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• Let 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,
Message 1 of 5 , Dec 2 3:04 AM
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Let 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
• ... ^ x(y^2 - z^2) (BC - sA) + [cyclically] = 0 (in trilinears) APH
Message 1 of 5 , Dec 2 3:09 AM
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Correction of a typo:

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

APH
• Let 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
Message 1 of 5 , Oct 19, 2001
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Let 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
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