## 7 Gs

Expand Messages
• Let ABC be a triangle, P = (x:y:z) a point, and A B C the cevian triangle of P. Gab := the centroid of A BP, Gac := the centroid of A CP Gbc := the centroid
Message 1 of 18 , Sep 6, 2004
• 0 Attachment
Let ABC be a triangle, P = (x:y:z) a point, and A'B'C'
the cevian triangle of P.

Gab := the centroid of A'BP, Gac := the centroid of A'CP
Gbc := the centroid of B'CP, Gba := the centroid of B'AP
Gca := the centroid of C'AP, Gcb := the centroid of C'BP

G1 := the centroid of GabGbcGca
G2 := the centroid of GbaGcbGac

We have G1 = G2 := P*.

Which are the h. coordinates of P* ?

APH
--
• Dear Antreas, ... I think barys are P* = (y + z) (6 x^3 + 8 x^2 y + 2 x y^2 + 8 x^2 z + 8 x y z + y^2 z + 2 x z^2 + y z^2) : : Also, let Ga = midpoint[Gba,
Message 2 of 18 , Sep 6, 2004
• 0 Attachment
Dear Antreas,

> Let ABC be a triangle, P = (x:y:z) a point, and A'B'C'
> the cevian triangle of P.
>
> Gab := the centroid of A'BP, Gac := the centroid of A'CP
> Gbc := the centroid of B'CP, Gba := the centroid of B'AP
> Gca := the centroid of C'AP, Gcb := the centroid of C'BP
>
> G1 := the centroid of GabGbcGca
> G2 := the centroid of GbaGcbGac
>
> We have G1 = G2 := P*.
>
> Which are the h. coordinates of P* ?
>

I think barys are

P* = (y + z) (6 x^3 + 8 x^2 y + 2 x y^2 + 8 x^2 z + 8 x y z + y^2 z
+ 2 x z^2 + y z^2) : :

Also, let

Ga = midpoint[Gba, Gca]
Gb = midpoint[Gcb, Gab]
Ga = midpoint[Gac, Gbc]

Ga' = midpoint[Gab, Gac]
Gb' = midpoint[Gbc, Gba]
Ga' = midpoint[Gca, Gcb]

P* is the centroid of GaGbGc & Ga'Gb'Gc'

Also Ga,Gb,Gc is perspective to ABC at
x (y + z) / (x + 3 y + 3 z) : :

Ga'' = Gba, Gbc /\ Gca, Gcb
Gb'' = Gcb, Gca /\ Gab, Gac
Gc'' = Gac, Gab /\ Gbc, Gba

Ga''Gb''Gc'' perspective to ABC at P.

Best regards

Peter J. C .Moses.
• Dear Peter, ... [PM] ... Thanks !! Now, Fab := the centroid of A BA, Fac := the centroid of A CA Fbc := the centroid of B CB, Fba := the centroid of B AB Fca
Message 3 of 18 , Sep 7, 2004
• 0 Attachment
Dear Peter,

[APH]:
>> Let ABC be a triangle, P = (x:y:z) a point, and A'B'C'
>> the cevian triangle of P.
>>
>> Gab := the centroid of A'BP, Gac := the centroid of A'CP
>> Gbc := the centroid of B'CP, Gba := the centroid of B'AP
>> Gca := the centroid of C'AP, Gcb := the centroid of C'BP
>>
>> G1 := the centroid of GabGbcGca
>> G2 := the centroid of GbaGcbGac
>>
>> We have G1 = G2 := P*.
>>
>> Which are the h. coordinates of P* ?
>>

