- Friends of Hyacinthos:

Four problems are offered

in

http://personal.us.es/rbarroso/trianguloscabri/

One of the inradius and

circumradius, of Vcente Vicario

another on the

equilateral, proposed for Ricard Peiró

and two of Paul Halmos

Greetings from

Sevilla

Ricardo

Barroso

[Non-text portions of this message have been removed] - Dear Hauke, Barry and Jean-Pierre!

> Some computation shows that for any circle that meets the 2 sides AB and AC harmonically, its center lies on the perpendicular bisector of BC.

I don't agree with you. Your circles are member of the pencil with Poncelet points (circles with radius 0) A and the projection of A upon BC.

The conic meets AB harmonically iff A lyes on the polar of B. Thus triangle ABC is autopolar wrt Hauke's conic. But in this case all six common points of this conic and the sidelines can't be real.

Sincerely Alexey

[Non-text portions of this message have been removed] > From: jpehrmfr <jean-pierre.ehrmann@...>

I finally realized that I used an incorrect formula for harmonic conjugates, namely (BP/PC)*(BQ/QC)=-1, instead of the correct (BP/PC)/(BQ/QC)=-1 when P,Q are harmonic conjugates wrt B,C. So my results apply to this strange kind of "conjugates," but not to harmonics.

>

> Dear Barry

> [Barry Wolk]

> > Some computation shows that for any circle that meets

> the 2 sides AB and AC harmonically, its center lies on the

> perpendicular bisector of BC.

>

> I don't agree with you. Your circles are member of the

> pencil with Poncelet points (circles with radius 0) A and

> the projection of A upon BC.

> Friendly. Jean-Pierre

--

Barry Wolk- Dear Hyacinthians,

Finally catching up on my mail. At the risk of having missed part of the

discussion, I only partially agree with Alexey (if I understand him

correctly). Using his observation that ABC is autopolar with respect to

Hauke's conic, it is a straightforward exercise in barycentric calculus

to show that any circle with the property that Hauke is looking for

always has its center at the orthocenter of ABC. Also, with a little

more work it can be shown that the radius of such a circle has to be

equal to 2*R*sqrt(-cos(aA)cos(aB)cos(aC)), with R the radius of the

circumcircle of ABC and aA, aB, aC the angles at A, B, C. It follows

that there is no such (real) circle for acute triangles and exactly one

for obtuse triangles.

Eisso

On 7/2/10 3:42 AM, Alexey Zaslavsky wrote:

>

> Dear Hauke, Barry and Jean-Pierre!

>

> > Some computation shows that for any circle that meets the 2 sides AB

> and AC harmonically, its center lies on the perpendicular bisector of BC.

>

> I don't agree with you. Your circles are member of the pencil with

> Poncelet points (circles with radius 0) A and the projection of A upon BC.

>

> The conic meets AB harmonically iff A lyes on the polar of B. Thus

> triangle ABC is autopolar wrt Hauke's conic. But in this case all six

> common points of this conic and the sidelines can't be real.

>

> Sincerely Alexey

>

>

--

========================================

Eisso J. Atzema, Ph.D.

Department of Mathematics& Statistics

University of Maine

Orono, ME 04469

Tel.: (207) 581-3928 (office)

(207) 866-3871 (home)

Fax.: (207) 581-3902

E-mail: atzema@...

========================================

[Non-text portions of this message have been removed] - I just remember the Faure theorem.

A triangle ABC is autopolar wrt the conic Gamma iff the ABC-circumcircle is

orthogonal to the orthoptic circle of Gamma.

Friendly

Francois

On Fri, Jul 2, 2010 at 8:15 PM, Barry Wolk <wolkbarry@...> wrote:

>

>

> > From: jpehrmfr <jean-pierre.ehrmann@...<jean-pierre.ehrmann%40wanadoo.fr>

> >

>

> >

> > Dear Barry

> > [Barry Wolk]

> > > Some computation shows that for any circle that meets

> > the 2 sides AB and AC harmonically, its center lies on the

> > perpendicular bisector of BC.

> >

> > I don't agree with you. Your circles are member of the

> > pencil with Poncelet points (circles with radius 0) A and

> > the projection of A upon BC.

> > Friendly. Jean-Pierre

>

> I finally realized that I used an incorrect formula for harmonic

> conjugates, namely (BP/PC)*(BQ/QC)=-1, instead of the correct

> (BP/PC)/(BQ/QC)=-1 when P,Q are harmonic conjugates wrt B,C. So my results

> apply to this strange kind of "conjugates," but not to harmonics.

> --

> Barry Wolk

>

>

>

[Non-text portions of this message have been removed]