## Re: PERP. BISECTORS (correction)

Expand Messages
• Dear Antreas, ... labelling ... barycentric ... of ... perpendicular ... These coordinates [of the inner Vecten point] should of course be (1/(SA-S) : 1/(SB-S)
Message 1 of 21 , Dec 7, 2001
Dear Antreas,

>
> [APH]: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three
> squares based on BC,CA,AB outwardly ABC.
>
> Ba = BC /\ AA2, Ca = BC /\ AA3, Similarly Cb, Ab; Bc, Ac.
> >
> >
> > \ /
> > Ab Ac
> > \ /
> > \ /
> > \ /
> > A3 \ / A2
> > \ /
> > A
> > /\
> > / \
> > / \
> > / \
> > / \
> > B3 / \ C2
> > / \
> > ---Ba------------------B--------------C------------------Ca-----
> > / \
> > / \
> > / \
> > / \
> > Bc Cb
> > / B1 C1 \
> >
> OLD PROBLEMS:
> ...
> [APH, Hyacinthos, message #2118, 18 Dec 2000]:
> Are the perpendicular bisectors of A2A3, B3B1, C1C2 concurrent?
>
> [FvL, Hyacinthos, message #2123, 19 Dec 2000]:
> These perpendicular bisectors are concurrent at:
> ( ... : 2SS + 2SSB - bbS + 4SAC + bbSB - aaSA - ccSC : ... )
> [in barycentrics]
>
> [APH]: NEW PROBLEMS:
> 1. Are the perpendicular bisectors of B3C2, C1A3, A2B1 concurrent?
> 2. Let A' = (perp. bis of BC) /\ (perp. bis. of A2A3)
> Similarly B', C'.
> Are the triangles ABC, A'B'C' perspective?
>
> ***
> I shall follow the notation in Floor's Friendship paper by
labelling
> these squares BA_bA_cC, CB_cB_aA, and AC_aC_bB [so that A2, A3 are
> the points B_a, C_a respectively]. These have homogeneous
barycentric
> coordinates
>
> B_a = (SC+S : -b^2 : S_A),
> C_a = (SB+S : S_A : -c^2).
>
> The midpoint of the segment is A' = (a^2+2S : -SC : -SB).
>
> Similarly,
> B_c = (SC : -b^2 : SA+S),
> C_b = (SB : SA+S : -c^2).
>
> The midpoint of the segment B_cC_b is A'' = (-a^2 : SC-S : SB-S),
> which is the center of the square erected on BC, on the same side
of
> A. This point is also equidistant from B_a and C_a.
>
> This gives an immediate solution to New Problem 2: the
perpendicular
> bisectors of BC and B_aC_a intersect at A''. It follows that
> A''B''C'' is perspective with ABC at the inner Vecten point
>
> (SA-S : SB-S : SC-S).

These coordinates [of the inner Vecten point] should of course be

(1/(SA-S) : 1/(SB-S) : 1/(SC-S))

Best regards
Sincerely
Paul
• Dear Antreas and Floor, [APH]: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three ... ... [PY](4498): I shall follow the notation in Floor s
Message 2 of 21 , Dec 7, 2001
Dear Antreas and Floor,

[APH]: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three
> squares based on BC,CA,AB outwardly ABC.
>
> Ba = BC /\ AA2, Ca = BC /\ AA3, Similarly Cb, Ab; Bc, Ac.
> >
> >
> > \ /
> > Ab Ac
> > \ /
> > \ /
> > \ /
> > A3 \ / A2
> > \ /
> > A
> > /\
> > / \
> > / \
> > / \
> > / \
> > B3 / \ C2
> > / \
> > ---Ba------------------B--------------C------------------Ca-----
> > / \
> > / \
> > / \
> > / \
> > Bc Cb
> > / B1 C1 \
> >
> OLD PROBLEMS:
> ...
> [APH, Hyacinthos, message #2118, 18 Dec 2000]:
> Are the perpendicular bisectors of A2A3, B3B1, C1C2 concurrent?
>
> [FvL, Hyacinthos, message #2123, 19 Dec 2000]:
> These perpendicular bisectors are concurrent at:
> ( ... : 2SS + 2SSB - bbS + 4SAC + bbSB - aaSA - ccSC : ... )
> [in barycentrics]
...

[PY](4498): I shall follow the notation in Floor's Friendship paper
by labelling these squares BA_bA_cC, CB_cB_aA, and AC_aC_bB [so that
A2, A3 are the points B_a, C_a respectively]. These have homogeneous
barycentric coordinates

B_a = (SC+S : -b^2 : S_A),
C_a = (SB+S : S_A : -c^2).

