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• ... [long formula shipped] If a point P is defined geometrically in a symmetric way (usually as a perspector), then there is a general method of getting
Message 1 of 9 , Aug 30, 2010
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--- On Sat, 8/28/10, Francisco Javier <garciacapitan@...> wrote:
> The perspector is not simple.
>
> In its coordinates, shown below, we can see the symbols Ra,
> Rb, Rc, that stand for the square root of 4(s-b)(s-c),
> 4(s-c)(s-a), 4(s-a)(s-b), respectively.
>
> It is remarkable that these coordinates appear to be non
> symmetric, as we can think from the problem. I think that
> the formula can be transformed in some way into a symmetric form.
>
> I've checked that this point is on lines FaF'a, FbF'b, FcF'c.
>
> Coordinates of the perspector:

[long formula shipped]

If a point P is defined geometrically in a symmetric way (usually as a perspector), then there is a general method of getting symmetric coordinates for P. If x,y,z are complicated functions of a,b,c, and a symmetrically defined P has non-symmetric coordinates
P = (x(a,b,c),y(a,b,c),z(a,b,c) ) = (x1,y1,z1), then also
P = (z(b,c,a),x(b,c,a),y(b,c,a) ) = (z2,x2,y2), and
P = (y(c,a,b)),z(c,a,b),x(c,a,b) ) = (y3,z3,x3). Then symmetric coordinates for P are (x1+y3+z2,x2+y1+z3,x3+y2+z1).
--
Barry Wolk
• Variation: Let ABC be a triangle and F1,F2,F3 the ex-Feuerbach points of the triangles HBC,HCA,HAB corrsponding to angles (HBC),(HCA),(HAB), resp. Let (K1) be
Message 2 of 9 , Aug 31, 2010
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Variation:

Let ABC be a triangle and F1,F2,F3 the ex-Feuerbach points
of the triangles HBC,HCA,HAB corrsponding to angles
(HBC),(HCA),(HAB), resp.

Let (K1) be the circle touching externally the common Nine Point
Circle (N) of HBC,HCA,HAB [and ABC] and the excircles (Iba),
(Ica) of the triangles HCA,HAB corresponding to angles (HCA),
(HAB), resp. Let F'1 be the point of contact of (N) and (K1).
Similarly F'2,F'3.

The triangles F1F2F3, F'1F'2F'3 are perspective.

Perspector?

APH

--- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@...> wrote:
>
> Let ABC be a triangle and Fa,Fb,Fc the three ex-Feuerbach points.
>
> Let (Ka) be the circle touching externally the Nine point circle
> (N) and the excircles (Ib) and (Ic), and F'a the point of contact of
> (N) and (Ka). Similarly the points F'b and F'c.
>
> By the Seven Circles Theorem (*), the triangles FaFbFc and F'aF'bF'c
> are perspective.
>
> Which is the perspector?
>
> (*) See:
> http://mathworld.wolfram.com/SevenCirclesTheorem.html
>
> http://www.cut-the-knot.org/Curriculum/Geometry/SevenCirclesTheorem.shtml
>
> APH
>
• Dear Antreas, A drawing suggests not. An easy variation is an inversion in the Speiker radical circle. Best regards, Peter. ... From: Antreas To:
Message 3 of 9 , Sep 1, 2010
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Dear Antreas,

A drawing suggests not.

An "easy" variation is an inversion in the Speiker radical circle.

Best regards,
Peter.

----- Original Message -----
From: Antreas
To: Hyacinthos@yahoogroups.com
Sent: Monday, August 30, 2010 10:00 PM
Subject: [EMHL] Re: 7 circles theorem

Dear Francisco, Angel and Peter

Thanks!

I am now wondering what happens to 7 CIRCLES THEOREM
if we replace the points of contact with their antipodes in
the corresponding circles.

Applying it in triangle ABC:

Let ABC be a triangle, Fa,Fb,Fc the three ex-Feuerbach points,
and Ga,Gb,Gc their antipodes in the excircles (Ia),(Ib),(Ic),
respectively.

Let (Ka) be the circle touching externally the Nine point circle
(N) and the excircles (Ib) and (Ic), and F'a the point of contact of
(N) and (Ka). Similarly the points F'b and F'c.

Let G'a be the antipode of F'a in (Ka). Similarly G'b, G'c.

Are the triangles GaGbGc and G'aG'bG'c perspective?

