- Dear All My Friends,

Following is my detail construction of point Ab on BC (by direction BAC)

Select X = B then X1 = X2 = B. We will get only 4 points

B1 = reflection of B in IA

B2 = symmetry of A in B1

B3 = reflection of B2 in IC

B4 = symmetry of C in B3

Lp = line perpendicular with IB at B

A1 = intersection of B2B4 with Lp

Select X = C. We will get 6 points

C1 = reflection of C in IB

C2 = symmetry of B in C1

C3 = reflection of C2 in IA

C4 = symmetry of A in C3

C5 = reflection of C4 in IC

C6 = symmetry of C in C5

A2 = intersection of CC1 with C4C6

Ab = intersection of A1A2 with BC

(Construction of Ac is similarly but by direction CAB.)

These procedures generate two points Ab, Ac on BC with simple barycentrics:

Ab = {0 : 2*(4*a + b - 2*c) : (a - 2*b + 4*c)}

Ac = {0 : (a - 2*c + 4*b) : 2*(4*a + c - 2*b)}

Best regards,

Bui Quang Tuan--- On Sun, 8/1/10, Quang Tuan Bui <bqtuan1962@...> wrote:

> From: Quang Tuan Bui <bqtuan1962@...>

> Subject: Re: [EMHL] A Circle From Reflections In Incenter Cevians And Symmetries Of Vertices

> To: Hyacinthos@yahoogroups.com

> Date: Sunday, August 1, 2010, 7:58 AM

> Dear All My Friend,

> Please note that we can easy to find synthetic proof for

> the fact that six points Ab, Ac, Bc, Ba, Ca, Cb are

> concyclic.

> After construction one point, say Ab by direction BAC, we

> start with X=Ab and construct X1, X2, X3, X4, X5, X6 for Ab.

> Construct a circumcircle (CC) of AbX2X4. Now we construct Ac

> and with X=Ac we construct X1, X2, X3, X4, X5, X6 for Ac by

> direction CAB. By reflections and symmetries, easy to show

> that Ac, X2, X4 of Ac are on (CC) too. So (CC) is our

> concyclic circle.

> Best regards,

> Bui Quang Tuan

>

> --- On Sun, 8/1/10, Quang Tuan Bui <bqtuan1962@...>

> wrote:

>

> > From: Quang Tuan Bui <bqtuan1962@...>

> > Subject: [EMHL] A Circle From Reflections In Incenter

> Cevians And Symmetries Of Vertices

> > To: Hyacinthos@yahoogroups.com

> > Cc: bqtuan1962@...

> > Date: Sunday, August 1, 2010, 5:33 AM

> > Dear All My Friends,

> >

> > Given triangle ABC with incenter I. Select X is

> variable

> > point on line BC. We do some reflections and

> symmetries as

> > following:

> > By dicrection BAC

> > X1 = reflection of X in BI

> > X2 = symmetry of B in X1

> > X3 = reflection of X2 in AI

> > X4 = symmetry of A in X3

> > X5 = reflection of X4 in CI

> > X6 = symmetry of C in X5

> >

> > There is only one X on BC such that X=X6. We name

> this

> > point as Ab.

> > Do the same procedures but with direction CAB, we get

> > another point on BC and name it as Ac

> >

> > It is easy to get barycentrics of X1, X2... X6 from

> > barycentrics of X. If we calculate intersection point

> Z of

> > two lines XX1 and X4X6 then Z is always on a line. By

> choose

> > X=B and X=C and do procedures as above, we can easy

> to

> > construct this line. Intersection of this line with BC

> is Ab

> > or Ac depend on direction we choosen. So Ab, Ac are

> easy to

> > construct by ruler and compass.

> >

> > We can do the same procedures to get another similar

> points

> > on lines CA, AB and we have total six points: Ab, Ac,

> Bc,

> > Ba, Ca, Cb.

> >

> > It is interesting that these six points are concyclic

> on

> > one circle with center T.

> > Barycentrics of T:

> > a*(3*a*(a^2 - b^2 - c^2) - 2*(b + c)*(a - b + c)*(a +

> b -

> > c) + 4*a*b*c) : :

> > Search value: +3.3876599710368

> > May be not in current ETC.

> > Some lines contain T center: X(1)X(3), X(2)X(952),

> > X(5)X(944), X(8)X(140), X(104)X(1621), X(145)X(631),

> > X(182)X(3242)...

