## 3rd diagonals

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• Let A,B,C,D be four points in general position. Let s call third diagonal of the complete quadrilateral (A,B,C,D) the diagonal which doesn t pass through
Message 1 of 6 , Oct 23, 2010
Let A,B,C,D be four points in general position. Let's call
"third diagonal" of the complete quadrilateral (A,B,C,D)
the diagonal which doesn't pass through vertices of (A,B,C,D).

Now, let A,B,C,D,E be five points in general position.

Is the following true?

The "third diagonals" of the complete quadrilaterals

(D,E,A,B),(D,E,B,C) and (D,E,C,A) are concurrent.

Application in the Triangle Geometry:

Let ABC be a triangle and D = (x:y:z), E = (u:v:w) two points.

Denote La the line (BD /\ CE)(BE/\CD) (ie the line passing
through the points BD /\ CE and BE /\ CD)

Similarly Lb := (CD /\ AE)(CE/\AD)

Are the Lines La,Lb,Lc concurrent?

Point of concurrence?
And in the special case D,E = isogonal conjugate points.

APH
• La Lb and Lc intersect at Q={u x (w y + v z) : v y (w x + u z) : w (v x + u y) z} When (u:v:w) is the isogonal conjugate of (x:y:z), then Q is the point (a^2
Message 2 of 6 , Oct 24, 2010
La Lb and Lc intersect at

Q={u x (w y + v z) : v y (w x + u z) : w (v x + u y) z}

When (u:v:w) is the isogonal conjugate of (x:y:z), then Q is the point

(a^2 x (c^2 y^2 + b^2 z^2): b^2 y (c^2 x^2 + a^2 z^2) :
c^2 (b^2 x^2 + a^2 y^2) z})

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
>
> Let A,B,C,D be four points in general position. Let's call
> "third diagonal" of the complete quadrilateral (A,B,C,D)
> the diagonal which doesn't pass through vertices of (A,B,C,D).
>
> Now, let A,B,C,D,E be five points in general position.
>
> Is the following true?
>
> The "third diagonals" of the complete quadrilaterals
>
> (D,E,A,B),(D,E,B,C) and (D,E,C,A) are concurrent.
>
> Application in the Triangle Geometry:
>
> Let ABC be a triangle and D = (x:y:z), E = (u:v:w) two points.
>
> Denote La the line (BD /\ CE)(BE/\CD) (ie the line passing
> through the points BD /\ CE and BE /\ CD)
>
> Similarly Lb := (CD /\ AE)(CE/\AD)
>
> Lc := (AD /\ BE)(AE/\BD)
>
> Are the Lines La,Lb,Lc concurrent?
>
> Point of concurrence?
> And in the special case D,E = isogonal conjugate points.
>
> APH
>
• So if E is a fixed point, the locus of D such that D,E,Q are collinear is some conic. And if D, E are isogonal points, the locus is some cubic. Are they
Message 3 of 6 , Oct 25, 2010
So if E is a fixed point, the locus of D such that D,E,Q are collinear is
some conic.

And if D, E are isogonal points, the locus is some cubic.

Are they interesting?

APH

On Sun, Oct 24, 2010 at 10:35 PM, Francisco Javier
<garciacapitan@...>wrote:

>
>
> La Lb and Lc intersect at
>
> Q={u x (w y + v z) : v y (w x + u z) : w (v x + u y) z}
>
> When (u:v:w) is the isogonal conjugate of (x:y:z), then Q is the point
>
> (a^2 x (c^2 y^2 + b^2 z^2): b^2 y (c^2 x^2 + a^2 z^2) :
> c^2 (b^2 x^2 + a^2 y^2) z})
>
>
> --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
> "Antreas" <anopolis72@...> wrote:
> >
> >
> > Let A,B,C,D be four points in general position. Let's call
> > "third diagonal" of the complete quadrilateral (A,B,C,D)
> > the diagonal which doesn't pass through vertices of (A,B,C,D).
> >
> > Now, let A,B,C,D,E be five points in general position.
> >
> > Is the following true?
> >
> > The "third diagonals" of the complete quadrilaterals
> >
> > (D,E,A,B),(D,E,B,C) and (D,E,C,A) are concurrent.
> >
> > Application in the Triangle Geometry:
> >
> > Let ABC be a triangle and D = (x:y:z), E = (u:v:w) two points.
> >
> > Denote La the line (BD /\ CE)(BE/\CD) (ie the line passing
> > through the points BD /\ CE and BE /\ CD)
> >
> > Similarly Lb := (CD /\ AE)(CE/\AD)
> >
> > Lc := (AD /\ BE)(AE/\BD)
> >
> > Are the Lines La,Lb,Lc concurrent?
> >
> > Point of concurrence?
> > And in the special case D,E = isogonal conjugate points.
> >
> > APH
>

[Non-text portions of this message have been removed]
• [APH] So if E is a fixed point, the locus of D such that D,E,Q are collinear is some conic. No, if I understand it correctly, then D, E Q are collinear if some
Message 4 of 6 , Oct 25, 2010
[APH]
So if E is a fixed point, the locus of D such that D,E,Q are collinear is
some conic.

No, if I understand it correctly, then D, E Q are collinear if some of D, E
is on a other's cevian

2010/10/25 Antreas Hatzipolakis <anopolis72@...>

>
>
> So if E is a fixed point, the locus of D such that D,E,Q are collinear is
> some conic.
>
> And if D, E are isogonal points, the locus is some cubic.
>
> Are they interesting?
>
> APH
>
> On Sun, Oct 24, 2010 at 10:35 PM, Francisco Javier
> <garciacapitan@... <garciacapitan%40gmail.com>>wrote:
>
>
> >
> >
> > La Lb and Lc intersect at
> >
> > Q={u x (w y + v z) : v y (w x + u z) : w (v x + u y) z}
> >
> > When (u:v:w) is the isogonal conjugate of (x:y:z), then Q is the point
> >
> > (a^2 x (c^2 y^2 + b^2 z^2): b^2 y (c^2 x^2 + a^2 z^2) :
> > c^2 (b^2 x^2 + a^2 y^2) z})
> >
> >
> > --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com><Hyacinthos%
> 40yahoogroups.com>,
>
> > "Antreas" <anopolis72@...> wrote:
> > >
> > >
> > > Let A,B,C,D be four points in general position. Let's call
> > > "third diagonal" of the complete quadrilateral (A,B,C,D)
> > > the diagonal which doesn't pass through vertices of (A,B,C,D).
> > >
> > > Now, let A,B,C,D,E be five points in general position.
> > >
> > > Is the following true?
> > >
> > > The "third diagonals" of the complete quadrilaterals
> > >
> > > (D,E,A,B),(D,E,B,C) and (D,E,C,A) are concurrent.
> > >
> > > Application in the Triangle Geometry:
> > >
> > > Let ABC be a triangle and D = (x:y:z), E = (u:v:w) two points.
> > >
> > > Denote La the line (BD /\ CE)(BE/\CD) (ie the line passing
> > > through the points BD /\ CE and BE /\ CD)
> > >
> > > Similarly Lb := (CD /\ AE)(CE/\AD)
> > >
> > > Lc := (AD /\ BE)(AE/\BD)
> > >
> > > Are the Lines La,Lb,Lc concurrent?
> > >
> > > Point of concurrence?
> > > And in the special case D,E = isogonal conjugate points.
> > >
> > > APH
> >
>
> [Non-text portions of this message have been removed]
>
>
>