The midpoint of the segment is A' = (a^2+2S : -SC : -SB).

Similarly,
B_c = (SC : -b^2 : SA+S),
C_b = (SB : SA+S : -c^2).

The midpoint of the segment B_cC_b is A'' = (-a^2 : SC-S : SB-S),
which is the center of the square erected on BC, on the same side of
A. This point is also equidistant from B_a and C_a.

...

[For the old problem]:
The perpendicular bisector of B_aC_a is therefore the line A'A''.
This has equation

(SB-SC)x + (SB-SC-2S)y + (SB-SC+2S)z = 0.

The equations of the other two perpendicular bisectors can be written
down similarly:

(SC-SA+2S)x + (SC-SA)y + (SC-SA-2S)z = 0,
(SA-SB-2S)x + (SA-SB+2S)y + (SA-SB)z = 0.

The three lines intersect at the point

(2S+2SA-SB-SC : 2S+2SB-SC-SA : 2S+2SC-SA-SB)
=(2S+b^2+c^2-a^2 : ... : ...).

This clearly lies on the line GK. It is not in the current edition of
ETC. [The coordinates apparently do not agree with those Floor gave].

*****
After writing the above, I found in the most recent issue of
Mathematics Magazine (which I just received yesterday) the article

L. Hoehn, Extriangles and excevians, Math. Mag. 74 (2001) 384--388

which is very similar to

F.M. van Lamoen, Friendship among triangle centers, Forum Geom. 1
(2001) 1--6.

Theorem 4. The three experpendicular bisectors of any triangle ABC
are concurrent.

This is exactly the ``old problem'' here. At the end of the article,
the author addresses the question ``what collinearity might be
possible among the points of concurrency for the exaltitudes, G,
exomedians, H, exangle bisectors, I, and experpendicular bisector[s]
J?'' The point J, according to the above calculation, lies on GK.

Best regards
Sincerely
Paul
• Dear Antreas and all Hyqcinthists, ... By Thales theorem CC 2 BC 2 ... AA 2 BA 2 and similarly with the other points, thus CB 3 CC 2 AC 1 AA 3
Message 3 of 21 , Dec 8, 2001
Dear Antreas and all Hyqcinthists,

> Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares
> based on BC,CA,AB, resp., outwardly ABC.
>
> A
> /\
> / \
> / \
> / \
> / \
> / \
> / \
> B--------------C
> /| |\
> / | | \
> / | | \
> / | | \
> / | | \
> B'1---B1--------------C1----C'1
>
> Let B'1 = AB /\ B1C1, C'1 = AC /\ B1C1.
>
> Similarly A'2, C'2, B'3, A'3
>
> Are the perp. bisectors of B'1C'1, C'2A'2, A'3B'3 concurrent?
>
> Antreas

By Thales theorem

CC'2 BC'2
------- = --------
AA'2 BA'2

and similarly with the other points, thus

CB'3 CC'2 AC'1 AA'3 BA'2 BB'1
------- ------- ------ ------- ------ ------- = 1
BB'3 BC'2 CC'1 CA'3 AA'2 AB'1

By Carnot theorem, the 6 points B'1, C'1, C'2, A'2, A'3, B'3 lie on a conic
(E).

Let B' and C' the orthogonal-projection points, on BC, of B'1 and C'1
respectively.

So BB'3 = c / sin B , BB' = a / tan B and

c - a cosB
B'B'3 = -------------
sinB

Similarly

b - a cosC
C'C'2 = ------------- and B'B'3 = C'C'2.
sinC

Thus B'1C'1 and B'3C'2 have the same perpendicular bisector.

So B'1C'1 and its perpendicular bisector are axis of the conic (E).

Similarly with C'2A'2, A'3B'3 and their perpendicular bisectors.

Thus (E) has too many axis, this is a circle.

The perpendicular bisectiors of B'1C'1, C'2A'2, A'3B'3 are concurrent at the
center of this circle

Friendly

Gilles
• Dear Sir ... Yes,they are. And just one more trivial thing,the triangle bounded by A 1B 1 , B 1C 1 and C 1A 1 is homothetic with ABC. Yours faithfully
Message 4 of 21 , Dec 8, 2001
Dear Sir

>Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares
>based on BC,CA,AB, resp., outwardly ABC.
>
> A
> /\
> / \
> / \
> / \
> / \
> / \
> / \
> B--------------C
> /| |\
> / | | \
> / | | \
> / | | \
> / | | \
> B'1---B1--------------C1----C'1
>
>Let B'1 = AB /\ B1C1, C'1 = AC /\ B1C1.
>
>Similarly A'2, C'2, B'3, A'3
>
>Are the perp. bisectors of B'1C'1, C'2A'2, A'3B'3 concurrent?
>
>
>Antreas

Yes,they are.

