ANOPOLIS 20

Let ABC be a triangle and P a point.

The parallel from P to BC intersects BA, CA at Ba, Ca, resp.
The parallel from P to CA intersects CB, AB at Cb, Ab, resp.
The parallel from P to AB intersects AC, BC at Ac, Bc, resp.

For which P's:

1. BaCa = CbAb = AcBc

2. A-altitude of ABaCa = B-altitude of AbBCb = C-altitude of AcBcC

3. ABa + ACa = BCb + BAb = CAc + CBc

4. (Ba-altitude + Ca-altitude) of ABaCa =
(Cb-altitude + Ab-altitude) of AbBCb =
(Ac-altitude + Bc-altitude) of AcBcC ?

Answers (Points in normals):

1. P = (1/a * (-1/a + 1/b + 1/c) ::)

Note: This is The Equal Parallelian Point (X_192 in ETC)

2. P = (1/a * (-a + b + c) ::) = Nagel Point.

3. P = (1/a * (-a^2 + b^2 + c^2 + ab + bc + ca) ::)

4. P = (1/a * (-1/a^2 + 1/b^2 + 1/c^2 + 1/ab + 1/bc + 1/ca) ::)

Loci Problems:
The perp. bisector l1 of BaCa intersects BC at A'
The perp. bisector l2 of CbAb intersects BC at B'
The perp. bisector l3 of AcBc intersects BC at C'

For which points P:

1. The l1,l2,l3 concur.

2. The AA', BB', CC' concur ?
[APH, Hyacinthos, msg #4667, 18 Jan 2002]

Answers:

1. The locus is the line GK. Furthermore, as P traverses this line, the
three perpendicular bisectors intersect at a point Q on the circle
through
the midpoints of the three segments BaCa, CbAb and AcBc, which also
passes
through P. Indeed, PQ is a diameter of the circle.
[Paul Yiu, Hyacinthos, msg #4670, 18 Jan 2002]

2. The locus is the cubic with equation in normals:
(xsinA + ysinB + zsinC)^2 * (xsin(B-C) + ysin(C-A) + zsin(A-B)) +
+ xyzsin(B-C)sin(C-A)sin(A-B) = 0
[APH, Hyacinthos, msg #4671, 19 Jan 2002]

Generalizations (Parametrizations):

Let ABC be a triangle, P a point and PaPbPc the pedal triangle of P.

Let P'a, P'b, P'c be points on PPa, PPb, PPc, resp. such that:

PP'a PP'b PP'c
---- = ---- = ---- = t, t=/= 1
PPa PPb PPc

The parallel from P'a to BC intersects BA, CA at Ba, Ca, resp.
The parallel from P'b to CA intersects CB, AB at Cb, Ab, resp.
The parallel from P'c to AB intersects AC, BC at Ac, Bc, resp.

The point P such that BaCa = CbAb = AcBc is:
P = (1/a * (-(1+t)/a + 1/b + 1/c ::), in normals.

Locus:
Which is the locus of P such that the perp. bisectors l1, l2, l3
of BaCa,CbAb,AcBc (resp.) concur?


Antreas P. Hatzipolakis

2nd ed. 20 Jan 2002