 Hyacinthos

 Restricted Group,
 3 members
Foci of steiner inelipse
Expand Messages
 0 Attachment
Dear friends,
by using of barycentrics I have proved the following
Theorem.
The unique points that are the Lemoine points
of their pedal triangles are the foci of Steiner inellipse of triangle.
Remark 1.
In case of a>b>c we have the above points Q1 and Q2
Q1(1+(b^2c^2)/E1:1+(a^2c^2)/E2:1(a^2b^2)/E3),
Q2(1(b^2c^2)/E1:1(a^2c^2)/E2:1+(a^2b^2)/E3),
where
E1=sqrt[(a^2+m)^2(b*c)^2],
E2=sqrt[(b^2+m)^2(c*a)^2],
E3=sqrt[(c^2+m)^2(a*b)^2]
and
m^2=a^4+b^4+c^4[(a*b)^2+(b*c)^2+(c*a)^2].
Remark 2.
As we know the Lemoine point is centroid of its pedal triangle,
and centroid of triangle is the center of homothety for
triangle ABC and its medial triangle, and also for excribed
and for inscribed Steiner ellipses of triangle ABC, and between
focii of these ellipses( i.e. between Bickhart points and these
pointsQ1 and Q2) we can see that centroid, Lemoine point,
pedal triangles and these two ellipses are interconnected.
I feel that there exist more undiscoverd relations here.
Best regards
Milorad R.Stevanovic
[Nontext portions of this message have been removed] 0 Attachment
Dear Milorad> by using of barycentrics I have proved the following
triangle.
>
> Theorem.
> The unique points that are the Lemoine points
> of their pedal triangles are the foci of Steiner inellipse of
>
This is an immediate consequence of the three following remarks :
> Remark 1.
> In case of a>b>c we have the above points Q1 and Q2
> Q1(1+(b^2c^2)/E1:1+(a^2c^2)/E2:1(a^2b^2)/E3),
> Q2(1(b^2c^2)/E1:1(a^2c^2)/E2:1+(a^2b^2)/E3),
> where
> E1=sqrt[(a^2+m)^2(b*c)^2],
> E2=sqrt[(b^2+m)^2(c*a)^2],
> E3=sqrt[(c^2+m)^2(a*b)^2]
> and
> m^2=a^4+b^4+c^4[(a*b)^2+(b*c)^2+(c*a)^2].
>
> Remark 2.
> As we know the Lemoine point is centroid of its pedal triangle,
> and centroid of triangle is the center of homothety for
> triangle ABC and its medial triangle, and also for excribed
> and for inscribed Steiner ellipses of triangle ABC, and between
> focii of these ellipses( i.e. between Bickhart points and these
> pointsQ1 and Q2) we can see that centroid, Lemoine point,
> pedal triangles and these two ellipses are interconnected.
> I feel that there exist more undiscoverd relations here.
1) If f is the affine map A,B,C > vertices of the pedal triangle of
P, then f(P*) = P where P* = isogonal conjugate of P
2) The sidelines of the pedal triangle of P are in ratio
a.AP : b.BP : c.CP
3) P is a focus of the Steiner inellipse <=> P is barycentric
1/AP^2 : 1/BP^2 : 1/CP^2
Now,
P is the Lemoine point of his pedal triangle <=>
P = f[(a.AP)^2 : (b.BP)^2 : (c.CP)^2]) <=>
P* = (a.AP)^2 : (b.BP)^2 : (c.CP)^2 <=> P = 1/AP^2 : 1/BP^2 : 1/CP^2
Friendly.JeanPierre 0 Attachment
Dear Milorad and JeanPierre!
Arsenij Akopjan proved another property of these foci. Let a, b, c be the complex numbers corresponding to the vertex of given triangle. Consider the polynom P(z)=(za)(zb)(zc). Then P'(z)=3(zf_1)(zf_2), where f_1, f_2 correspond to the foci of Steiner inellipse. The proof is in our book published (in Russian) by Moscow center of mathematical education. Also it will be published in English by AMS.
Sincerely Alexey
Antivirus scanning: Symantec Mail Security for SMTP.
[Nontext portions of this message have been removed] 0 Attachment
Dear Alexei
> Arsenij Akopjan proved another property of these foci. Let a, b, c be
the complex numbers corresponding to the vertex of given triangle.