[PM]
(with little editing):
>I think barys are
>
>P* = (y + z) (6 x^3 + 8 x^2 y + 2 x y^2 + 8 x^2 z + 8 x y z + y^2 z
>+ 2 x z^2 + y z^2) : :
>
>Also, let
>
>Ga = midpoint[Gba, Gca]
>Gb = midpoint[Gcb, Gab]
>Gc = midpoint[Gac, Gbc]
>
>Ga' = midpoint[Gab, Gac]
>Gb' = midpoint[Gbc, Gba]
>Gc' = midpoint[Gca, Gcb]
>
>P* is the centroid of GaGbGc & Ga'Gb'Gc'
>
>Also GaGbGc is perspective to ABC at
>x (y + z) / (x + 3 y + 3 z) : :
>
>Ga'' = GbaGbc /\ GcaGcb
>Gb'' = GcbGca /\ GabGac
>Gc'' = GacGab /\ GbcGba
>
>Ga''Gb''Gc'' is perspective to ABC at P.

Thanks !!

Now,

Fab := the centroid of A'BA, Fac := the centroid of A'CA
Fbc := the centroid of B'CB, Fba := the centroid of B'AB
Fca := the centroid of C'AC, Fcb := the centroid of C'BC

F1 := the centroid of FabFbcFca
F2 := the centroid of FbaFcbFac

We have F1 = F2 := F#

Coordinates?

I guess that we have analogous properties for Fa, Fa'
(defined analogously to Ga, Ga')

Also, for which Ps the points Gs or Fs are lying on conics ?

Greetings

Antreas

--
• Dear Antreas, [APH] ... F# = (y + z) (4 x^2 + 3 x y + 3 x z + 2 y z) : : some ETC points, {2,2},{4,373},{30,1651},{99,1649} Fa = midpoint[Fba, Fca] Fb =
Message 4 of 18 , Sep 7, 2004
• 0 Attachment
Dear Antreas,

[APH]
> Now,
>
> Fab := the centroid of A'BA, Fac := the centroid of A'CA
> Fbc := the centroid of B'CB, Fba := the centroid of B'AB
> Fca := the centroid of C'AC, Fcb := the centroid of C'BC
>
> F1 := the centroid of FabFbcFca
> F2 := the centroid of FbaFcbFac
>
> We have F1 = F2 := F#
>
> Coordinates?
>
> I guess that we have analogous properties for Fa, Fa'
> (defined analogously to Ga, Ga')
>
> Also, for which Ps the points Gs or Fs are lying on conics ?

F# = (y + z) (4 x^2 + 3 x y + 3 x z + 2 y z) : :

some ETC points, {2,2},{4,373},{30,1651},{99,1649}

Fa = midpoint[Fba, Fca]
Fb = midpoint[Fcb, Fab]
Fa = midpoint[Fac, Fbc]

Fa' = midpoint[Fab, Fac]
Fb' = midpoint[Fbc, Fba]
Fa' = midpoint[Fca, Fcb]

Fa'' = midpoint[Fcb, Fbc]
Fb'' = midpoint[Fca, Fac]
Fc'' = midpoint[Fab, Fba]

I reckon that
F# is the centroid of FaFbFc, Fa'Fb'Fc' & Fa''Fb''Fc''

and if

Fa* = Fba, Fbc /\ Fca, Fcb
Fb* = Fcb, Fca /\ Fab, Fac
Fc* = Fac, Fab /\ Fbc, Fba

then
Fa*= Fb*= Fc* = G.

Does not seem that the F's have exactly analogous properties to the
G's.

Best regards
Peter
• Dear Peter We have seen Gs,Fs; now how about Ds? Defined as follows: A B C := the cevian triangle of P = (x:y:z) Dab := the centroid of B C B Dac := the
Message 5 of 18 , Sep 7, 2004
• 0 Attachment
Dear Peter

We have seen Gs,Fs; now how about Ds?
Defined as follows:

A'B'C' := the cevian triangle of P = (x:y:z)

Dab := the centroid of B'C'B
Dac := the centroid of C'B'C

Dbc := the centroid of C'A'C
Dba := the centroid of A'C'A

Dca := the centroid of A'B'A
Dcb := the centroid of B'A'B

D1 := the centroid of DabDbcDca
D2 := the centroid of DbaDcbDac

I think that D1 = D2
If so, then let D1 = D2 := P'

Coordinates of P'?

