- View SourceLet ABC be a triangle and P a point on its plane.

We construct [the how, and the number of solutions is another story !]

three similar triangles PAbAc, PBaBc, PCaCb, whose the vertices Ai, Bi, Ci

lie on the sides of ABC.

A

/\

/ \

/ \

/ \

Ac Ma Ab

/ q f \

/ w \

/ P \

Bc w w Cb

/ f q \

/ Mb Mc \

/ q f \

B-----Ba--------Ca-------C

Angles of the similar triangles PAbAc, PBaBc, PCaCb:

P [= AcPAb = BcPBa = CaPCb] := omega [w; not be confused with

Brocard's!]

Ab [=PAbAc] = Bc = Ca := phi (f)

Ac [=PAcAb] = Ba = Cb := theta (q)

Let now Ma,Mb,Mc be the midpoints of AbAc, BcBa, CaCb.

The locus of the points P such that MaMbMc be in perspective with ABC

is a cubic [actually a diparametric family of cubics, since the

equation has two parameters, namely f,q [the third, w, is expressd by the

twos : w = Pi - (f+q)]).

If the similar triangles are isosceles with f = q = (Pi - w)/2, then the

cubic, according to my computations, is an isogonal one (actually a pencil

of isogonal cubics) with equation:

x(y^2 + z^2) (BC - sA) + [cyclically] = 0 (in trilinears)

where A,B,C are shortcuts for sin(A+w) - sinA, B =..., C = ...., s = sinw.

The pivot (sin(B+w) - sinB)(sin(C+w) - sinC) - sinw(sin(A+w) - sinA) ::)

can be simplified, but I didn't make the calculations.

The most interesting cases are of course for w = 60 d. or 90 d.

If w = 60 d. then the three similar triangles are equilateral, and a

natural question is: How about their centers K1,K2,K3 ? That is:

Which is the locus of P such that K1K2K3 be in perspective with ABC?

(I think that the locus is a sextic, but I don't know whether reduces or not)

If w = 90 d. then the midpoints Ma,Mb,Mc are centers of three squares

(Kenmotu configuration), whose the fourth vertices let be P1,P2,P3:

A

/\

/P1\

/ \

/ \

Ac Ma Ab

/ \

/ \

/ P \

Bc Cb

/ \

/ Mb Mc \

/ \ P3

P2 B-----Ba--------Ca-------C

And now the question is about the P1,P2,P3:

For which points P the triangle P1P2P3 is in perspective with ABC?

(in general, we can also consider P1,P2,P3 as the forth vertices of

three similar rhombi - FvL's configuration).

(I haven't worked on this locus.)

Antreas - View SourceLet ABC be a triangle, and PaPbPc the pedal triangle of P.

The circle (P, PPa) intersects the bisector of ang(PbPPc) at A', A"

[A' near to A]

Similary we define the points B', C'; B", C".

Which are the loci of P such that:

1. ABC, A'B'C'

2. ABC, A"B"C"

are perspective?

Let A1, A2 be the orth. proj. of A', A" (resp.) on BC.

Similarly we define the points B1, C1; B2, C2.

Which are the loci of P such that:

3. ABC, A1B1C1

4. ABC, A2B2C2

are perspective?

Antreas