APH

--- In Hyacinthos@yahoogroups.com, "Moses, Peter J. C." <mows@...> wrote:
>
> Dear Antreas, Fransisco & Angel,
>
> Following Fransisco let Ra = Sqrt[(a + b - c) (a - b + c)] = 2 Sqrt[sb sc] = 2 Sqrt[(s - b)(s - c)].
> The (symmetrical) perspector is then ..
>
> (b + c) (Rb + Rc) sb sc ((b + c) S^2 + 2 sa ((b + c) Ra sb sc + (b - c) s (Rb sb - Rc sc))) ::
>
> SEARCH = 1.51056102112873398780....which is not in ETC, nor on any of its lines.
>
> Various Ra type relations
> a = Ra (Rb^2 + Rc^2) / (2 Rb Rc)
> Ra^2 Rb^2 + Rb^2 Rc^2 + Rc^2 Ra^2 = 4 S^2
> rA = radius A - excircle = Ra S / (2 Rb Rc)
> Ra Rb Rc = 4 r S
> Rb Rc / Ra = 2 sa
> Rb^2 + Rc^2 = 4 a sa
> Ra^3 = 4 Ra sb sc
> Ra^5 = 16 Ra sb^2 sc^2
> .....
>
> Best regards,
> Peter.
>
>
> ----- Original Message -----
> From: Angel
> To: Hyacinthos@yahoogroups.com
> Sent: Monday, August 30, 2010 3:14 PM
> Subject: [EMHL] Re: 7 circles theorem
>
>
>
> Dear Antreas and Francisco
>
> F'a and F''a are the points of intersection of the polar of the point D= IbIc /\ BC, with respect to the nine-points circle. Similarly, F'b, F''b, F'c, F''c.
>
> F'a={(b^2-c^2)^2*Sqrt[(s-b)(s-c)],
> a*b*c(a^2+b*c-c^2)-(a^2b^2+2c^2SC)*Sqrt[(s-b)(s-c)],
> a*b*c(a^2-b^2+b*c)-(2b^2SB+a^2c^2)*Sqrt[(s-b)(s-c)]}
>
> F''a= {-(b^2-c^2)^2*Sqrt[(s-b)(s-c)],
> a*b*c(a^2+b*c-c^2)+(a^2b^2+2c^2SC)*Sqrt[(s-b)(s-c)],
> a*b*c(a^2-b^2+b*c)+(2b^2SB+a^2c^2)*Sqrt[(s-b)(s-c)]}
>
> Fa the ex-Feuerbach points:
> Fa = {(b-c)^2s, -(a+c)^2(s-c), -(a+b)^2(s-b)}
>
> Perspector of FaFbFc and F'aF'bF'c (NO triangle center):
>
> X= {(b^2-c^2)
> (-4(s-c)*c^2(a^2+b^2+a*b+a*c+b*c)*Sqrt[(s-a)(s-b)]
> -4(s-c)*s*Sqrt[(s-a)(s-c)](a*b^2-b^2c+a*c^2+c^3)
> +2(s-c)s(a^3b-a*b^3+a^3c+b^3c+2a*b*c^2-a*c^3-b*c^3)
> +Sqrt[(s-b)(s-c)](a^4b+2a^3b^2+a^2b^3+a^4c+2a^3b*c+
> 3a^2b^2c+2a*b^3c+2a^2b*c^2-b^3c^2-2a^2c^3-
> 2a*b*c^3-b^2c^3+b*c^4+c^5)),
> (a^2-c^2)
> (-4(s-c)c^2(a^2+a*b+b^2+a*c+b*c)Sqrt[(s-a)(s-b)]
> -4(s-c)s*Sqrt[(s-b)(s-c)](a^2b-a^2c+b*c^2+c^3)
> -2(s-c)s(a^3b-a*b^3-a^3c-b^3c-2a*b*c^2+a*c^3+b*c^3)+
> Sqrt[(s-a)(s-c)](a^3b^2+2a^2b^3+a*b^4+2a^3b*c+
> 3a^2b^2c+2a*b^3c+b^4c-a^3c^2+2a*b^2c^2-a^2c^3-
> 2a*b*c^3-2b^2c^3+a*c^4+c^5)),
> -(a+b)
> (4(s-c)c^2(a^3+b^3+a^2c+b^2c)Sqrt[(s-a)(s-b)]
> +(s-c)(a^5b-2a^3b^3+a*b^5-a^5c+a^4b*c+a*b^4c-
> b^5c-a^4c^2-a^3b*c^2+4a^2b^2c^2-a*b^3c^2-b^4c^2+a^3c^3+a^2b*c^3+a*b^2c^3+b^3c^3+
> a^2c^4+2a*b*c^4+b^2c^4)+
> Sqrt[(s-b)(s-c)](a^5b+a^4b^2-a^3b^3-a^2b^4-
> a^5c-a^4b*c+a^3b^2c-a^2b^3c-2a*b^4c-2a^2b^2c^2-
> a*b^3c^2+b^4c^2+2a^3c^3-2a^2b*c^3+a*b^2c^3+
> b^3c^3-a*b*c^4-b^2c^4-a*c^5-b*c^5)
> -Sqrt[(s-a)(s-c)](a^4b^2+a^3b^3-a^2b^4-a*b^5+
> 2a^4b*c+a^3b^2c-a^2b^3c+a*b^4c+b^5c-a^4c^2+
> a^3b*c^2+2a^2b^2c^2-a^3c^3-a^2b*c^3+2a*b^2c^3-
> 2b^3c^3+a^2c^4+a*b*c^4+a*c^5+b*c^5))}
>
> Perspector of F'aF'bF'c and F''aF''bF''c (triangle center):
>
> Y=((b+c)(a^3(b^2+c^2)-a(b-c)^2(b^2+b*c+c^2)-b*c(b-c)^2(b+c)): ... : ... ).
>
> Angel
>
> --- In Hyacinthos@yahoogroups.com, "Antreas" <anopolis72@> wrote:
> >
> > Let ABC be a triangle and Fa,Fb,Fc the three ex-Feuerbach points.
> >
> > Let (Ka) be the circle touching externally the Nine point circle
> > (N) and the excircles (Ib) and (Ic), and F'a the point of contact of
> > (N) and (Ka). Similarly the points F'b and F'c.
> >
> > By the Seven Circles Theorem (*), the triangles FaFbFc and F'aF'bF'c
> > are perspective.
> >
> > Which is the perspector?
> >
> > (*) See:
> > http://mathworld.wolfram.com/SevenCirclesTheorem.html
> >
> > http://www.cut-the-knot.org/Curriculum/Geometry/SevenCirclesTheorem.shtml
> >
> > APH
> >
>
>
>
>
>
> [Non-text portions of this message have been removed]
>