And just one more trivial thing,the triangle bounded by A'1B'1 , B'1C'1 and
C'1A'1 is homothetic with ABC.

Yours faithfully
Atul.A.Dixit

_________________________________________________________________
• Dear Antreas, [APH}: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares based on BC,CA,AB outwardly ABC. Ba = BC / AA2, Ca = BC / AA3,
Message 5 of 21 , Dec 8, 2001
Dear Antreas,

[APH}: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three
squares based on BC,CA,AB outwardly ABC.

Ba = BC /\ AA2, Ca = BC /\ AA3, Similarly Cb, Ab; Bc, Ac.
>
>
> \ /
> Ab Ac
> \ /
> \ /
> \ /
> A3 \ / A2
> \ /
> A
> /\
> / \
> / \
> / \
> / \
> B3 / \ C2
> / \
> ---Ba------------------B--------------C------------------Ca-----
> / \
> / \
> / \
> / \
> Bc Cb
> / B1 C1 \
>
OLD PROBLEMS:

[APH, Hyacinthos, message #8, 26 Dec 1999]:
Are the perpendicular bisectors of BaCa, AcBc, CbAb are concurrent?

[PY, Hyacinthos, messages ##75,76, 4 Jan 2000]:
Yes. The homogeneous barycentric coordinates of the intersection
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)) : ... : ...)

The tripolar coordinates: (cotA : cotB : cotC).

***
This point now appears as X(1078) in ETC.

[APH]: NEW PROBLEMS:
3. Let A" = (perp. bis. BaCa) /\ (perp. bis. of A2A3).
Similarly B", C".
Are the triangles ABC, A"B"C" perspective?

***
This intersection has barycentric coordinates

(aa(SBC - 2AS), SSS-SBCC, SSS-SBBC).

With the other two counterparts, it does not form a triangle
persective with ABC.

Best regards
Sincerely
Paul
• Dear Antreas and Gilles, [APH]: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares based on BC,CA,AB, resp., outwardly ABC. ... Let B 1 = AB
Message 6 of 21 , Dec 8, 2001
Dear Antreas and Gilles,

[APH]: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the
three squares based on BC,CA,AB, resp., outwardly ABC.
>> >
>> > A
>> > /\
>> > / \
>> > / \
>> > / \
>> > / \
>> > / \
>> > / \
>> > B--------------C
>> > /| |\
>> > / | | \
>> > / | | \
>> > / | | \
>> > / | | \
>> > B'1---B1--------------C1----C'1
>> >
Let B'1 = AB /\ B1C1, C'1 = AC /\ B1C1.
Similarly A'2, C'2, B'3, A'3
Are the perp. bisectors of B'1C'1, C'2A'2, A'3B'3 concurrent?

[GB]: By Thales theorem
>>
>> CC'2 BC'2
>> ------- = --------
>> AA'2 BA'2
>>
>> and similarly with the other points, thus
>>
>>
>> CB'3 CC'2 AC'1 AA'3 BA'2 BB'1
>> ------- ------- ------ ------- ------ ------- = 1
>> BB'3 BC'2 CC'1 CA'3 AA'2 AB'1
>>
>>
By Carnot theorem, the 6 points B'1, C'1, C'2, A'2, A'3, B'3 lie on
a conic (E).

Let B' and C' the orthogonal-projection points, on BC, of B'1
and C'1 respectively.
>
>> So BB'3 = c / sin B , BB' = a / tan B and
>>
c - a cosB
B'B'3 = -------------
sinB

Similarly

b - a cosC
C'C'2 = ------------- and B'B'3 = C'C'2.
sinC
Thus B'1C'1 and B'3C'2 have the same perpendicular bisector.

So B'1C'1 and its perpendicular bisector are axis of the conic
(E). Similarly with C'2A'2, A'3B'3 and their perpendicular bisectors.
Thus (E) has too many axis, this is a circle.
The perpendicular bisectiors of B'1C'1, C'2A'2, A'3B'3 are
concurrent at the center of this circle.

[PY]: In homogeneous barycentric coordinates,

B'1 = (-a^2 : 0 : S+a^2),
C'1 = (-a^2 : S+a^2 : 0).