We can also define Da, etc, analogously to Ga, etc.

Greetings

Antreas

--
• Dear friends, let Q be the centroid of the cevian triangle of P. to construct P with given point Q. Best regards Bernard
Message 6 of 18 , Sep 7, 2004
• 0 Attachment
Dear friends,

let Q be the centroid of the cevian triangle of P.

to construct P with given point Q.

Best regards

Bernard
• Dear Bernard! ... Let P has homogenius barycentric coordinates P(x:y:z) (in base of ABC), then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where
Message 7 of 18 , Sep 7, 2004
• 0 Attachment
Dear Bernard!
>
> let Q be the centroid of the cevian triangle of P.
>
> to construct P with given point Q.
>
>
Let P has homogenius barycentric coordinates P(x:y:z) (in base of ABC),
then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where s=x+y+z.
So, we have equation of 3-d degree to find x,y,z and I think that in
general we can't construct P by ruler and compass.
Best regards,
Alex
• Dear Antreas, [APH] ... I also think D1 = D2. P = (y + z) (5 x^2 + 3 x y + 3 x z + y z) : : Da = midpoint[Dba, Dca] Db = midpoint[Dcb, Dab] Da = midpoint[Dac,
Message 8 of 18 , Sep 7, 2004
• 0 Attachment
Dear Antreas,

[APH]
>
> A'B'C' := the cevian triangle of P = (x:y:z)
>
> Dab := the centroid of B'C'B
> Dac := the centroid of C'B'C
>
> Dbc := the centroid of C'A'C
> Dba := the centroid of A'C'A
>
> Dca := the centroid of A'B'A
> Dcb := the centroid of B'A'B
>
>
> D1 := the centroid of DabDbcDca
> D2 := the centroid of DbaDcbDac
>
> I think that D1 = D2
> If so, then let D1 = D2 := P'
>
> Coordinates of P'?
>
> We can also define Da, etc, analogously to Ga, etc.
>

I also think D1 = D2.

P' = (y + z) (5 x^2 + 3 x y + 3 x z + y z) : :

Da = midpoint[Dba, Dca]
Db = midpoint[Dcb, Dab]
Da = midpoint[Dac, Dbc]

Da' = midpoint[Dab, Dac]
Db' = midpoint[Dbc, Dba]
Da' = midpoint[Dca, Dcb]

Da'' = midpoint[Dcb, Dbc]
Db'' = midpoint[Dca, Dac]
Dc'' = midpoint[Dab, Dba]

P'is the centroid of DaDbDc, Da'Db'Dc' & Da''Db''Dc''

Da* = Dba, Dbc /\ Dca, Dcb
Db* = Dcb, Dca /\ Dab, Dac
Dc* = Dac, Dab /\ Dbc, Dba

Da*= Db*= Dc* = The centroid of the cevian of P.
= x (y + z) (2 x + y + z) : :

Some ETC points,
{1,1962},{2,2},{4,51},{7,354},{8,210},{20,154},
{144,165},{253,1853}

looks like
Dac, Dab, Da', Da* are colinear and parallel to BC.

Similiarily
Dbc, Dba, Db', Db* // to CA
Dca, Dcb, Dc', Dc* // to AB

Best regards
Peter.
• ... [alex_geom@mtu-net.ru] ... Which is the locus of P such that G, P, Q are collinear? G = (1:1:1) P = (x:y:z) Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the
Message 9 of 18 , Sep 7, 2004
• 0 Attachment
[BG]:
>> let Q be the centroid of the cevian triangle of P.

[alex_geom@...]
>Let P has homogenius barycentric coordinates P(x:y:z) (in base of ABC),
>then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where s=x+y+z.

Which is the locus of P such that G, P, Q are collinear?

