## Conjectures (was: F.G. - M.)

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• ... The conjecture was: Let ABC be a triangle. From A we draw parallels to b- and c- altitudes intersecting the BC at Ba, Ca respectively. Similarly, we define
Message 1 of 5 , Jan 2, 2000
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John Conway wrote:

>I've filled all the available space in my copy of today's New York Times
>with calculations in an attempt to verify Steve's assertion that the
>ex- and extra- Gergonne triangles are in perspective.

The conjecture was:

Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
are concurrent.

Synthetically, we can prove it by finding a point which belongs to
each one of the lines. But this is not always easy! Sometimes it is, as
in the following example:
Let A'B'C' be the orthic of a given triangle ABC. From B', C' draw
oaralles to the lines AC'B, AB'C, respectively, intersecting at A".
Similarly define B", C". Then the Euler lines of the triangles ABC,
A"B'C', B"C'A', C"A'B' are concurrent.
They are concurrent because the four triangles share the same orthocenter.
But which is the analogous common point in the conjecture?

Analytically, using trilinears, we can prove it this way:

Let H, H_a, H_b, H_c be the orthocenters, and O, O_a, O_b, O_c the
circumcenters of the triangles ABC, ABaCa, BCbAb, CBcAc, respectively.

Trilinears of Ha, Hb, Hc (in respect to triangle ABC):

(-sinBsinC : cosAcosC : cosAcosB)

(-sinCsinA : cosBcosA : cosBcosC)

(-sinAsinB : cosCcosB : cosCcosA)

Trilinears of Oa, Ob, Oc (in respect to triangle ABC):

(-cosA : cosB : cosC)

(cosA : -cosB : cosC)

(cosA : cosB : -cosC)

Notation:
sinA := D, cosA := d; sinB := E, cosB := e; sinC := F, cosC := f

Trilinears of Oa, Ha:

(-EF : df : de)

(-d : e : f)

Equation of Line OaHa (Euler line of ABaCa):

|df de| |-EF de| |-EF df|
x*| | - y*| | + z*| | = 0
|e f | |-d f| |-d f |

Similarly, Euler lines of BCbAb, CBcAc:

|ed ef| |-FD ef| |-FD ed|
x*| | - y*| | + z*| | = 0
|f d| |-e d| |-e d |

|fe fd| |-DE fd| |-DE fe|
x*| | - y*| | + z*| | = 0
|d e| |-f e| |-f e |

The three Euler lines concur if:

| |df de| |-EF de| |-EF df| |
| | | -| | | | |
| |e f | |-d f| |-d f | |
| |
| |ed ef| |-FD ef| |-FD ed| |
| | | -| | | | | = 0
| |f d| |-e d| |-e d | |
| |
| |fe fd| |-DE fd| |-DE fe| |
| | | -| | | | |
| |d e| |-f e| |-f e | |

Hmmm..... We need a whole month for all algebraic-trigonometrical
calculations!

> It occurred to me that the triangle got by reflecting ABC in its
>altitudes might be as interesting as the ones your conjectures were

The three triangles got by reflecting ABC about its altitudes have
the same orthocenter as the reference triangle.
So the four Euler lines of these triangles concur.

Antreas
• ... According to my calculations, no three of those four lines are concurrent. I used Maple to do the messy algebra. Incidentally, onelist.com suppresses email
Message 2 of 5 , Jan 4, 2000
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Antreas wrote:

> The conjecture was:
>
> Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
> intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
> and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
> are concurrent.

According to my calculations, no three of those four lines are concurrent.
I used Maple to do the messy algebra.

articles. I suggest that posters include their address in their sig-file.
--
Barry Wolk <wolkb@...>
• ... Thanks, Barry. Hope that the other conjecture (about the perpendicular bisectors of BaCa, CbAb, AcBc) is true! Antreas
Message 3 of 5 , Jan 4, 2000
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Barry Wolk wrote:

>Antreas wrote:
>
>> The conjecture was:
>>
>> Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
>> intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
>> and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
>> are concurrent.
>
>According to my calculations, no three of those four lines are concurrent.
>I used Maple to do the messy algebra.

Thanks, Barry.
Hope that the other conjecture (about the perpendicular bisectors of
BaCa, CbAb, AcBc) is true!

Antreas
• Dear Antreas, As Barry just pointed out, these Euler lines do not concur. The same is true for the Brocard axes. However, your original conjecture [Message 3
Message 4 of 5 , Jan 4, 2000
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Dear Antreas,

As Barry just pointed out, these Euler lines do not concur.
The same is true for the Brocard axes. However, your
original conjecture [Message 3 in Digest 3, Saturday,
26/12/99] is true.

The perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.

The homogeneous barycentric coordinates of the intersection
is the point

a^2(a^8-4a^6(b^2+c^2)
+2a^4(3b^4-b^2c^2+3c^4)
-4a^2(b^2-c^2)^2(b^2+c^2)
+(b^2-c^2)^2(b^4+6b^2c^2+c^4))
: ... : ...

In P-perpendicularity, the coordinates are
much more elegant, simply

f(g+h)(g^2+h^2-f^2) : ... : ...

Best regards.
Sincerely,
Paul

----------
From: Antreas P. Hatzipolakis[SMTP:xpolakis@...]
Sent: Sunday, January 02, 2000 8:43 AM
To: Hyacinthos@onelist.com
Subject: [EMHL] Conjectures (was: F.G. - M.)

From: xpolakis@... (Antreas P. Hatzipolakis)

John Conway wrote:

>I've filled all the available space in my copy of today's New York Times
>with calculations in an attempt to verify Steve's assertion that the
>ex- and extra- Gergonne triangles are in perspective.

The conjecture was:

Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
are concurrent.

Synthetically, we can prove it by finding a point which belongs to
each one of the lines. But this is not always easy! Sometimes it is, as
in the following example:
Let A'B'C' be the orthic of a given triangle ABC. From B', C' draw
oaralles to the lines AC'B, AB'C, respectively, intersecting at A".
Similarly define B", C". Then the Euler lines of the triangles ABC,
A"B'C', B"C'A', C"A'B' are concurrent.
They are concurrent because the four triangles share the same orthocenter.
But which is the analogous common point in the conjecture?

Analytically, using trilinears, we can prove it this way:

Let H, H_a, H_b, H_c be the orthocenters, and O, O_a, O_b, O_c the
circumcenters of the triangles ABC, ABaCa, BCbAb, CBcAc, respectively.

Trilinears of Ha, Hb, Hc (in respect to triangle ABC):

(-sinBsinC : cosAcosC : cosAcosB)

(-sinCsinA : cosBcosA : cosBcosC)

(-sinAsinB : cosCcosB : cosCcosA)

Trilinears of Oa, Ob, Oc (in respect to triangle ABC):

(-cosA : cosB : cosC)

(cosA : -cosB : cosC)

(cosA : cosB : -cosC)

Notation:
sinA := D, cosA := d; sinB := E, cosB := e; sinC := F, cosC := f

Trilinears of Oa, Ha:

(-EF : df : de)

(-d : e : f)

Equation of Line OaHa (Euler line of ABaCa):

|df de| |-EF de| |-EF df|
x*| | - y*| | + z*| | = 0
|e f | |-d f| |-d f |

Similarly, Euler lines of BCbAb, CBcAc:

|ed ef| |-FD ef| |-FD ed|
x*| | - y*| | + z*| | = 0
|f d| |-e d| |-e d |

|fe fd| |-DE fd| |-DE fe|
x*| | - y*| | + z*| | = 0
|d e| |-f e| |-f e |

The three Euler lines concur if:

| |df de| |-EF de| |-EF df| |
| | | -| | | | |
| |e f | |-d f| |-d f | |
| |
| |ed ef| |-FD ef| |-FD ed| |
| | | -| | | | | = 0
| |f d| |-e d| |-e d | |
| |
| |fe fd| |-DE fd| |-DE fe| |
| | | -| | | | |
| |d e| |-f e| |-f e | |

Hmmm..... We need a whole month for all algebraic-trigonometrical
calculations!

> It occurred to me that the triangle got by reflecting ABC in its
>altitudes might be as interesting as the ones your conjectures were

The three triangles got by reflecting ABC about its altitudes have
the same orthocenter as the reference triangle.
So the four Euler lines of these triangles concur.

Antreas

Cell Biology Reference Set--3 books + 2 CDs-for \$7.99! A \$127.40 value,
yours with membership in Library of Science, the leading science book club
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------------------------------------------------------------------------
• Dear Antreas, I should have remarked a simple geometric property of this intersection of the ``first perpendicular bisectors of the orthial triangles. Its
Message 5 of 5 , Jan 4, 2000
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Dear Antreas,

I should have remarked a simple geometric
property of this intersection of the ``first''
perpendicular bisectors of the orthial triangles.
Its distances from the vertices A, B, C are in
the proportions of

cot A : cot B : cot C.

This follows from a remark in your second message
[12/26/99] that

The "bases" BaCa, CbAb, BcAc of these triangles are equal to:

atanBtanC, btanCtanA, ctanAtanB.