Their midpoint is

(-2a^2 : S+a^2 : S+a^2)

and the perpendicular bisector is the line

(b^2-c^2)(S+a^2)x + a^2(S+b^2-c^2)y - a^2(S-b^2+c^2)z = 0.

Similarly, by writing down the equations of the other two
perpendicular bisectors, we find the three lines intersecting at

(a^2(S.SA + SAA-SBC) : ... : ...).

This is a point on the Brocard axis, not found in the current edition
of ETC. Indeed, this is the point dividing KO externally in the ratio

KO : OP = 1 : cot w.

Best regards
Sincerely,
Paul
• Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares based on BC,CA,AB, resp., outwardly ABC. A / / / / / /
Message 7 of 21 , Dec 8, 2001
Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares
based on BC,CA,AB, resp., outwardly ABC.

A
/\
/ \
/ \
/ \
/ \
/ \
/ \
B--------------C
/| |\
/ | | \
/ | | \
/ | | \
/ | | \
B'1---B1--------------C1----C'1

Let B'1 = AB /\ B1C1, C'1 = AC /\ B1C1.

Similarly A'2, C'2, B'3, A'3

Are the perp. bisectors of B'1C'1, C'2A'2, A'3B'3 concurrent?

Antreas
• ... However, it was an significant discovery by Grebe: the homothetic center is the symmedian point. APH
Message 8 of 21 , Dec 8, 2001
On Saturday, December 8, 2001, at 06:22 AM, Atul Dixit wrote:

> Dear Sir
>
>> Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares
>> based on BC,CA,AB, resp., outwardly ABC.
>>
>> A
>> /\
>> / \
>> / \
>> / \
>> / \
>> / \
>> / \
>> B--------------C
>> /| |\
>> / | | \
>> / | | \
>> / | | \
>> / | | \
>> B'1---B1--------------C1----C'1
>>
>> Let B'1 = AB /\ B1C1, C'1 = AC /\ B1C1.
>>
>> Similarly A'2, C'2, B'3, A'3
>>
>> Are the perp. bisectors of B'1C'1, C'2A'2, A'3B'3 concurrent?
>>
>>
>> Antreas
>
> Yes,they are.
>
> And just one more trivial thing,the triangle bounded by A'1B'1 , B'1C'1
> and
> C'1A'1 is homothetic with ABC.
>
>

However, it was an significant discovery by Grebe:
the homothetic center is the symmedian point.

APH
• Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares based on BC,CA,AB, resp., outwardly ABC. A2 A3 A / / / / /
Message 9 of 21 , Dec 8, 2001
Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares
based on BC,CA,AB, resp., outwardly ABC.

A2
A3
A
/\
/ \
/ \
/ \
/ \ C2
B3 / \
/ \
-----Ba-------------------B--------------C-----------------------Ca----

Let Ba = A3B3 /\ BC, Ca = BC /\ A2C2.

Similarly Cb, Ab, Ac, Bc

1. Are the perp. bisectors of BaCa, CbAb, AcBc concurrent?

2. Are the triangles ABC, GaGbGc in perspective, where Ga,Gb,Gc
are the centroids of ABaCa, BCbAb, CAcBc, resp. ?
[Equivalently: Let Ma, Mb, Mc be the midpoints of BaCa, CbAb, AcBc, resp.
Are the lines AMa, BMb, CMc concurrent?]

Antreas
• Dear Antreas, [APH]: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares based on BC,CA,AB, resp., outwardly ABC. ... ********** Let S denote
Message 10 of 21 , Dec 9, 2001
Dear Antreas,

[APH]: Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three
squares based on BC,CA,AB, resp., outwardly ABC.
>
> A'
>
>
>
>
>
>
>
> A3
> A2
> A
> /\
> / \
> / \
> / \
> B3 / \
> / \ C2
> / \
> B--------------C
> | |
> | |
> | |
> | |
> | |
> B' B1--------------C1 C'
>
>
> Let A'B'C' be the triangle bounded by the lines B1C1, C2A2, A3B3
> (Grebe triangle).
>
> Denote 1 = ABC, 2 = medial ABC, 3 = antimedial ABC,
> 4 = A'B'C', 5 = medial A'B'C', 6 = antimedial A'B'C'
>
> (x,y) := homothetic center of x, y.
>
>
> (1,2) = G | --- | --- |
> | | |
> (1,3) = G | (2,3) = G | --- |
> | | |
> (1,4) = K | (2,4) = ? | (3,4) = ? |
> | | |
> (1,5) = ? | (2,5) = ?