[Non-text portions of this message have been removed]
• ... [snip] ... I now have an update to that post. At the time I didn t know how to handle certain points (such as the circular points at infinity) where the
Message 4 of 9 , May 27
View Source
--- In Hyacinthos@yahoogroups.com, Barry Wolk <wolkbarry@...> wrote:
>
> --- On Sat, 8/28/10, Francisco Javier <garciacapitan@...> wrote:
> > The perspector is not simple.
> >
> > In its coordinates, shown below, we can see the symbols Ra,
> > Rb, Rc, that stand for the square root of 4(s-b)(s-c),
> > 4(s-c)(s-a), 4(s-a)(s-b), respectively.
> >
> > It is remarkable that these coordinates appear to be non
> > symmetric, as we can think from the problem. I think that
> > the formula can be transformed in some way into a symmetric form.
> >
> > I've checked that this point is on lines FaF'a, FbF'b, FcF'c.
> >
> > Coordinates of the perspector:
>
> [long formula shipped]
>
> If a point P is defined geometrically in a symmetric way (usually as a perspector), then there is a general method of getting symmetric coordinates for P. If x,y,z are complicated functions of a,b,c, and a symmetrically defined P has non-symmetric coordinates
> P = (x(a,b,c),y(a,b,c),z(a,b,c) ) = (x1,y1,z1), then also
> P = (z(b,c,a),x(b,c,a),y(b,c,a) ) = (z2,x2,y2), and
> P = (y(c,a,b)),z(c,a,b),x(c,a,b) ) = (y3,z3,x3). Then symmetric coordinates for P are (x1+y3+z2,x2+y1+z3,x3+y2+z1).
> --
> Barry Wolk

> Note: The symmetric coordinates for Q are obtained using the method described by Barry Wolk in the post # 19239 of Hyacinthos.
> (http://tech.groups.yahoo.com/group/Hyacinthos/message/19239)
[snip]
> Angel Montesdeoca

I now have an update to that post. At the time I didn't know how to handle certain points (such as the circular points at infinity) where the above method gave the meaningless result (0:0:0).

For such points P, apply the symmetrization method to the isotomic conjugate of P. Symmetric coordinates for that conjugate lead immediately to symmetric coordimates for P.

This idea gave the following result: A cyclically symmetric parametrization of the circumcircle is:

P(t) = ( 1/(S_A*S_A - S_B*S_C + t(S_B-S_C)) :
1/(S_B*S_B - S_C*S_A + t(S_C-S_A)) :
1/(S_C*S_C - S_A*S_B + t(S_A-S_B)) )

This just has to be useful for some calculations. The circular points at infinity are at t = +i S and t = -i S.
--
Barry Wolk
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