## Re: Feuerbach and Miquel

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• Yes, in this case we have something: all lines m(d) pass through X(56) = exsimilicenter of (O) and (I). Francisco Javier.
Message 1 of 9 , Mar 10, 2009
Yes, in this case we have something: all lines m(d) pass through X(56) = exsimilicenter of (O) and (I).

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:
>
>
> [APH]
> > > > > Let ABC be a triangle, F the Feuerbach point, F* the antipodal
> > > > > of F on the incircle, and f,f* the tangents to incircle at > > > > > F,F*, respectively.
> > > > >
> > > > > Which points (on the circumcircle of ABC) are the Miquel > > > > > > points of the complete quadrilaterals (AB,BC,CA, f) and
> > > > > (AB,BC,CA,f*)?
>
> [FJ]
> > > > These are X(106) band X(109), respectively.
>
> [APH]
> > > Let's call this line (ie X(106)X(109)) as Feuerbach-Miquel
> > > line of the incircle (I).
> > > I am wondering if the similar Feuerbach - Miquel lines
> > > of the Excircles (Ia),(Ib),(Ic) have some interesting properties.
> > > (For exampe: Are they concurrent? Bound a triangle
> > > in perspective with ABC or some other simple triangle?)
>
> [FJ]
> > Dear Antreas, these three lines are not concurrent, nor bound
> > a triangle perspective with ABC.
>
> Dear Francisco
>
> Hmmmmm... It seems 3G [= Great God of Geometry] did not make
> everything perfect in the Euclidean Space :-)
>
> I hope that the following was at least made correctly :-)
>
> Let D be a point on the Incircle, D* its antipodal point, and
> d,d* the tangents to Incircle at D,D*, resp.
>
> Let m(d) be the line joining the Miquel points (on the
> circumcircle) of the complete quadrilaterals (AB,BC,CA, d)
> and (AB,BC,CA, d*).
>
> As D moves on the Incircle, which is the envelope of the line m(d)?
> Is it a zero circle? (ie are they concurrent? If yes, at which
> point?)
>
> APH
>
• [APH] ... [FJ] ... Dear Francisco I think that the general is true: Let p be a line (trilinear polar of a point P = (x:y:z)) and Q a point on the line. Let
Message 2 of 9 , Mar 11, 2009
[APH]
> > Let D be a point on the Incircle, D* its antipodal point, and
> > d,d* the tangents to Incircle at D,D*, resp.
> >
> > Let m(d) be the line joining the Miquel points (on the
> > circumcircle) of the complete quadrilaterals (AB,BC,CA, d)
> > and (AB,BC,CA, d*).
> >
> > As D moves on the Incircle, which is the envelope of the
> > line m(d)?
> > Is it a zero circle? (ie are they concurrent? If yes, at which
> > point?)

[FJ]
> Yes, in this case we have something: all lines m(d) pass through
> X(56) = exsimilicenter of (O) and (I).

Dear Francisco

I think that the general is true:

Let p be a line (trilinear polar of a point P = (x:y:z)) and Q
a point on the line.
Let q,q* be the tangents to incircle from Q, and m(q) the
line joining the Miquel points of the complete
4laterals (AB,BC,CA,q) and (AB,BC,CA, q*).
As Q moves on the line p, the lines m(q) concur at some point.
[In the case of the tangents at antipodal points of the incircle,
the line p is the Linf]

Another problem is:
If Q moves on some circle (instead of line), which is the envelope
of m(q)? Special Case: Q moves on the circumcircle.

APH
• Dear Antreas: here we have the following mapping: If Q=(u:v:w), then the line m(q) is (s-a)((s-a)u - a (v+w)) x / a^2 + (s-b)((s-b)v - b (w+u)) y / b^2 +
Message 3 of 9 , Mar 11, 2009
Dear Antreas: here we have the following mapping:

If Q=(u:v:w), then the line m(q) is

(s-a)((s-a)u - a (v+w)) x / a^2
+ (s-b)((s-b)v - b (w+u)) y / b^2
+ (s-c)((s-c)w - c (u+v)) z / c^2 = 0

Hence, if Q is on the line px + qy + rz = 0 then all lines m(q) pass through the point

(a^2((s - a) p - b q - c r)/(s - a) :
b^2((s - b) q - c r - a p)/(s - b) :
c^2((s - c) r - a p - b q)/(s - c)).

--- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@...> wrote:
>
> [APH]
> > > Let D be a point on the Incircle, D* its antipodal point, and
> > > d,d* the tangents to Incircle at D,D*, resp.
> > >
> > > Let m(d) be the line joining the Miquel points (on the
> > > circumcircle) of the complete quadrilaterals (AB,BC,CA, d)
> > > and (AB,BC,CA, d*).
> > >
> > > As D moves on the Incircle, which is the envelope of the
> > > line m(d)?
> > > Is it a zero circle? (ie are they concurrent? If yes, at which
> > > point?)
>
> [FJ]
> > Yes, in this case we have something: all lines m(d) pass through
> > X(56) = exsimilicenter of (O) and (I).
>
> Dear Francisco
>
> I think that the general is true:
>
> Let p be a line (trilinear polar of a point P = (x:y:z)) and Q
> a point on the line.
> Let q,q* be the tangents to incircle from Q, and m(q) the
> line joining the Miquel points of the complete
> 4laterals (AB,BC,CA,q) and (AB,BC,CA, q*).
> As Q moves on the line p, the lines m(q) concur at some point.
> [In the case of the tangents at antipodal points of the incircle,
> the line p is the Linf]
>
> Another problem is:
> If Q moves on some circle (instead of line), which is the envelope
> of m(q)? Special Case: Q moves on the circumcircle.
>
> APH
>
• And, if Q is on the circumcircle, then the line m(q) evolves an ellipse centererd on the line OI. The center of this ellipse is the point (a (a+b-c) (a-b+c) (4
Message 4 of 9 , Mar 11, 2009
And, if Q is on the circumcircle, then the line m(q) evolves an ellipse centererd on the line OI. The center of this ellipse is the point

(a (a+b-c) (a-b+c) (4 a^4-5 a^3 b-3 a^2 b^2+5 a b^3-b^4-5 a^3 c+15 a^2 b c-6 a b^2 c-3 a^2 c^2-6 a b c^2+2 b^2 c^2+5 a c^3-c^4) :

b (a+b-c) (-a+b+c) (-a^4+5 a^3 b-3 a^2 b^2-5 a b^3+4 b^4-6 a^2 b c+15 a b^2 c-5 b^3 c+2 a^2 c^2-6 a b c^2-3 b^2 c^2+5 b c^3-c^4) :

c (a-b+c) (-a+b+c) (-a^4+2 a^2 b^2-b^4+5 a^3 c-6 a^2 b c-6 a b^2 c+5 b^3 c-3 a^2 c^2+15 a b c^2-3 b^2 c^2-5 a c^3-5 b c^3+4 c^4)).

--- In Hyacinthos@yahoogroups.com, "garciacapitan" <garciacapitan@...> wrote:
>
> Dear Antreas: here we have the following mapping:
>
> If Q=(u:v:w), then the line m(q) is
>
> (s-a)((s-a)u - a (v+w)) x / a^2
> + (s-b)((s-b)v - b (w+u)) y / b^2
> + (s-c)((s-c)w - c (u+v)) z / c^2 = 0
>
> Hence, if Q is on the line px + qy + rz = 0 then all lines m(q) pass through the point
>
> (a^2((s - a) p - b q - c r)/(s - a) :
> b^2((s - b) q - c r - a p)/(s - b) :
> c^2((s - c) r - a p - b q)/(s - c)).
>
>
>
>
> --- In Hyacinthos@yahoogroups.com, "xpolakis" <xpolakis@> wrote:
> >
> > [APH]
> > > > Let D be a point on the Incircle, D* its antipodal point, and
> > > > d,d* the tangents to Incircle at D,D*, resp.
> > > >
> > > > Let m(d) be the line joining the Miquel points (on the
> > > > circumcircle) of the complete quadrilaterals (AB,BC,CA, d)
> > > > and (AB,BC,CA, d*).
> > > >
> > > > As D moves on the Incircle, which is the envelope of the
> > > > line m(d)?
> > > > Is it a zero circle? (ie are they concurrent? If yes, at which
> > > > point?)
> >
> > [FJ]
> > > Yes, in this case we have something: all lines m(d) pass through
> > > X(56) = exsimilicenter of (O) and (I).
> >
> > Dear Francisco
> >
> > I think that the general is true:
> >
> > Let p be a line (trilinear polar of a point P = (x:y:z)) and Q
> > a point on the line.
> > Let q,q* be the tangents to incircle from Q, and m(q) the
> > line joining the Miquel points of the complete
> > 4laterals (AB,BC,CA,q) and (AB,BC,CA, q*).
> > As Q moves on the line p, the lines m(q) concur at some point.
> > [In the case of the tangents at antipodal points of the incircle,
> > the line p is the Linf]
> >
> > Another problem is:
> > If Q moves on some circle (instead of line), which is the envelope
> > of m(q)? Special Case: Q moves on the circumcircle.
> >
> > APH
> >
>
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