>(3,5) = ? |
> | | |
> (1,6) = ? | (2,6) = ? | (3,6) = ? |
>
> (4,5) = (4,6) = (5,6) = G' = ?
>
>
> Coordinates?
>

**********
Let S denote twice the area of triangle ABC.
The barycentic coordinates of A' can be easily written down.
The oriented areas A'CA and A'AB are respectively -bb/2 and -cc/2.
That of A'BC is (S+bb+cc)/2. It follows that

A' = (S+bb+cc : -bb : -cc).

Similarly,

B' = (-aa : S+cc+aa : -cc),
C' = (-aa : -bb : S+aa+bb).

The centroid of A'B'C' is

G' = (S+bb+cc-2aa : S+cc+aa-2bb : S+aa+bb-2cc).

This is the same for the medial and antimedial of A'B'C'.
G' clearly lies on the line GK.
It is not in the current edition of ETC.

The vertices of the medial triangle of A'B'C' are

(-2aa : S+2SB : S+2SC),
(S+2SA : -2bb : S+2SC),
(S+2SA : S+2SB : -2cc).

Those of the antimedial triangle of A'B'C' are

(-(S+2aa+bb+cc) : S+cc+aa : S+aa+bb),
(S+bb+cc : -(S+aa+2bb+cc) : S+aa+bb),
(S+bb+cc : S+cc+aa : -(S+aa+bb+2cc)).

Here is a table of the perspectors, listing only the A-coordinates.
These perspectors are all on the line GK.

1=ABC 2=medial ABC 3=antimedial ABC
------------------------------------------------------
4=A'B'C' (aa:bb:cc) (S+bb+cc::) (S+bb+cc-aa::)
Grebe not in ETC not in ETC
-------------------------------------------------------
5=medial (S+bb+cc-aa::) (aa:bb:cc) (S+bb+cc-2aa::)
of A'B'C' not in ETC =G', not in ETC
--------------------------------------------------------
6=antimedial (S+bb+cc::) (S+2aa+bb+cc) (aa:bb:cc)
of A'B'C' not in ETC not in ETC

Best regards
Sincerely
Paul
• Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares based on BC,CA,AB, resp., outwardly ABC. A A3 A2 A / / / / B3
Message 11 of 21 , Dec 9, 2001
Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares
based on BC,CA,AB, resp., outwardly ABC.

A'

A3
A2
A
/\
/ \
/ \
/ \
B3 / \
/ \ C2
/ \
B--------------C
| |
| |
| |
| |
| |
B' B1--------------C1 C'

Let A'B'C' be the triangle bounded by the lines B1C1, C2A2, A3B3
(Grebe triangle).

Denote 1 = ABC, 2 = medial ABC, 3 = antimedial ABC,
4 = A'B'C', 5 = medial A'B'C', 6 = antimedial A'B'C'

(x,y) := homothetic center of x, y.

(1,2) = G | --- | --- |
| | |
(1,3) = G | (2,3) = G | --- |
| | |
(1,4) = K | (2,4) = ? | (3,4) = ? |
| | |
(1,5) = ? | (2,5) = ? | (3,5) = ? |
| | |
(1,6) = ? | (2,6) = ? | (3,6) = ? |

(4,5) = (4,6) = (5,6) = G' = ?

Coordinates?

APH
• Dear Antreas ... The 6 points Ba, Ca, Cb, Ab, Ac, Bc are concyclic (cf. my previous message 4507) So the perpendicular bisectors of BaCa, CbAb, AcBc are
Message 12 of 21 , Dec 9, 2001
Dear Antreas

> On Saturday, December 8, 2001, at 04:55 PM, Antreas P. Hatzipolakis
> wrote:
>
> > Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares
> > based on BC,CA,AB, resp., outwardly ABC.
> >
> >
> > A2
> > A3
> > A
> > /\
> > / \
> > / \
> > / \
> > / \ C2
> > B3 / \
> > / \
> > -----Ba-------------------B--------------C-----------------------Ca----
> >
> > Let Ba = A3B3 /\ BC, Ca = BC /\ A2C2.
> >
> > Similarly Cb, Ab, Ac, Bc
> >
> > 1. Are the perp. bisectors of BaCa, CbAb, AcBc concurrent?

The 6 points Ba, Ca, Cb, Ab, Ac, Bc are concyclic (cf. my previous message
4507)

So the perpendicular bisectors of BaCa, CbAb, AcBc are concurrent.

The perpendicular bisectors of AbAc, BaBc, CaCb are concurrent at the same
point !