> >

> > Denote constant:

> > K = (5*b*c - 4*SA)*(5*c*a - 4*SB)*(5*a*b - 4*SC)

> > then radius of this circle is: Sqrt(K)/(36*Area)

> > Here Area is area of triangle ABC

> >

> > If this circle is well-known before and how is it

> related

> > with other well-known circles?

> >

> > Thank you and best regards,

> > Bui Quang Tuan - Dear Quang Tuan Bui
> Given triangle ABC with incenter I. Select X is variable point on line BC. We do some reflections and symmetries as following:

We have IT = IO/3 Friendly. Jean-Pierre

> By dicrection BAC

> X1 = reflection of X in BI

> X2 = symmetry of B in X1

> X3 = reflection of X2 in AI

> X4 = symmetry of A in X3

> X5 = reflection of X4 in CI

> X6 = symmetry of C in X5

>

> There is only one X on BC such that X=X6. We name this point as Ab.

> Do the same procedures but with direction CAB, we get another point on BC and name it as Ac

>

> It is easy to get barycentrics of X1, X2... X6 from barycentrics of X. If we calculate intersection point Z of two lines XX1 and X4X6 then Z is always on a line. By choose X=B and X=C and do procedures as above, we can easy to construct this line. Intersection of this line with BC is Ab or Ac depend on direction we choosen. So Ab, Ac are easy to construct by ruler and compass.

>

> We can do the same procedures to get another similar points on lines CA, AB and we have total six points: Ab, Ac, Bc, Ba, Ca, Cb.

>

> It is interesting that these six points are concyclic on one circle with center T.

> Barycentrics of T:

> a*(3*a*(a^2 - b^2 - c^2) - 2*(b + c)*(a - b + c)*(a + b - c) + 4*a*b*c) : :

> Search value: +3.3876599710368

- Dear Jean-Pierre,

Thank you for interesting remark! It explains why a lot of central lines contain center T.

Other than my quite complicated original construction, may be there are some elegant smart constructions for these six concyclic points?

I am very happy if any one can find it!

Thank you and best regards,

Bui Quang Tuan

--- On Sun, 8/1/10, jpehrmfr <jean-pierre.ehrmann@...> wrote:

> From: jpehrmfr <jean-pierre.ehrmann@...>

> Subject: [EMHL] Re: A Circle From Reflections In Incenter Cevians And Symmetries Of Vertices

> To: Hyacinthos@yahoogroups.com

> Date: Sunday, August 1, 2010, 9:17 PM

> Dear Quang Tuan Bui

> >

> > It is interesting that these six points are concyclic

> on one circle with center T.

> > Barycentrics of T:

> > a*(3*a*(a^2 - b^2 - c^2) - 2*(b + c)*(a - b + c)*(a +

> b - c) + 4*a*b*c) : :

> > Search value: +3.3876599710368

>

> We have IT = IO/3 Friendly. Jean-Pierre

> - Dear All My Friends,

If p, q are two real numbers and Ab, Ac are two points on BC with barycentrics:

Ab={0: p*(a*p^2 + c*p*q + b*q^2) : -q*(c*p^2 + b*p*q + a*q^2) }

Ac={0: -q*(b*p^2 + c*p*q + a*q^2) : p*(a*p^2 + b*p*q + c*q^2) }

Cyclically define Bc, Ba on CA; Ca, Cb on AB then these six points are always concyclic on one circle centered on line X(1)X(3).

Our circle here is one case when p=2, q=-1.

When p=0 (or q=0), we get circumcircle.

When p=1, q=-1, we get incircle.

Best regards,

Bui Quang Tuan

--- On Sun, 8/1/10, Quang Tuan Bui <bqtuan1962@...> wrote:

> From: Quang Tuan Bui <bqtuan1962@...>

> Subject: Re: [EMHL] A Circle From Reflections In Incenter Cevians And Symmetries Of Vertices

> To: Hyacinthos@yahoogroups.com

> Date: Sunday, August 1, 2010, 11:35 AM

> Dear All My Friends,

>

> Following is my detail construction of point Ab on BC (by

> direction BAC)

>

> Select X = B then X1 = X2 = B. We will get only 4 points

> B1 = reflection of B in IA

> B2 = symmetry of A in B1

> B3 = reflection of B2 in IC

> B4 = symmetry of C in B3

> Lp = line perpendicular with IB at B

>

> A1 = intersection of B2B4 with Lp

>

> Select X = C. We will get 6 points

> C1 = reflection of C in IB

> C2 = symmetry of B in C1

> C3 = reflection of C2 in IA

> C4 = symmetry of A in C3

> C5 = reflection of C4 in IC

> C6 = symmetry of C in C5

>

> A2 = intersection of CC1 with C4C6

>

> Ab = intersection of A1A2 with BC

>

> (Construction of Ac is similarly but by direction CAB.)