Consider the polynom P(z)=(za)(zb)(zc). Then P'(z)=3(zf_1)(zf_2),
where f_1, f_2 correspond to the foci of Steiner inellipse. The proof
is in our book published (in Russian) by Moscow center of mathematical
education. Also it will be published in English by AMS.
If M as affix z
1) P'(z)=0 <=> 1/(za)+1/(zb)+1/(zc) = 0
<=> (za)/za^2 + (zb)/zb^2 +(zc)/zc^2 = 0
<=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
2) As the roots of P'(z) = 0 give a pair of isogonal conjugate points
with midpoint G, they are the affixes of the focii of the Steiner
inellipse
I think that 1) and 2) are old and classical results.
Have you any reference?
Friendly. JeanPierre 0 Attachment
Dear JeanPierre,
> If M as affix z
Is there something similar for the two other imaginary foci ?
> 1) P'(z)=0 <=> 1/(za)+1/(zb)+1/(zc) = 0
> <=> (za)/za^2 + (zb)/zb^2 +(zc)/zc^2 = 0
> <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
> 2) As the roots of P'(z) = 0 give a pair of isogonal conjugate points
> with midpoint G, they are the affixes of the focii of the Steiner
> inellipse
Best regards
Bernard
[Nontext portions of this message have been removed] 0 Attachment
Dear Bernard
> > If M as affix z
points
> > 1) P'(z)=0 <=> 1/(za)+1/(zb)+1/(zc) = 0
> > <=> (za)/za^2 + (zb)/zb^2 +(zc)/zc^2 = 0
> > <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
> > 2) As the roots of P'(z) = 0 give a pair of isogonal conjugate
> > with midpoint G, they are the affixes of the focii of the Steiner
The problem is that we cannot represent an imaginary point of the plane
> > inellipse
>
>
> Is there something similar for the two other imaginary foci ?
with a complex number; we must use 2 complex numbers or a quaternion or
something like that.
As any one an idea?
JeanPierre 0 Attachment
Alexey.A.Zaslavsky wrote:> Dear Milorad and JeanPierre!
> Arsenij Akopjan proved another property of these foci. Let a, b, c be
> the complex numbers corresponding to the vertex of given triangle.
> Consider the polynom P(z)=(za)(zb)(zc). Then
> P'(z)=3(zf_1)(zf_2), where f_1, f_2 correspond to the foci of
> Steiner inellipse. The proof is in our book published (in Russian) by
> Moscow center of mathematical education. Also it will be published in
> English by AMS.
>
> Sincerely Alexey
On 8/11/2000 Jim Parish wrote:
> Ignacio Larrosa Cañestro asked:
>> A theorem of complex variable (Lucas' theorem?) say that if {a_1,
>> ... ,a_n}are the roots of a polinomal P(z) of degree n, then the n1
>> roots {r_1, ...,r_(n1)} of P'(z) are inside of the convex capsule
>> of {a_1, ...,a_n}.
>>
>> The configuration {a_1, ..., a_n, r_1, ..., r_(n1)} is invariant
>> under isometries or dilatations (similitudes?).
>>
>> If n=3, the points r_1, r_2 are
>>
>> {r_1, r_2} = {(a+b+c)/3 +/ sqrt((a^2+b^2+c^2)/6 
>> ((a+b+c)/3)^2/2)}
>>
>> where {a, b, c} are the complex numbers correspond to the vertices.
>>
>> Are there {r_1, r_2} known points of the triangle?
>
> Yes; they are the foci of the Steiner inellipse.
> Measure everything from the centroid; then a+b+c=0. Let z, w be
> such that a=z+w, b=oz+oow, c=ooz+ow, where o is a primitive
> cube root of unity. Then r_1 and r_2 work out to be the square
> roots of zw.
> The midpoint G1 of the side BC is (b+c)/2 = 1/2(z+w). Therefore
> G1r_1=1/2(Sqrt[z]Sqrt[w])^2 and G1r_2=1/2(Sqrt[z]+Sqrt[w])^2.