The circumradii of these triangles are therefore

(a tan B tan C)/(sin (pi-A)) = 2R tan B tan C

etc,

later.

Best regards.
Sincerely,
Paul
----------
From: Paul Yiu[SMTP:yiu@...]
Sent: Tuesday, January 04, 2000 3:18 PM
To: 'Hyacinthos@onelist.com'
Subject: RE: [EMHL] Conjectures (was: F.G. - M.)

Dear Antreas,

As Barry just pointed out, these Euler lines do not concur.
The same is true for the Brocard axes. However, your
original conjecture [Message 3 in Digest 3, Saturday,
26/12/99] is true.

The perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.

The homogeneous barycentric coordinates of the intersection
is the point

a^2(a^8-4a^6(b^2+c^2)
+2a^4(3b^4-b^2c^2+3c^4)
-4a^2(b^2-c^2)^2(b^2+c^2)
+(b^2-c^2)^2(b^4+6b^2c^2+c^4))
: ... : ...

In P-perpendicularity, the coordinates are
much more elegant, simply

f(g+h)(g^2+h^2-f^2) : ... : ...

Best regards.
Sincerely,
Paul

----------
From: Antreas P. Hatzipolakis[SMTP:xpolakis@...]
Sent: Sunday, January 02, 2000 8:43 AM
To: Hyacinthos@onelist.com
Subject: [EMHL] Conjectures (was: F.G. - M.)

From: xpolakis@... (Antreas P. Hatzipolakis)

John Conway wrote:

>I've filled all the available space in my copy of today's New York Times
>with calculations in an attempt to verify Steve's assertion that the
>ex- and extra- Gergonne triangles are in perspective.

The conjecture was:

Let ABC be a triangle. From A we draw parallels to b- and c- altitudes
intersecting the BC at Ba, Ca respectively. Similarly, we define Cb, Ab,
and Ac, Bc. The Euler lines of the triangles ABC, ABaCa, BCbAb, CAcBc,
are concurrent.

Synthetically, we can prove it by finding a point which belongs to
each one of the lines. But this is not always easy! Sometimes it is, as
in the following example:
Let A'B'C' be the orthic of a given triangle ABC. From B', C' draw
oaralles to the lines AC'B, AB'C, respectively, intersecting at A".
Similarly define B", C". Then the Euler lines of the triangles ABC,
A"B'C', B"C'A', C"A'B' are concurrent.
They are concurrent because the four triangles share the same orthocenter.
But which is the analogous common point in the conjecture?

Analytically, using trilinears, we can prove it this way:

Let H, H_a, H_b, H_c be the orthocenters, and O, O_a, O_b, O_c the
circumcenters of the triangles ABC, ABaCa, BCbAb, CBcAc, respectively.

Trilinears of Ha, Hb, Hc (in respect to triangle ABC):

(-sinBsinC : cosAcosC : cosAcosB)

(-sinCsinA : cosBcosA : cosBcosC)

(-sinAsinB : cosCcosB : cosCcosA)

Trilinears of Oa, Ob, Oc (in respect to triangle ABC):

(-cosA : cosB : cosC)

(cosA : -cosB : cosC)

(cosA : cosB : -cosC)

Notation:
sinA := D, cosA := d; sinB := E, cosB := e; sinC := F, cosC := f

Trilinears of Oa, Ha:

(-EF : df : de)

(-d : e : f)

Equation of Line OaHa (Euler line of ABaCa):

|df de| |-EF de| |-EF df|
x*| | - y*| | + z*| | = 0
|e f | |-d f| |-d f |

Similarly, Euler lines of BCbAb, CBcAc:

|ed ef| |-FD ef| |-FD ed|
x*| | - y*| | + z*| | = 0
|f d| |-e d| |-e d |

|fe fd| |-DE fd| |-DE fe|
x*| | - y*| | + z*| | = 0
|d e| |-f e| |-f e |

The three Euler lines concur if:

| |df de| |-EF de| |-EF df| |
| | | -| | | | |
| |e f | |-d f| |-d f | |
| |
| |ed ef| |-FD ef| |-FD ed| |
| | | -| | | | | = 0
| |f d| |-e d| |-e d | |
| |
| |fe fd| |-DE fd| |-DE fe| |
| | | -| | | | |
| |d e| |-f e| |-f e | |

Hmmm..... We need a whole month for all algebraic-trigonometrical
calculations!

> It occurred to me that the triangle got by reflecting ABC in its
>altitudes might be as interesting as the ones your conjectures were

The three triangles got by reflecting ABC about its altitudes have
the same orthocenter as the reference triangle.
So the four Euler lines of these triangles concur.

Antreas

--

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