The perpendicular bisectors of BcCb, CaAc, AbBa are concurrent at the same
point (was B'1C'1, C'2A'2, A'3B'3 in your message 4506

> According to my calculations, they are concurrent IFF
>
> [where, to save space, I used the abbreviacions:
> a = sinA, A = cosA, etc; P = sinAsinBsinC]
>
> the det
>
> | -Bc(P-b^2+c^2)+Cb(P+b^2-c^2) b(P+b^2-c^2) -c(P-b^2+c^2) |
> | |
> | -a(P-c^2+a^2) -Ca(P-c^2+a^2)+Ac(P+c^2-a^2) c(P+c^2-a^2) |
> | |
> | a(P+a^2-b^2) -b(P-a^2+b^2) -Ab(P-a^2+b^2)+Ba(P+a^2-b^2) |
>
> is equal to 0. Is it???

If your calculations arre right, then this determinant vanishes.

Friendly

Gilles
• On Saturday, December 8, 2001, at 04:55 PM, Antreas P. Hatzipolakis ... According to my calculations, they are concurrent IFF [where, to save space, I used the
Message 13 of 21 , Dec 9, 2001
On Saturday, December 8, 2001, at 04:55 PM, Antreas P. Hatzipolakis
wrote:

> Let ABC be a triangle, and BCC1B1, CAA2C2, ABB3A3 the three squares
> based on BC,CA,AB, resp., outwardly ABC.
>
>
> A2
> A3
> A
> /\
> / \
> / \
> / \
> / \ C2
> B3 / \
> / \
> -----Ba-------------------B--------------C-----------------------Ca----
>
> Let Ba = A3B3 /\ BC, Ca = BC /\ A2C2.
>
> Similarly Cb, Ab, Ac, Bc
>
> 1. Are the perp. bisectors of BaCa, CbAb, AcBc concurrent?
>

According to my calculations, they are concurrent IFF

[where, to save space, I used the abbreviacions:
a = sinA, A = cosA, etc; P = sinAsinBsinC]

the det

| -Bc(P-b^2+c^2)+Cb(P+b^2-c^2) b(P+b^2-c^2) -c(P-b^2+c^2) |
| |
| -a(P-c^2+a^2) -Ca(P-c^2+a^2)+Ac(P+c^2-a^2) c(P+c^2-a^2) |
| |
| a(P+a^2-b^2) -b(P-a^2+b^2) -Ab(P-a^2+b^2)+Ba(P+a^2-b^2) |

is equal to 0. Is it???

> 2. Are the triangles ABC, GaGbGc in perspective, where Ga,Gb,Gc
> are the centroids of ABaCa, BCbAb, CAcBc, resp. ?
> [Equivalently: Let Ma, Mb, Mc be the midpoints of BaCa, CbAb, AcBc,
> resp.
> Are the lines AMa, BMb, CMc concurrent?]
>

In general, they are not.

Antreas
• In this message, we treat the Grebe triangle of a triangle. Erect squares BBaCaC, CCbAbA, AAcBcB on the sides BC, CA, AB of triangle ABC (outwards). Then, the
Message 14 of 21 , Jan 28, 2003
In this message, we treat the Grebe triangle of a triangle.

Erect squares BBaCaC, CCbAbA, AAcBcB on the sides BC, CA, AB of
triangle ABC (outwards). Then, the triangle A'B'C' enclosed by the
lines BaCa, CbAb, AcBc is called the Grebe triangle of ABC. It is
perspective (and homothetic) with triangle ABC, the perspector being
the symmedian point of ABC (this is a well-known fact).

With the help of computer reducing, I have found out that the Grebe
triangle is also in perspective with the orthic triangle of ABC. The
perspector, which I call "Grebe-orthic" perspector, has homogeneous
trilinears

/ (D+a²)(2D+b²+c²) (D+b²)(2D+c²+a²) (D+c²)(2D+a²+b²) \
( ---------------- : ---------------- : ---------------- ),
\ cos A cos B cos C /

where D is the area of triangle ABC. Not in ETC (*).

By the way, for A' we have trilinears

A' ( 2D + b² + c² : -ab : ca )

The circumcenter T of the Grebe triangle A'B'C' lies on the Brocard
axis of ABC. The trilinears of T are

( (1 + 2 cot w) cos A - 2 sin A
: (1 + 2 cot w) cos B - 2 sin B
: (1 + 2 cot w) cos C - 2 sin C ),

where w is the Brocard angle of triangle of triangle ABC.

(Note that 1 + 2 cot w is the homothetic factor of triangles ABC and
A'B'C': for example, B'C' = (1 + 2 cot w) a.)

