- John Conway wrote:

> I'm afraid I did not return to your conjectures yesterday. Instead,

The conjecture was:

>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.

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

The three triangles got by reflecting ABC about its altitudes have

>altitudes might be as interesting as the ones your conjectures were

>about.

the same orthocenter as the reference triangle.

So the four Euler lines of these triangles concur.

Antreas - Antreas wrote:

> The conjecture was:

According to my calculations, no three of those four lines are concurrent.

>

> 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.

I used Maple to do the messy algebra.

Incidentally, onelist.com suppresses email addresses in the headers of posted

articles. I suggest that posters include their address in their sig-file.

--

Barry Wolk <wolkb@...> - Barry Wolk wrote:

>Antreas wrote:

Thanks, Barry.

>

>> 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.

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 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@...]

Reply To: Hyacinthos@onelist.com

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'm afraid I did not return to your conjectures yesterday. Instead,

The conjecture was:

>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.

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

The three triangles got by reflecting ABC about its altitudes have

>altitudes might be as interesting as the ones your conjectures were

>about.

the same orthocenter as the reference triangle.

So the four Euler lines of these triangles concur.

Antreas

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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,

This result has generalizations. I shall write about this

later.

Best regards.

Sincerely,

Paul

----------

From: Paul Yiu[SMTP:yiu@...]

Reply To: Hyacinthos@onelist.com

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@...]

Reply To: Hyacinthos@onelist.com

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'm afraid I did not return to your conjectures yesterday. Instead,

The conjecture was:

>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.

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

The three triangles got by reflecting ABC about its altitudes have

>altitudes might be as interesting as the ones your conjectures were

>about.

the same orthocenter as the reference triangle.

So the four Euler lines of these triangles concur.

Antreas

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