- Dear Nikos

[APH]:>> Let ABC be a triangle, and P a point.

[ND]:

>>

>> Ab : = (Perpendicular Bisector of AB) /\ AP

>>

>> Ac : = (Perpendicular Bisector of AC) /\ AP

>>

>> A' : = BAb /\ CAc. Similarly B', C'.

>>

>>

>> Which is the locus of P such that ABC, A'B'C' are perspective?

>if my computations are correct

And I found a complete cubic (with xyz term)

>I think that the locus is an ugly cubic.

Any remarkable points on it?

APH - Dear Antreas sorry.

I had made a mistake.

You are right in your conjecture.

The locus is the circumcircle and the Euler line.

Best regards

Nikolaos Dergiades

> -----Original Message-----

> From: Hyacinthos@yahoogroups.com

> [mailto:Hyacinthos@yahoogroups.com]On Behalf Of Antreas P . Hatzipolakis

> Sent: Monday, May 30, 2005 7:15 PM

> To: undisclosed-recipients:

> Subject: RE: [EMHL] Locus

>

>

> Dear Nikos

>

> [APH]:

> >> Let ABC be a triangle, and P a point.

> >>

> >> Ab : = (Perpendicular Bisector of AB) /\ AP

> >>

> >> Ac : = (Perpendicular Bisector of AC) /\ AP

> >>

> >> A' : = BAb /\ CAc. Similarly B', C'.

> >>

> >>

> >> Which is the locus of P such that ABC, A'B'C' are perspective?

>

> [ND]:

> >if my computations are correct

> >I think that the locus is an ugly cubic.

>

> And I found a complete cubic (with xyz term)

> Any remarkable points on it?

>

> APH

>

>

>

>

>

>

> Yahoo! Groups Links

>

>

>

>

>

> - Dear Nikos

>> [APH]:

[ND]:

>> >> Let ABC be a triangle, and P a point.

>> >>

>> >> Ab : = (Perpendicular Bisector of AB) /\ AP

>> >>

>> >> Ac : = (Perpendicular Bisector of AC) /\ AP

>> >>

>> >> A' : = BAb /\ CAc. Similarly B', C'.

>> >>

>> >>

>> >> Which is the locus of P such that ABC, A'B'C' are perspective?

>>

>> [ND]:

>> >if my computations are correct

>> >I think that the locus is an ugly cubic.

>I had made a mistake.

In general:

>You are right in your conjecture.

>The locus is the circumcircle and the Euler line.

Let ABC be a triangle, Q a fixed point, P a variable point

and QaQbQc the pedal triangle of Q.

Ab : = QQc /\ AP

Ac : = QQb /\ AP

A' : = BAb /\ CAc. Similarly B', C'.

The locus of P such that ABC, A'B'C' are perspective

is a Cubic.

Variation :

Q is the variable and P the fixed point.

We can also take both P,Q as variable points:

Q = Isogonal (or Isotomic) Conjugate of P

Now the locus should be a Sextic, I think.

Greetings

Antreas

-- - Dear Antreas

>

Yes the locus is a cubic.

> >> [APH]:

> >> >> Let ABC be a triangle, and P a point.

> >> >>

> >> >> Ab : = (Perpendicular Bisector of AB) /\ AP

> >> >>

> >> >> Ac : = (Perpendicular Bisector of AC) /\ AP

> >> >>

> >> >> A' : = BAb /\ CAc. Similarly B', C'.

> >> >>

> >> >>

> >> >> Which is the locus of P such that ABC, A'B'C' are perspective?

> >>

> In general:

>

> Let ABC be a triangle, Q a fixed point, P a variable point

> and QaQbQc the pedal triangle of Q.

>

> Ab : = QQc /\ AP

>

> Ac : = QQb /\ AP

>

> A' : = BAb /\ CAc. Similarly B', C'.

>

> The locus of P such that ABC, A'B'C' are perspective

> is a Cubic.

In the special case where Q is the orthocenter of ABC then the locus is all

the plane,

that is the triangles ABC, A'B'C' are perspective

and the perspector in barycentrics is the point

(xxS_A : yyS_B : zzS_C) where

(x : y : z) are the barycentrics of P.

Best regards

Nikolaos Dergiades