HYACINTHOS 4205

Antreas P. Hatzipolakis

 

Dear Paul Yiu,


[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.
>>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 ?

[PY]:

>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.

Probably a solution by normals is simpler. Let's see:


Ac
|\
| \A
| /\
|/ \
Ab \
/| \
Pc | Pb
/ P \
/ | \
/ | \
B-----Pa----------C
|
| l1
|


The lines l1, l2, l3 are the lines PPa, PPb, PPc, where PaPbPc is
the pedal triangle of P = (x:y:z) in actual normals.

Triangle PPcAb: angl(AbPcP) = 90 d., ang(PcPAb) = B

==> PAb = zsecB

Similarly PAc = ysecC

==> PAb + PAc = zsecB + ysecC

and cyclically:

PBc + PBa = xsecC + zsecA

PCa + PCb = ysecA + xsecB

Therefore

PAb + PAc = PBc + PBa = PCa + PCb <==>

zsecB + ysecC = xsecC + zsecA = ysecA + xsecB

OR, as a system:

xsecB + ysecA + z0 =

0x + ysecC + zsecB =

xsecC + 0y + zsecA


==>

x y z
--------------------- = --------------------- = --------------------
| 1 secA 0 | | secB 1 0 | | secB secA 1 |
| | | | | |
| 1 secC secB | | 0 1 secB | | 0 secC 1 |
| | | | | |
| 1 0 secA | | secC 1 secA | | secC 0 1 |


==>

x y z
--------------------- = --------------------- = -------------------
secA(-secA+secB+secC) secB(secA-secB+secC) secC(secA+secB-secC)


So, the point P such that PAb + PAc = PBc + PBa = PCa + PCb
is the point: (secA(-secA + secB + secC) ::) in normals.


Antreas