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## RE: [EMHL] Locus

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• Dear Nikos ... And I found a complete cubic (with xyz term) Any remarkable points on it? APH
Message 1 of 4 , May 30, 2005
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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
• 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
Message 2 of 4 , May 30, 2005
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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 ... 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
Message 3 of 4 , May 30, 2005
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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.

[ND]:
>I had made a mistake.
>You are right in your conjecture.
>The locus is the circumcircle and the Euler line.

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.

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. In the special case where Q is the orthocenter of ABC then the locus is all the plane, that is the triangles ABC,
Message 4 of 4 , May 31, 2005
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Dear Antreas

>
> >> [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.

Yes the locus 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
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