--
---
Francisco Javier Garc�a Capit�n
http://garciacapitan.auna.com

[Non-text portions of this message have been removed]
• ... [small snip] ... [big snip] ... I have a shorter symmetric formula. Let p=(b+c-a)(c+a-b)(a+b-c), and let k be a root of (p-abc)k^2+abc(p+2abc)k-(abc)^3=0.
Message 5 of 6 , Nov 5, 2010
--- In Hyacinthos@yahoogroups.com, "Francisco Javier" <garciacapitan@...> wrote:
>
> Dear Nikos: Your A-circle is the locus of points P such that PB^2/PC^2 = AB/AC; this easily implies that if P is on the A-circle and on the B-circle the P also is on the C-circle, and the three circles have two common points.
>
> The barycentrics are complicate. In an asymmetric form one of the points is:
>
> {{2 a^2 (a^4 - 2 a^2 b^2 + b^4 - a^2 b c + 2 a b^2 c - b^3 c -

[small snip]

>
> We can put this point in symmetric form, but the result is now really huge:
>
> {-a^2 (a^4 - 2 a^2 b^2 + b^4 - a^2 b c + 2 a b^2 c - b^3 c -

[big snip]

>
> Best regards,
>
> Francisco Javier.

I have a shorter symmetric formula.

Let p=(b+c-a)(c+a-b)(a+b-c),
and let k be a root of (p-abc)k^2+abc(p+2abc)k-(abc)^3=0.

Let x=(2k/a)(k/b+k/c-a^2)-(k/c+k/a-b^2)(k/a+k/b-c^2),
y=..., z=... cyclically. Then P=(1/x,1/y,1/z) satisfies
a*PA^2 = b*PB^2 = c*PC^2 = k
--
Barry Wolk
• ... A different approach gave a symmetric formula with simpler coordinates. We want the points P1 and P2 that satisfy a*PA^2 = b*PB^2 = c*PC^2. The midpoint of
Message 6 of 6 , Nov 6, 2010
> --- In Hyacinthos@yahoogroups.com,
> "Francisco Javier" <garciacapitan@...> wrote:
> >
> > Dear Nikos: Your A-circle is the locus of points P
> such that PB^2/PC^2 = AB/AC; this easily implies that if P
> is on the A-circle and on the B-circle the P also is on the
> C-circle, and the three circles have two common points.
> >
> > The barycentrics are complicate. In an asymmetric
> form  one of the points is:
> >
> > {{2 a^2 (a^4 - 2 a^2 b^2 + b^4 - a^2 b c + 2 a b^2 c -
> b^3 c -
>
> [small snip]
>
> >
> > We can put this point in symmetric form, but the
> result is now really huge:
> >
> > {-a^2 (a^4 - 2 a^2 b^2 + b^4 - a^2 b c + 2 a b^2 c -
> b^3 c -
>
> [big snip]
>
> >
> > Best regards,
> >
> > Francisco Javier.
>
> I have a shorter symmetric formula.
>
> Let p=(b+c-a)(c+a-b)(a+b-c),
> and let k be a root of (p-abc)k^2+abc(p+2abc)k-(abc)^3=0.
>
> Let x=(2k/a)(k/b+k/c-a^2)-(k/c+k/a-b^2)(k/a+k/b-c^2),
> y=..., z=... cyclically. Then P=(1/x,1/y,1/z) satisfies
> a*PA^2 = b*PB^2 = c*PC^2 = k

A different approach gave a symmetric formula with
simpler coordinates. We want the points P1 and P2
that satisfy a*PA^2 = b*PB^2 = c*PC^2.

The midpoint of P1P2 is
( a(b-c)^2-a^2(b+c-2a) , ... , ... ) = (x1,y1,z1)

The infinite point on the line P1P2 is
( -2bca^2+a(b+c)(a+b-c)(a-b+c) , ... , ... ) = (x2,y2,z2)

Let f(x,y,z)=2yz+2zx+2xy-x^2-y^2-z^2.
Let t be either square root of ( f(a,b,c)/f(a^2,b^2,c^2) ).
Then P1 and P2 are (x1 + t x2, y1 + t y2, z1 + t z2)
--
Barry Wolk
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