G = (1:1:1)
P = (x:y:z)
Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the above is correct

G,P,Q are collinear ==> Det. of the matrix = 0

==> etc

Antreas
--
• Dear Antreas! ... So, the equation is x^3(y-z)+y^3(z-x)+z^3(x-y)=0 or, another form: (y^2-z^2)/x+(z^2-x^2)/y+(x^2-y^2)/z=0 I don,t know what it is, even I
Message 10 of 18 , Sep 7, 2004
• 0 Attachment
Dear Antreas!
> [BG]:
>>> let Q be the centroid of the cevian triangle of P.
>
> [alex_geom@...]
>> Let P has homogenius barycentric coordinates P(x:y:z) (in base of
>> ABC),
>> then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where
>> s=x+y+z.
>
> Which is the locus of P such that G, P, Q are collinear?
>
> G = (1:1:1)
> P = (x:y:z)
> Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the above is correct
>
> G,P,Q are collinear ==> Det. of the matrix = 0
>
> ==> etc
>
>
So, the equation is
x^3(y-z)+y^3(z-x)+z^3(x-y)=0
or, another form:
(y^2-z^2)/x+(z^2-x^2)/y+(x^2-y^2)/z=0
I don,t know what it is, even I don't know what interesting points lay
on it (exsept of course centroid).
Yours,
Alex (Alexei Myakishev)
And sorry for so silly nikname "Teacher" - that is simply because I am
using now the comp from the school where I'm working.
• Dear Alexei ... This quartic is not listed in Bernard s curves: (unless I didn t see it!) http://perso.wanadoo.fr/bernard.gibert/relatedcurves.html We can get
Message 11 of 18 , Sep 7, 2004
• 0 Attachment
Dear Alexei

>> [BG]:
>>>> let Q be the centroid of the cevian triangle of P.
>>
>> [Alexei Myakishev]
>>> Let P has homogenius barycentric coordinates P(x:y:z) (in base of
>>> ABC),
>>> then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where
>>> s=x+y+z.
>>
>> Which is the locus of P such that G, P, Q are collinear?
>>
>> G = (1:1:1)
>> P = (x:y:z)
>> Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the above is correct
>>
>> G,P,Q are collinear ==> Det. of the matrix = 0
>>
>> ==> etc

[Alexei Myakishev]:
>So, the equation is
>x^3(y-z)+y^3(z-x)+z^3(x-y)=0
>or, another form:
>(y^2-z^2)/x+(z^2-x^2)/y+(x^2-y^2)/z=0
>I don,t know what it is, even I don't know what interesting points lay
>on it (exsept of course centroid).

This quartic is not listed in Bernard's curves:
(unless I didn't see it!)

We can get more curves if we consider the isotomic points tP, tQ
of the above ones and make C(5,3) = 10 combinations, namely:

G = (1:1:1)
P = (x:y:z)
Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2))

tP = (1/x:1/y:1/z)
tQ = (1/x(s^2-x^2):1/y(s^2-y^2):1/z(s^2-z^2))

Loci of P such that be collinear the points:

1. G, P, Q (the above one)

2. G, P, tP (isotomic cubic with pivot G)

3. G, P, tQ

4. G, Q, tP

5. G, Q, tQ

6. G, tP, tQ

7. P, Q, tP

8. P, Q, tQ

9. P, tP, tQ

10. Q, tP, tQ

[Alexei Myakishev]:
>And sorry for so silly nikname "Teacher" - that is simply because I am
>using now the comp from the school where I'm working.

Well... you may give the above to your students, as exercises in
the Linear Algebra :-)