> An easy computation then shows that the sum dist(r_1,G1)+dist(r_2,G1)
> is z+w, and the same result obtains for the other midpoints G2,
> G3. Thus r_1, r_2 are the foci of the Steiner inellipse.
>
> Jim Parish
>
Best regards,
Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa@... 0 Attachment
I've proven a generalization of this result to any polygon which is affine
equivalent to a regular polygon:
http://forumgeom.fau.edu/FG2006volume6/FG200633index.html
Jim Parish
Ignacio Larrosa Cañestro wrote:> Alexey.A.Zaslavsky wrote:

> > Dear Milorad and JeanPierre!
> > Arsenij Akopjan proved another property of these foci. Let a, b, c be
> > the complex numbers corresponding to the vertex of given triangle.
> > Consider the polynom P(z)=(za)(zb)(zc). Then
> > P'(z)=3(zf_1)(zf_2), where f_1, f_2 correspond to the foci of
> > Steiner inellipse. The proof is in our book published (in Russian) by
> > Moscow center of mathematical education. Also it will be published in
> > English by AMS.
> >
> > Sincerely Alexey
>
> On 8/11/2000 Jim Parish wrote:
> > Ignacio Larrosa Cañestro asked:
> >> A theorem of complex variable (Lucas' theorem?) say that if {a_1,
> >> ... ,a_n}are the roots of a polinomal P(z) of degree n, then the n1
> >> roots {r_1, ...,r_(n1)} of P'(z) are inside of the convex capsule
> >> of {a_1, ...,a_n}.
> >>
> >> The configuration {a_1, ..., a_n, r_1, ..., r_(n1)} is invariant
> >> under isometries or dilatations (similitudes?).
> >>
> >> If n=3, the points r_1, r_2 are
> >>
> >> {r_1, r_2} = {(a+b+c)/3 +/ sqrt((a^2+b^2+c^2)/6 
> >> ((a+b+c)/3)^2/2)}
> >>
> >> where {a, b, c} are the complex numbers correspond to the vertices.
> >>
> >> Are there {r_1, r_2} known points of the triangle?
> >
> > Yes; they are the foci of the Steiner inellipse.
> > Measure everything from the centroid; then a+b+c=0. Let z, w be
> > such that a=z+w, b=oz+oow, c=ooz+ow, where o is a primitive
> > cube root of unity. Then r_1 and r_2 work out to be the square
> > roots of zw.
> > The midpoint G1 of the side BC is (b+c)/2 = 1/2(z+w). Therefore
> > G1r_1=1/2(Sqrt[z]Sqrt[w])^2 and G1r_2=1/2(Sqrt[z]+Sqrt[w])^2.
> > An easy computation then shows that the sum dist(r_1,G1)+dist(r_2,G1)
> > is z+w, and the same result obtains for the other midpoints G2,
> > G3. Thus r_1, r_2 are the foci of the Steiner inellipse.
> >
> > Jim Parish
> >
>
> Best regards,
>
> Ignacio Larrosa Cañestro
> A Coruña (España)
> ilarrosa@...
>
SIUE Web Mail 0 Attachment
Dear Friends,
thanks to JeanPierre's nice proof,
I've learned why the barycentrics of the focii
of Steiner inellipse are 1/AM^2 : 1/BM^2 : 1/CM^2.
For a generalization to any inconic in ABC
(the conic that is tangent to the sides of ABC
at the vertices of the cevian triangle A'B'C'
of P=(x : y : z) in barycentrics)
let p,a',b',c' be the affixes of the perspector P
and the tangency points A',B',C' then
b'=(xa + zc)/(x + z)
c'=(xa + yb)/(x + y)
Let f be the affix of a focus F of the inconic.
From Poncelet's theorem that
a tangent to a conic segment XY
between two other tangents, is seen from F with a
constant angle XFY, we get that
angle(AFC')  angle(B'FA) = 0 (mod pi) or
arg((c'f)/(af))  arg((af)/(b'f)) = 0 (mod pi) or
the number Q = ((c'f)/(af))/((af)/(b'f)) is real.
Substituting b', c' we get that the number
[xy(fa)(fb)+yz(fb)(fc)+zx(fc)(fa)]/(af)^2
must be real. Similarly the numbers
[xy(fa)(fb)+yz(fb)(fc)+zx(fc)(fa)]/(bf)^2
[xy(fa)(fb)+yz(fb)(fc)+zx(fc)(fa)]/(cf)^2
must be real.
