HYACINTHOS 3001

• Antreas P. Hatzipolakis
  
  Let ABC be a triangle and PaPbPc the pedal triangle of a point P.
  
  
  
  Ac A Ab
  \ /\ /
  y \ / \ /
  \ / \ / z
  Pc Pb
  / \ / \
  / z P y \
  / | \
  / |x \
  / | \
  B-------Pa---------C
  
  Let Ab, Ac be on (the extensions of) PPb, PPc, resp., such that:
  PbAb = PPc := z, PcAc = PPb := y
  
  Similarly:
  
  Bc, Ba on PPc, PPa, resp. such, that:
  PcBc = PPa := x, PaBa = PPc := z
  
  Ca, Cb on PPa, PPb, resp., such that:
  PaCa = PPb := y, PbCb = PPa := x
  
  For which P's the orth. proj. ("shadows") A'bA'c of AbAc on BC,
  B'cB'a of BcBa on CA, and C'aC'b of CaCb on AB are equal?
  
  Answer: P = (-a^2 + b^2 + c^2 + bc + ca + ab ::) in normals
  (for P inside ABC)
  
  LOCI:
  
  
  
  Ac A Ab
  |\ /\ /|
  | \y / \ z/ |
  | \ / \ / |
  | Pc Pb |
  | / \z y/ \ |
  | / P \ |
  |/ | \ |
  | |x \ |
  /| | \ |
  B-A'c----Pa--------C-A'b----
  
  1.
  Let A1 = BAb /\ CAc, and similarly B1 = CBc /\ ABa, C1 = ACa /\ BCb
  
  Which is the locus of P such that AA1, BB1, CC1 are concurrent?
  
  2.
  
  Let A2 = BAc /\ CAb, and similarly B2 = CBa /\ ABc, C2 = ACb /\ BCa
  
  Which is the locus of P such that AA2, BB2, CC2 are concurrent?
  
  The loci are "not nice" cubics with no "xyz" term.
  (Probably they can be decomposed into line + conic.)
  
  
  Ac A Ab
  |\ /\ /|
  | \y / \ z/ |
  | \ / \ / |
  | Pc Pb |
  | / \z y/ \ |
  A"c------P----------A"b
  |/ | \ |
  | |x \ |
  /| | \ |
  B-A'c----Pa--------C-A'b----
  
  
  Let A"b, A"c be the orth. proj, of P on AbA'b, AcA'c, resp.
  Similarly we define the points B"c, B"a; C"a, C"b.
  
  3.
  
  Let A3 = BA"b /\ CA"c, and similarly B3 = CB"c /\ AB"a, C3 = AC"a /\ BC"b
  
  Which is the locus of P such that AA3, BB3, CC3 are concurrent?
  
  I found that the locus is the isogonal cubic with pivot in normals:
  
  sinBsinC
  (-------------- ::)
  1 + sinBsinC
  
  4.
  
  Let A4 = BA"c /\ CA"b, and similarly B4 = CB"a /\ AB"c, C4 = AC"b /\ BC"a
  
  Which is the locus of P such that AA4, BB4, CC4 are concurrent?
  
  I found that this locus is the cubic with equation in normals:
  
  x((ysinBcosC)^2 - (zcosBsinC)^2) + cyclic = 0
  
  
  Antreas