Greetings

Antreas

--
• Dear Antreas! ... We can also get some lines, for example: 1. P^3=(x^3:y^3:z^3);P;Q (Ofcourse, P^3=P^2*P in sence of multiplication of points according Paul
Message 12 of 18 , Sep 7, 2004
• 0 Attachment
Dear Antreas!
>
>>> [BG]:
>>>>> let Q be the centroid of the cevian triangle of P.
>>>
>>> [Alexei Myakishev]
>>>> Let P has homogenius barycentric coordinates P(x:y:z) (in base of
>>>> ABC),
>>>> then Q has coordinates x(s^2-x^2):y(s^2-y^2):z(s^2-z^2), where
>>>> s=x+y+z.
>>>
>>> Which is the locus of P such that G, P, Q are collinear?
>>>
>>> G = (1:1:1)
>>> P = (x:y:z)
>>> Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2)), if the above is correct
>>>
>>> G,P,Q are collinear ==> Det. of the matrix = 0
>>>
>>> ==> etc
>
> We can get more curves if we consider the isotomic points tP, tQ
> of the above ones and make C(5,3) = 10 combinations, namely:
>
> G = (1:1:1)
> P = (x:y:z)
> Q = (x(s^2-x^2):y(s^2-y^2):z(s^2-z^2))
>
> tP = (1/x:1/y:1/z)
> tQ = (1/x(s^2-x^2):1/y(s^2-y^2):1/z(s^2-z^2))

We can also get some lines, for example:
1. P^3=(x^3:y^3:z^3);P;Q
(Ofcourse, P^3=P^2*P in sence of multiplication of points according
Paul Yiu - but may be somebody can invent "direct" way of construction
of cube of a point?)
2. Let R=(y+z,x+z,y+x) - homotety image of P with center G and ratio
k=2.
So we have collineary points P*tR;P;Q
3. And, moreover, P^2*tR;P;Q
-->P*tR;P^2*tR;P;Q are collinear
(* means barycentric multiplication)

Best regards,
sincerely,
Alexei
• Dear Alexei, ... here is one : tX is the isotomic conjugate of X, cX is the complement of X G is the centroid of ABC P is a point [P,Q] is the P-Ceva conjugate
Message 13 of 18 , Sep 8, 2004
• 0 Attachment
Dear Alexei,

> but may be somebody can invent "direct" way of construction
> of cube of a point?

here is one :

tX is the isotomic conjugate of X,
cX is the complement of X
G is the centroid of ABC
P is a point
[P,Q] is the P-Ceva conjugate of Q.

the lines P, ctP and G, tcP intersect at the (barycentric) square P^2
of P.

P^3 is the harmonic conjugate of [P,ctP] with respect to P^2 and
[P,P^2].

I'm sure there's a simpler construction.

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Ricardo! ... G, a point N on AC, a point L on AB and the straight r in which it is BC ========== The desired construction can be performed in 2 steps:
Message 14 of 18 , Dec 27, 2004
• 0 Attachment
Dear Ricardo!

> To construct ABC knowing:
G,
a point N on AC,
a point L on AB
and the straight r in which it is BC
==========

The desired construction can be performed in 2 steps:

(1) we draw an arbitrary triangle A'B'C' with centroid G' and construct the
points N' and M' on sides A'C' and A'B' that are images of N and M under the
affine map f that takes ABC onto A'B'C';

(2) we find the vertex A of the unknown triangle as the image of A' under
the map inverse to f, which is uniquely defined by the given images G, N, M
of G', N', M'.

Namely, the equation |N', B'C'| : |G', B'C'| = |N, r| : |G, r|, where |X, l|
denotes the distance from X to line l, yields the distance |N', B'C'|, which
enables us to construct N' on A'C'. M' is constructed similarly.
Then, let A'G' meet N'M' at P'. We construct P on NM such that NP:PM =
N'P':P'M' and then A (on the line GP) such that GA : GP = G'A' : G'P'.
AN and AM meet r at C and B.

With best New Year wishes,

• Dear friends, coming back again to Ricardo s problem -- ... -- let me notice that since we can easily find the ratios in which points L and N divide their
Message 15 of 18 , Dec 29, 2004
• 0 Attachment
Dear friends,

coming back again to Ricardo's problem --

> To construct ABC knowing:
> G,
> a point N on AC,
> a point L on AB
> and the straight r in which it is BC

-- let me notice that since we can easily find the ratios in which points L
and N divide their respective sides (they are the u and v used in Francois's
solution), we know the barycentrics of these points. So the problem
naturally generalizes to the following one:

construct triangle ABC given three points X,Y,Z in its plane and their
barycentric coordinates wrt to ABC.