Hence
xy(fa)(fb)+yz(fb)(fc)+zx(fc)(fa) = 0 (1)
because otherwise two of the numbers
(af)^2, (bf)^2, (cf)^2 would have the same
argument or two of the numbers
af, bf, cf would have arguments that differ by pi
or that F lies on one side of ABC that is impossible.
From (1) we get that f is a root of the equation
Kff  Lf + N = 0 where
K = xy + yz + zx
L = x(y+z)a + y(z+x)b + z(x+y)c
N = xyab + yzbc + zxca
If P is on the Steiner circumellipse K = xy+yz+zx = 0
then we have only one focus and the inconic is
a parabola. If K <> 0 the equation gives two roots
f, f' hence two focii F, F' the projections of which
on the tangent sides of ABC are lying on the same
circle and hence F, F' are isogonal conjugate.
The center O of the inconic
that is the mid point of the focii has affix
L/2K = [(b+c)/x+(c+a)/y+(a+b)/z]/2(1/x+1/y+1/z)
that is the affix of the complement of the
isotomic conjugate of P.
From (1) we get also that
(fa)/x(fa)^2)+(fb)/y(fb)^2+(fc)/z(fc)^2 = 0
or
(fa)/x.AF^2+(fb)/y.BF^2+(fc)/z.CF^2 = 0
or that F has barycentrics
1/x.AF^2 : 1/y.BF^2 : 1/z.CF^2.
I don't like my proof. Is there a simpler?
Best regards
Nikos Dergiades
[JPE]> Let a, b, c be
___________________________________________________________
> the complex numbers corresponding to the vertex of
> given triangle.
> Consider the polynom P(z)=(za)(zb)(zc). Then
> P'(z)=3(zf_1)(zf_2),
> where f_1, f_2 correspond to the foci of Steiner
> inellipse. The proof
> is in our book published (in Russian) by Moscow
> center of mathematical
> education. Also it will be published in English by
> AMS.
>
> If M as affix z
> 1) P'(z)=0 <=> 1/(za)+1/(zb)+1/(zc) = 0
> <=> (za)/za^2 + (zb)/zb^2 +(zc)/zc^2 = 0
> <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
> 2) As the roots of P'(z) = 0 give a pair of isogonal
> conjugate points
> with midpoint G, they are the affixes of the focii
> of the Steiner
> inellipse
> I think that 1) and 2) are old and classical
> results.
> Have you any reference?
> Friendly. JeanPierre
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr 0 Attachment
Oops. I had to go to bed.
I misunderstood why ?
P'(z)=0 <=> 1/(za)+1/(zb)+1/(zc) = 0
<=> (za)/za^2 + (zb)/zb^2 +(zc)/zc^2 = 0
So I repeat my previous message
in the part I think I understand it.
Best regards
Nikos Dergiades
> Dear Alexei
___________________________________________________________
>
> > Arsenij Akopjan proved another property of these
> foci. Let a, b, c be
> the complex numbers corresponding to the vertex of
> given triangle.
> Consider the polynom P(z)=(za)(zb)(zc). Then
> P'(z)=3(zf_1)(zf_2),
> where f_1, f_2 correspond to the foci of Steiner
> inellipse. The proof
> is in our book published (in Russian) by Moscow
> center of mathematical
> education. Also it will be published in English by
> AMS.
>
> If M as affix z
> 1) P'(z)=0 <=> 1/(za)+1/(zb)+1/(zc) = 0
> <=> (za)/za^2 + (zb)/zb^2 +(zc)/zc^2 = 0
> <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
> 2) As the roots of P'(z) = 0 give a pair of isogonal
> conjugate points
> with midpoint G, they are the affixes of the focii
> of the Steiner
> inellipse
> I think that 1) and 2) are old and classical
> results.
> Have you any reference?