>

> These procedures generate two points Ab, Ac on BC with

> simple barycentrics:

> Ab = {0 : 2*(4*a + b - 2*c) : (a - 2*b + 4*c)}

> Ac = {0 : (a - 2*c + 4*b) : 2*(4*a + c - 2*b)}

>

> Best regards,

> Bui Quang Tuan

> --- On Sun, 8/1/10, Quang Tuan Bui <bqtuan1962@...>

> wrote:

>

> > From: Quang Tuan Bui <bqtuan1962@...>

> > Subject: Re: [EMHL] A Circle From Reflections In

> Incenter Cevians And Symmetries Of Vertices

> > To: Hyacinthos@yahoogroups.com

> > Date: Sunday, August 1, 2010, 7:58 AM

> > Dear All My Friend,

> > Please note that we can easy to find synthetic proof

> for

> > the fact that six points Ab, Ac, Bc, Ba, Ca, Cb are

> > concyclic.

> > After construction one point, say Ab by direction BAC,

> we

> > start with X=Ab and construct X1, X2, X3, X4, X5, X6

> for Ab.

> > Construct a circumcircle (CC) of AbX2X4. Now we

> construct Ac

> > and with X=Ac we construct X1, X2, X3, X4, X5, X6 for

> Ac by

> > direction CAB. By reflections and symmetries, easy to

> show

> > that Ac, X2, X4 of Ac are on (CC) too. So (CC) is our

> > concyclic circle.

> > Best regards,

> > Bui Quang Tuan

> >

> > --- On Sun, 8/1/10, Quang Tuan Bui <bqtuan1962@...>

> > wrote:

> >

> > > From: Quang Tuan Bui <bqtuan1962@...>

> > > Subject: [EMHL] A Circle From Reflections In

> Incenter

> > Cevians And Symmetries Of Vertices

> > > To: Hyacinthos@yahoogroups.com

> > > Cc: bqtuan1962@...

> > > Date: Sunday, August 1, 2010, 5:33 AM

> > > Dear All My Friends,

> > >

> > > Given triangle ABC with incenter I. Select X is

> > variable

> > > point on line BC. We do some reflections and

> > symmetries as

> > > following:

> > > By dicrection BAC

> > > X1 = reflection of X in BI

> > > X2 = symmetry of B in X1

> > > X3 = reflection of X2 in AI

> > > X4 = symmetry of A in X3

> > > X5 = reflection of X4 in CI

> > > X6 = symmetry of C in X5

> > >

> > > There is only one X on BC such that X=X6. We

> name

> > this

> > > point as Ab.

> > > Do the same procedures but with direction CAB, we

> get

> > > another point on BC and name it as Ac

> > >

> > > It is easy to get barycentrics of X1, X2... X6

> from

> > > barycentrics of X. If we calculate intersection

> point

> > Z of

> > > two lines XX1 and X4X6 then Z is always on a

> line. By

> > choose

> > > X=B and X=C and do procedures as above, we can

> easy

> > to

> > > construct this line. Intersection of this line

> with BC

> > is Ab

> > > or Ac depend on direction we choosen. So Ab, Ac

> are

> > easy to

> > > construct by ruler and compass.

> > >

> > > We can do the same procedures to get another

> similar

> > points

> > > on lines CA, AB and we have total six points: Ab,

> Ac,

> > Bc,

> > > Ba, Ca, Cb.

> > >

> > > It is interesting that these six points are

> concyclic

> > on

> > > one circle with center T.

> > > Barycentrics of T:

> > > a*(3*a*(a^2 - b^2 - c^2) - 2*(b + c)*(a - b +

> c)*(a +

> > b -

> > > c) + 4*a*b*c) : :

> > > Search value: +3.3876599710368

> > > May be not in current ETC.

> > > Some lines contain T center: X(1)X(3),

> X(2)X(952),

> > > X(5)X(944), X(8)X(140), X(104)X(1621),

> X(145)X(631),

> > > X(182)X(3242)...

> > >

> > > Denote constant:

> > > K = (5*b*c - 4*SA)*(5*c*a - 4*SB)*(5*a*b - 4*SC)

> > > then radius of this circle is: Sqrt(K)/(36*Area)

> > > Here Area is area of triangle ABC

> > >

> > > If this circle is well-known before and how is

> it

> > related

> > > with other well-known circles?

> > >

> > > Thank you and best regards,

> > > Bui Quang Tuan