This boils down to solving a system of linear equations and consructing the
solutions.
A somewhat more geometric construction can be given in the spirit of my
solution of the original problem: we draw an arbitrary triangle A'B'C', then
find points X', Y', Z' with the given barycentrics wrt A'B'C', and then
affinely map these points on XYZ; ABC is the image of A'B'C' under this
mapping.
In practice, we find the ratios in which, say, line A'B' divides segments
X'Y' and X'Z', and construct points P and Q on lines XY and XZ that divide
XY and XZ in these ratios, respectively. The line PQ will be just the side
line AB. Two other side lines are constructed similarly.

Best wishes for 2005,

• Dear François, Vladimir, Jean. Pierre, Hyacinthos: In http://www.personal.us.es/rbarroso/trianguloGNLr2.htm is the solution of François with applet de
Message 16 of 18 , Dec 30, 2004
• 0 Attachment
Dear François, Vladimir, Jean. Pierre, Hyacinthos:

In

http://www.personal.us.es/rbarroso/trianguloGNLr2.htm

is the solution of François with applet de Cabri-Java.

Thanks for yours interest.

Ricardo

______________________________________________
Renovamos el Correo Yahoo!: ¡250 MB GRATIS!
http://correo.yahoo.es
• Bernard, Do you have a similar construction for PQ. the barycentric product of P and Q. I have searched for this with no success. From the (right now warm)
Message 17 of 18 , Jan 5, 2005
• 0 Attachment
Bernard,

Do you have a similar construction for PQ. the barycentric product of P
and Q. I have searched for this with no success.

From the (right now warm) south of USA,
Steve

On Sep 8, 2004, at 7:36 AM, Bernard Gibert wrote:

>
>> but may be somebody can invent "direct" way of construction
>> of cube of a point?
>
> here is one :
>
> tX is the isotomic conjugate of X,
> cX is the complement of X
> G is the centroid of ABC
> P is a point
> [P,Q] is the P-Ceva conjugate of Q.
>
> the lines P, ctP and G, tcP intersect at the (barycentric) square P^2
> of P.
>
> P^3 is the harmonic conjugate of [P,ctP] with respect to P^2 and
> [P,P^2].
>
> I'm sure there's a simpler construction.
>
> Best regards
>
> Bernard
• Dear Steve, nice to hear from you. ... I doubt there is a similar construction because of the disymmetry introduced by P and G, but maybe I m wrong... best
Message 18 of 18 , Jan 6, 2005
• 0 Attachment
Dear Steve,

nice to hear from you.

> Do you have a similar construction for PQ. the barycentric product of
> P
> and Q. I have searched for this with no success.

I doubt there is a similar construction because of the disymmetry
introduced by P and G, but maybe I'm wrong...

best regards

Bernard

>
> On Sep 8, 2004, at 7:36 AM, Bernard Gibert wrote:
>
> >
> >> but may be somebody can invent "direct" way of construction
> >> of cube of a point?
> >
> > here is one :
> >
> > tX is the isotomic conjugate of X,
> > cX is the complement of X
> > G is the centroid of ABC
> > P is a point
> > [P,Q] is the P-Ceva conjugate of Q.
> >
> > the lines P, ctP and G, tcP intersect at the (barycentric) square
> P^2
> > of P.
> >
> > P^3 is the harmonic conjugate of [P,ctP] with respect to P^2 and
> > [P,P^2].
> >
> > I'm sure there's a simpler construction.
> >
> > Best regards
> >
> > Bernard
>
>
>
>
> � To visit your group on the web, go to:
> http://groups.yahoo.com/group/Hyacinthos/
> �
> � To unsubscribe from this group, send an email to:
> Hyacinthos-unsubscribe@yahoogroups.com
> �
> � Your use of Yahoo! Groups is subject to the Yahoo! Terms of
> Service.
>
>
>

[Non-text portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.