## Re: [EMHL] Collinear Points

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• ... Variations: 1. Let A B C be the cevian triangle of P. Denote Ab = CC / Parallel to BB from A Ac = BB / Parallel to CC from A L1 = The line AbAc.
Message 1 of 10 , Jan 20, 2004
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[APH]:

>
>6. Let P be a point with traces A_P, B_P, C_P. Denote by L1 the line
>joining the perpendicular feet of A_P on the cevians BB_P and CC_P.
>Let L2, L3 be the two lines analogously defined. The triangle bounded
>by L1, L2, L3 is perspective with ABC if and only if P lies on the
>Darboux cubic (together with the line at infinity, the three circles
>with diameters BC, CA, AB, and the Yiu quintic Q006 ((Paul Yiu,
>Hyacinthos #1967).
>
>
>Now, denote
>
>A* = L1 /\ BC
>
>B* = L2 /\ CA
>
>C* = L3 /\ AB
>
>Which is the locus of P such that A*,B*,C* are collinear?

Variations:

1. Let A'B'C' be the cevian triangle of P.

Denote

Ab = CC' /\ Parallel to BB' from A'

Ac = BB' /\ Parallel to CC' from A'

L1 = The line AbAc. Similarly the lines L2, L3.

Which is the locus of P such that the triangle bounded by L1,L2,L3
is perspective with ABC?
[Special case: L1,L2,L3 are concurrent]

2. Let A'B'C' be the cevian triangle of P.

Denote

Ab = CC' /\ Parallel to AB from A'

Ac = BB' /\ Parallel to AC from A'

L1 = The line AbAc. Similarly the lines L2, L3.

Which is the locus of P such that the triangle bounded by L1,L2,L3
is perspective with ABC?
[Special case: L1,L2,L3 are concurrent]

Also same questions for A*,B*,C* as defined above.

Greetings from Athens

Antreas

--
• Let ABC be a triangle P a fixed point and Q a variable point Denote: A1 = (Perpendicular to AQ through P) / BC Similarly B1, C1. The locus of Q such that
Message 2 of 10 , Jan 3
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Let ABC be a triangle P a fixed point and Q a variable point

Denote:

A1 = (Perpendicular to AQ through P)  /\ BC
Similarly B1, C1.

The locus of Q such that A1,B1, C1 are collinear is
the rec. circumhyperbola through P. (+ ?)

Reference

Let P,P* be two isogonal conjugate points

A1 = (Perpendicular to AP* through P)  /\ BC
Similarly B1, C1.

Which is the locus of P such that A1,B1,C1 are collinear?

Happy New Year 2015

APH

• From: César Lozada Which is the locus of P such that A1,B1,C1 are collinear? Locus={Circumcircle} / {McCay Cubic} If P on circumcircle then A1,B1,C1 and P
Message 3 of 10 , Jan 3
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Which is the locus of P such that A1,B1,C1 are collinear?

Locus={Circumcircle} \/ {McCay Cubic}
If P on circumcircle then A1,B1,C1 and P are collinear.

De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com]
Para: anopolis@yahoogroups.com; Hyacinthos
Asunto: [EGML] Collinear Points

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

Denote:

A1 = (Perpendicular to AQ through P)  /\ BC

Similarly B1, C1.

The locus of Q such that A1,B1, C1 are collinear is

the rec. circumhyperbola through P. (+ ?)

Let P,P* be two isogonal conjugate points

A1 = (Perpendicular to AP* through P)  /\ BC
Similarly B1, C1.

Which is the locus of P such that A1,B1,C1 are collinear?

Happy New Year 2015

APH

--
• From: César Lozada [Attachment(s) from =?UTF-8?Q?C=C3=A9sar_Lozada?= included below] A conjectured relation with circumnormal
Message 4 of 10 , Jan 3
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A conjectured relation with circumnormal triangle.

See attached image.

De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com]
Para: Anopolis@yahoogroups.com
Asunto: RE: [EGML] Collinear Points

Which is the locus of P such that A1,B1,C1 are collinear?

Locus={Circumcircle} \/ {McCay Cubic}
If P on circumcircle then A1,B1,C1 and P are collinear.

