,

[APH]: Let ABC be a triangle and l1, l2, l3 three concurrent lines at a
point P,
such that: l1 _|_ BC, l2 _|_ CA, l3 _|_ AB.

>
> Ac
> |\
> | \A
> | /\
> |/ \
> Ab \
> /| \
> / | \
> / P \
> / | \
> / | \
> B-----|----------C
> |
> | l1
> |
>

Ab = l1 /\ AB, Ac = l1 /\ AC

Bc = l2 /\ BC, Ba = l2 /\ BA

Ca = l3 /\ CA, Cb = l3 /\ CB

For which point P we have that :

PAb + PAc = PBc + PBa = PCa + PCb ?

*********
Let me try.

Let P = (u:v:w) in homogeneous barycentric coordinates. We try to find
k such that

(u : v : w)/(u+v+w) + k(1/a^2)(a^2 : -S_C : -S_B)

lies on AC. This means kS_C/a^2 = v/(u+v+w), and the length of PAb [I
switch your Ab and Ac] is

(v/(u+v+w)).(a^2/S_C) times the A-altitude, namely,

(v/(u+v+w)).(a^2/S_C).(S/a)
= (v/(u+v+w)).(S/S_C).a

Similarly, the length of PAc is (w/(u+v+w))(S/S_B).a, and

PAb + PAc
= (aS)(v/S_C + w/S_B)/(u+v+w)
= (aS/(S_BC))(vS_B + wS_C)/(u+v+w)
= (S/S_ABC)aS_A(vSB + wS_C)/(u+v+w).

By writing down the corresponding expressions for the other two sums, and
equating,
we have

aS_A(vS_B+wS_C) = bS_B(wS_C+uS_A) = cS_C(uS_A+vS_B).

vS_B + wS_C : wS_C + uS_A : uS_A + vS_B = 1/(aS_A) : 1/(bS_B) : 1/(cS_C).

The inferior of (uS_A : vS_B : wS_C) is the point (1/(aS_A) : ... : ...).

(uS_A : vS_B : wS_C)
= superior of (1/(aS_A) : ... : ...)
= (-bcS_BC + caS_CA + abS_AB : ... : ...).

It follows that

(u:v:w) = ((-bcS_BC + caS_CA + abS_AB)/S_A : ... : ...)
= (f(a,b,c), f(b,c,a), f(c,a,b)),

where

f(a,b,c) =
(a^5(b+c)+a^4bc -
2a^3(b^3+c^3)+a(b+c)(b-c)^2(b^2+c^2)-bc(b^2-c^2)^2)/(b^2+c^2-a^2).

This point is not in the current edition of ETC.

Remark: The point (f(a,b,c) : ... : ...), the superior of X(92), is also
not in
ETC.

Best regards
Sincerely
Paul Yiu