> Friendly. JeanPierre
>
>
>
>
> Yahoo! Groups Links
>
>
>
>
>
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr 0 Attachment
Dear Friends,
For a generalization to any inconic in ABC
(the conic that is tangent to the sides of ABC
at the vertices of the cevian triangle A'B'C'
of P=(x : y : z) in barycentrics)
let p,a',b',c' be the affixes of the perspector P
and the tangency points A',B',C' then
b'=(xa + zc)/(x + z)
c'=(xa + yb)/(x + y)
Let f be the affix of a focus F of the inconic.
From Poncelet's theorem that
a tangent to a conic segment XY
between two other tangents, is seen from F with a
constant angle XFY, we get that
angle(AFC')  angle(B'FA) = 0 (mod pi) or
arg((c'f)/(af))  arg((af)/(b'f)) = 0 (mod pi) or
the number Q = ((c'f)/(af))/((af)/(b'f)) is real.
Substituting b', c' we get that the number
[xy(fa)(fb)+yz(fb)(fc)+zx(fc)(fa)]/(af)^2
must be real. Similarly the numbers
[xy(fa)(fb)+yz(fb)(fc)+zx(fc)(fa)]/(bf)^2
[xy(fa)(fb)+yz(fb)(fc)+zx(fc)(fa)]/(cf)^2
must be real.
Hence
xy(fa)(fb)+yz(fb)(fc)+zx(fc)(fa) = 0 (1)
because otherwise two of the numbers
(af)^2, (bf)^2, (cf)^2 would have the same
argument or two of the numbers
af, bf, cf would have arguments that differ by pi
or that F lies on one side of ABC that is impossible.
From (1) we get that f is a root of the equation
Kff  Lf + N = 0 where
K = xy + yz + zx
L = x(y+z)a + y(z+x)b + z(x+y)c
N = xyab + yzbc + zxca
If P is on the Steiner circumellipse K = xy+yz+zx = 0
then we have only one focus and the inconic is
a parabola. If K <> 0 the equation gives two roots
f, f' hence two focii F, F' the projections of which
on the tangent sides of ABC are lying on the same
circle and hence F, F' are isogonal conjugate.
The center O of the inconic
that is the mid point of the focii has affix
L/2K = [(b+c)/x+(c+a)/y+(a+b)/z]/2(1/x+1/y+1/z)
that is the affix of the complement of the
isotomic conjugate of P.
I don't like my proof. Is there a simpler?
Best regards
Nikos Dergiades
[JPE]> Let a, b, c be
___________________________________________________________
> the complex numbers corresponding to the vertex of
> given triangle.
> Consider the polynom P(z)=(za)(zb)(zc). Then
> P'(z)=3(zf_1)(zf_2),
> where f_1, f_2 correspond to the foci of Steiner
> inellipse. The proof
> is in our book published (in Russian) by Moscow
> center of mathematical
> education. Also it will be published in English by
> AMS.
>
> If M as affix z
> 1) P'(z)=0 <=> 1/(za)+1/(zb)+1/(zc) = 0
> <=> (za)/za^2 + (zb)/zb^2 +(zc)/zc^2 = 0
> <=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
> 2) As the roots of P'(z) = 0 give a pair of isogonal
> conjugate points
> with midpoint G, they are the affixes of the focii
> of the Steiner
> inellipse
> I think that 1) and 2) are old and classical
> results.
> Have you any reference?
> Friendly. JeanPierre
Χρησιμοποιείτε Yahoo!;
Βαρεθήκατε τα ενοχλητικά μηνύματα (spam); Το Yahoo! Mail
διαθέτει την καλύτερη δυνατή προστασία κατά των ενοχλητικών
μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr 0 Attachment
Dear JeanPierre!
I don't know first reference, may be Arsenij can help you. His mail: akopjan@.... But it is interesting that the result of Milorad immediately follows from your assertion 1)
Sincerely Alexey
Dear Alexei
> Arsenij Akopjan proved another property of these foci. Let a, b, c be
the complex numbers corresponding to the vertex of given triangle.
Consider the polynom P(z)=(za)(zb)(zc). Then P'(z)=3(zf_1)(zf_2),
where f_1, f_2 correspond to the foci of Steiner inellipse. The proof
is in our book published (in Russian) by Moscow center of mathematical
education. Also it will be published in English by AMS.