De: Anopolis@yahoogroups.com [mailto:Anopolis@yahoogroups.com]
Para: anopolis@yahoogroups.com; Hyacinthos
Asunto: [EGML] Collinear Points

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

Denote:

A1 = (Perpendicular to AQ through P)  /\ BC

Similarly B1, C1.

The locus of Q such that A1,B1, C1 are collinear is

the rec. circumhyperbola through P. (+ ?)

Let P,P* be two isogonal conjugate points

A1 = (Perpendicular to AP* through P)  /\ BC
Similarly B1, C1.

Which is the locus of P such that A1,B1,C1 are collinear?

Happy New Year 2015

APH

• Let ABC be a triangle and A B C the pedal triangle of I (aka Intouch triangle). Denote: N* = the NPC center of A B C O* = the circumcenter of the pedal
Message 5 of 10 , Feb 18
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Let ABC be a triangle and A'B'C' the pedal triangle of I (aka Intouch triangle).

Denote:

N* = the NPC center of A'B'C'

O* = the circumcenter of the pedal triangle of N* wrt A'B'C'

The points N*, O* and Feuerbach point of ABC are collinear.

References:
Buratino Giggle [=Tran Quang Hung]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&p=3751766#p3751766

APH

• [APH]: Let ABC be a triangle and A B C the pedal triangle of I (aka Intouch ... [Randy Hutson]: Dear Antreas, N* = X(942) = INVERSE-IN-INCIRCLE OF X(36). O* =
Message 6 of 10 , Feb 18
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[APH]:

Let ABC be a triangle and A'B'C' the pedal triangle of I (aka Intouch triangle).

Denote:

N* = the NPC center of A'B'C'

O* = the circumcenter of the pedal triangle of N* wrt A'B'C'

The points N*, O* and Feuerbach point of ABC are collinear.

References:
Buratino Giggle [=Tran Quang Hung]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&p=3751766#p3751766

[Randy Hutson]:

Dear Antreas,

N* = X(942) = INVERSE-IN-INCIRCLE OF X(36).
O* = midpoint of X(942) and [X(3649) = KS(INTOUCH TRIANGLE)]

The common line id X(11)X(113).

Best regards,
Randy

• ... Given that the Feuerbach point of ABC wrt Intouch triangle is X(110) of the intouch triangle, it is equivalent to (by taking the intouch triangle as
Message 7 of 10 , Feb 19
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[APH]:
Let ABC be a triangle and A'B'C' the pedal triangle of I (aka Intouch triangle).

Denote:

N* = the NPC center of A'B'C'

O* = the circumcenter of the pedal triangle of N* wrt A'B'C'

The points N*, O* and Feuerbach point of ABC are collinear.

References:
Buratino Giggle [=Tran Quang Hung]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&p=3751766#p3751766

Given that the Feuerbach point of ABC wrt Intouch triangle is X(110) of the intouch triangle,
it is equivalent to (by taking the intouch triangle as reference triangle):

Let ABC be a triangle, A'B'C' the pedal triangle of N and O' the circumcenter of A'B'C'
The line NO' passes through X(110)

Question: Is the other than X(110) intersection of NO' and circumcircle an interesting point?

APH
• ... *********************************************** [César Lozada] It is X(1141). [Peter Moses] Hi Antreas, X(1141). Best regards Peter. [APH]: [APH]: Let ABC
Message 8 of 10 , Feb 19
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[APH]:

[APH]:

Let ABC be a triangle and A'B'C' the pedal triangle of I (aka Intouch triangle).

Denote:

N* = the NPC center of A'B'C'

O* = the circumcenter of the pedal triangle of N* wrt A'B'C'

The points N*, O* and Feuerbach point of ABC are collinear.

References:
Buratino Giggle [=Tran Quang Hung]
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=48&p=3751766#p3751766

Given that the Feuerbach point of ABC wrt Intouch triangle is X(110) of the intouch triangle,
it is equivalent to (by taking the intouch triangle as reference triangle):

Let ABC be a triangle, A'B'C' the pedal triangle of N and O' the circumcenter of A'B'C'
The line NO' passes through X(110)

Question: Is the other than X(110) intersection of NO' and circumcircle an interesting point?

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