---------------------------

(*) PS for Clark Kimberling: If you are interested in including the
Grebe-orthic perspector

/ (D+a²)(2D+b²+c²) (D+b²)(2D+c²+a²) (D+c²)(2D+a²+b²) \
( ---------------- : ---------------- : ---------------- )
\ cos A cos B cos C /

in ETC, then you can also include the points

/ (D-a²)(2D+b²+c²) (D-b²)(2D+c²+a²) (D-c²)(2D+a²+b²) \
( ---------------- : ---------------- : ---------------- ),
\ cos A cos B cos C /

/ (D+a²)(2D-b²-c²) (D+b²)(2D-c²-a²) (D+c²)(2D-a²-b²) \
( ---------------- : ---------------- : ---------------- ),
\ cos A cos B cos C /

/ (D-a²)(2D-b²-c²) (D-b²)(2D-c²-a²) (D-c²)(2D-a²-b²) \
( ---------------- : ---------------- : ---------------- ).
\ cos A cos B cos C /

All four points are not in ETC.

Darij Grinberg
• Dear Darij ... being ... Grebe ... Consider any point P; Q = isotomic conjugate of the anticomplement of P; then any triangle homothetic at P with ABC will be
Message 15 of 21 , Jan 29, 2003
Dear Darij

> Erect squares BBaCaC, CCbAbA, AAcBcB on the sides BC, CA, AB of
> triangle ABC (outwards). Then, the triangle A'B'C' enclosed by the
> lines BaCa, CbAb, AcBc is called the Grebe triangle of ABC. It is
> perspective (and homothetic) with triangle ABC, the perspector
being
> the symmedian point of ABC (this is a well-known fact).
>
> With the help of computer reducing, I have found out that the
Grebe
> triangle is also in perspective with the orthic triangle of ABC.

Consider any point P; Q = isotomic conjugate of the anticomplement
of P; then any triangle homothetic at P with ABC will be perspective
with the cevian triangle of Q and the perspector lies on the conic
through P, Q and the feet of the cevian of Q.
Hence, your result remains true for any Kiepert-rectangle, the
perspector of the orthic triangle and the generalized Grebe triangle
lying on the rectangular hyperbola through K, H and the vertices of
the orthic triangle.
Friendly. Jean-Pierre
• Dear Jean-Pierre Ehrmann, Thank you very much for the reply. ... Now, this gives a new nice property of isotomcomplements (for P is the isotomcomplement of Q).
Message 16 of 21 , Jan 30, 2003
Dear Jean-Pierre Ehrmann,

Thank you very much for the reply.

>> Consider any point P; Q = isotomic conjugate
>> of the anticomplement of P; then any triangle
>> homothetic at P with ABC will be perspective
>> with the cevian triangle of Q and the
>> perspector lies on the conic through P, Q and
>> the feet of the cevian of Q.

Now, this gives a new nice property of isotomcomplements (for P is
the isotomcomplement of Q). As I have conjectured former (this was
proven by Paul Yiu), any triangle homothetic at P with the cevian
triangle of Q is perspective with ABC -- and now any triangle
homothetic at P with ABC is perspective with the cevian triangle of
Q. Very interesting!

Sincerely,
Darij Grinberg
• I apologize for a typo in Hyacinthos message #6445 (Grebe triangle): I wrote A ( 2D + b² + c² : -ab : ca ), but this should be A ( 2D + b² + c² : -ab :
Message 17 of 21 , Feb 1, 2003
I apologize for a typo in Hyacinthos message #6445 (Grebe triangle):
I wrote A' ( 2D + b² + c² : -ab : ca ), but this should be

A' ( 2D + b² + c² : -ab : -ca ),

or equivalently A' ( - 2D - b² - c² : ab : ca ).

Darij Grinberg
• THIS MESSAGE CONTAINS HYACINTHOS #6445, CORRECTED AFTER HYACINTHOS #6463, AND SOME NEW RESULTS. We will treat two Grebe triangles of a triangle: the outer
Message 18 of 21 , Feb 15, 2003
THIS MESSAGE CONTAINS HYACINTHOS #6445, CORRECTED AFTER HYACINTHOS
#6463, AND SOME NEW RESULTS.

We will treat two Grebe triangles of a triangle: the outer Grebe
triangle and the inner Grebe triangle.