If M as affix z
1) P'(z)=0 <=> 1/(za)+1/(zb)+1/(zc) = 0
<=> (za)/za^2 + (zb)/zb^2 +(zc)/zc^2 = 0
<=> M has barycentrics 1/AM^2 : 1/BM^2 : 1/CM^2
2) As the roots of P'(z) = 0 give a pair of isogonal conjugate points
with midpoint G, they are the affixes of the focii of the Steiner
inellipse
I think that 1) and 2) are old and classical results.
Have you any reference?
Friendly. JeanPierre
Antivirus scanning: Symantec Mail Security for SMTP.
[Nontext portions of this message have been removed] 0 Attachment
Dear JeanPierre,
and all other friends
1.
I have considered the problem
to find the points P(x:y:z) with property
x(AP^2)=y(BP^2)=z(CP^2).
The coordinate solution of this problem I have
found about 15 years ago.
2.
I have considered the problem
to find the points that are Lemoine points
of their pedal triangles.
I have found the coordinates of solutions of
this problem not long ago..
3.
I have worked on a known problem of equal cevian
points, ie of the points P with propertiy
AA1=BB1=CC1
where A1,B1,C1 are the points of intersections
of AP,BP and CP with sides of triangle ABC.
It has been discovered that the solutions
of the problem 3 were the focii of Steiner exellipse.
When I connected the sollutions of problems
2 and 3 I have found that the solutions of problem
2 are the focii of Steiner inellipse.
When I connected the solutions of problems 1.
and new solution of problem 2 I have found that
the solutions of problem 1 are the focii of Steiner
inellipse.
That I had for some other message.
It is interesting to see that by various roads,
in various times,guided by the same or different idea,
we came to the same place.
Friendly
All the best
Milorad R.Stevanovic
[Nontext portions of this message have been removed] 0 Attachment
Dear JeanPierre,
> The problem is that we cannot represent an imaginary point of the
I've wondered about using 'complicated numbers' like those defined in
> plane with a complex number; we must use 2 complex numbers or a
> quaternion or something like that.
> As any one an idea?
[1] below.
The author writes in the introduction "At first sight it may therefore
seem paradoxical that in the present paper a Cauchy formula will be
established for an algebra that has almost the inevitability of
quaternions and Clifford matrices, but which is commutative and has zero
divisors."
I'm not even close to presenting an argument for the use of the
complicated numbers. But, please consider this as an idea nevertheless.
[1] I.J. Good, "A simple generalization of analytic function theory",
Expositiones Mathematicae, Expo. Math 6 (1988), 289311, Published by
Bibliographisches Institut & F.A. Brockhaus AG 1988. 0 Attachment
Dear JeanPierre and all friends
As I said in earlier message,
I have considered the problem
to find the points P(x:y:z)(given
in barycentrics), with property
x(AP^2)=y(BP^2)=z(CP^2).
I have found in my papers, the following solutions
(under the condition a>b>c)
1.Q1(x1:y1:z1)
x1=c^2+msqrt(D3),
y1=b^2,
z1=a^2+msqrt(D1).
2.Q2(x2:y2:z2)
z2=c^2+m+sqrt(D3),
y2=b^2,
z2=a^2+m+sqrt(D1)
where
m^2=a^4+b^4+c^4(ab)^2(bc)^2(ca)^2,(m>0),
D1=(a^2+m)^2(bc)^2,
D3=(c^2+m)^2(ab)^2.
These two points are real and are focii of Steiner inellipse.
There are also two complex points
3.Q3(x3:y3:z3)
x3=c^2m+i sqrt(E3),
y3=b^2,
z3=a^2m+i sqrt(E1).
4.Q4(x4:y4:z4)
x4=c^2mi sqrt(E3),
y4=b^2,
z4=a^2mi sqrt(E1)
where
m^2=a^4+b^4+c^4(ab)^2(bc)^2(ca)^2,(m>0),
E1=(bc)^2(a^2m)^2,
E3=(ab)^2(c^2m)^2..
The points Q3 and Q4 are collinear
with centroid G of triangle ABC.
How these points Q3 and Q4 are related
to the focii of Steiner inellipse?
What is the geometric meaning of these points Q3 and Q4?
Friendly
Yours
Milorad R.Stevanovic
[Nontext portions of this message have been removed]
Your message has been successfully submitted and would be delivered to recipients shortly.