- 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 > --- In Hyacinthos@yahoogroups.com,

A different approach gave a symmetric formula with

> "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

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