--- THE OUTER GREBE TRIANGLE

Erect squares BBaCaC, CCbAbA, AAcBcB on the sides BC, CA, AB of
triangle ABC (outwards). Then, the triangle A'B'C' enclosed by the
lines BaCa, CbAb, AcBc is called the outer Grebe triangle of ABC. It
is perspective (and homothetic) with triangle ABC, the perspector
being the symmedian point of ABC (this is a well-known fact).

With the help of computer reducing, I have found out that the outer
Grebe triangle is also in perspective with the orthic triangle of
ABC. The perspector, which I call outer Grebe-orthic perspector, has
homogeneous trilinears

/ (D+a²)(2D+b²+c²) (D+b²)(2D+c²+a²) (D+c²)(2D+a²+b²) \
( ---------------- : ---------------- : ---------------- ),
\ cos A cos B cos C /

where D is the area of triangle ABC.

By the way, for A' we have trilinears

A' ( 2D + b² + c² : -ab : -ca ),

or equivalently A' ( - 2D - b² - c² : ab : ca ).

The circumcenter T of the outer Grebe triangle A'B'C' lies on the
Brocard axis of ABC. The trilinears of T are

( (1 + 2 cot w) cos A - 2 sin A
: (1 + 2 cot w) cos B - 2 sin B
: (1 + 2 cot w) cos C - 2 sin C ),

where w is the Brocard angle of triangle of triangle ABC.

(Note that 1 + 2 cot w is the homothetic factor of triangles ABC and
A'B'C': for example, B'C' = (1 + 2 cot w) a.)

We can also write the trilinears of T as

( (2area + a² + b² + c²)/(2area) cos A - 2 sin A : ... )

= ( (1 + 2 cot A + 2 cot B + 2 cot C) cos A - 2 sin A : ... ).

Neither the outer Grebe-orthic perspector, nor T is (yet) in ETC.

--- THE INNER GREBE TRIANGLE

Edward Brisse had pointed me to a modification of the outer Grebe
triangle:

The inner Grebe triangle results if we erect the squares BBaCaC,
CCbAbA, AAcBcB inwards instead of outwards. Then, we get the inner
Grebe triangle A"B"C" enclosed by the lines BaCa, CbAb, AcBc.

This inner Grebe triangle A"B"C" is also perspective to the orthic
triangle of ABC. The perspector is called the inner Grebe-orthic
perspector and has trilinears

/ (D-a²)(2D-b²-c²) (D-b²)(2D-c²-a²) (D-c²)(2D-a²-b²) \
( ---------------- : ---------------- : ---------------- ).
\ cos A cos B cos C /

The vertices of the inner Grebe triangle have trilinears

A" ( 2D - b² - c² : ab : ca ) etc..

The circumcenter T' of the inner Grebe triangle A"B"C" lies on the
Brocard axis of ABC. The trilinears of T' are

( (1 - 2 cot w) cos A + 2 sin A
: (1 - 2 cot w) cos B + 2 sin B
: (1 - 2 cot w) cos C + 2 sin C ),

where w is the Brocard angle of triangle of triangle ABC.

(Now 1 - 2 cot w is the homothetic factor of triangles ABC and
A"B"C": for example, B"C" = (1 - 2 cot w) a.)

We can also write the trilinears of T' as

( (2area - a² - b² - c²)/(2area) cos A + 2 sin A : ... )

= ( (1 - 2 cot A - 2 cot B - 2 cot C) cos A + 2 sin A : ... ).

Neither the inner Grebe-orthic perspector, nor T' is (yet) in ETC.

--- TWO POINTS WITHOUT GEOMETRICAL DESCRIPTION

We have identified the points with trilinears

/ (D+a²)(2D+b²+c²) (D+b²)(2D+c²+a²) (D+c²)(2D+a²+b²) \
( ---------------- : ---------------- : ---------------- )
\ cos A cos B cos C /

and

/ (D-a²)(2D-b²-c²) (D-b²)(2D-c²-a²) (D-c²)(2D-a²-b²) \
( ---------------- : ---------------- : ---------------- )
\ cos A cos B cos C /

as the outer and inner Grebe-orthic perspectors. But we can also
define two other points,

/ (D+a²)(2D-b²-c²) (D+b²)(2D-c²-a²) (D+c²)(2D-a²-b²) \
( ---------------- : ---------------- : ---------------- )
\ cos A cos B cos C /

and

/ (D-a²)(2D+b²+c²) (D-b²)(2D+c²+a²) (D-c²)(2D+a²+b²) \
( ---------------- : ---------------- : ---------------- ),
\ cos A cos B cos C /

which are also not in ETC. I don't know of a geometrical
signification of these points.

Darij Grinberg
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