- Dear Darij and Vladimir

For the case that the quadrilateral is not convex from a sketch

I concluded (perhaps this is not correct) that the diagonals

BD, AC of the quadrilateral must be parallel and the point O

must lie on the line joining the mid points of the diagonals.

For example if the points A,B,D,C are concyclic and the

quadrilateral ABDC is isosceles trapezium AB=DC, AC || BD

then the quadrilateral ABCD is not convex and if

the cartesian coordinates are

B(-a , 0) , D(a , 0) , O( -d, 0)

where O is the point satisfying the "halfside" condition

A(-p , q) , C( p , q), a, d, p, q > 0

then p , q can be found from p/d = q/a

(aa+dd)pp + 2ad(a-2d)p + ddaa = 0

Special Case :

d = 2a and p = 2a q = a or

d = 2a and p = 2a/5 q = a/5

Best regards

Nikolaos Dergiades - Dear Darij!

III.8. Let ABC be a triangle with circumradius R and

inradius r. It is well-known that there exist

infinitely many triangles having the same circumcircle

and the same incircle as triangle ABC.

(a) Show that the orthocenters of all of these

triangles lie on a common circle.

(b) What can be said about the locus of the nine-point

centers of all of these triangles in the case R = 4r ?

My pupils D.Kosov and M.Muzafarof examined the trajectories of some centers

of Ponsele's triangle. Their results are typed in "Kvant",

N 2, 2003 (www.kvant.mccme.ru).

Sincerely Alexey - Dear Alexey,

In Hyacinthos message #9659, you wrote:

>> My pupils D.Kosov and M.Muzafarof examined the

By a curious coincidence, I have found this

>> trajectories of some centers of Ponsele's

>> triangle. Their results are typed in "Kvant",

>> N 2, 2003 (www.kvant.mccme.ru).

paper yesterday before receiving your mail! It

was indeed nice to see my conjecture verified

that the poristic locus (or "trajectory") of

the symmedian point is an ellipse. However,

concerning the proofs of the rather simple

poristic loci (i. e., those of the Nagel point,

nine-point center, orthocenter, centroid), I

prefer using the Feuerbach theorem rather than

calculating the distances using barycentric

coordinates.

Sincerely,

Darij Grinberg - Dear Darij!
>

You are right, for this points there are a synthetical proof. But without

>By a curious coincidence, I have found this

>paper yesterday before receiving your mail! It

>was indeed nice to see my conjecture verified

>that the poristic locus (or "trajectory") of

>the symmedian point is an ellipse. However,

>concerning the proofs of the rather simple

>poristic loci (i. e., those of the Nagel point,

>nine-point center, orthocenter, centroid), I

>prefer using the Feuerbach theorem rather than

>calculating the distances using barycentric

>coordinates.

>

coordinates we could not prove the most interesting our result: the

trajectory of Gergonne point is the circle coaxial with the circumcircle and

the incircle.

Sincerely Alexey - Dear Darij and Vladimir

sorry a correction. In my 9658 message

I wrote:

> For the case that the quadrilateral is not convex from a sketch

********

> I concluded (perhaps this is not correct) that the diagonals

> BD, AC of the quadrilateral must be parallel and the point O

> must lie on the line joining the mid points of the diagonals.

>

> For example if the points A,B,D,C are concyclic and the

> quadrilateral ABDC is isosceles trapezium AB=DC, AC || BD

> then the quadrilateral ABCD is not convex and if

> the cartesian coordinates are

> B(-a , 0) , D(a , 0) , O( -d, 0)

> where O is the point satisfying the "halfside" condition

> A(-p , q) , C( p , q), a, d, p, q > 0

> then p , q can be found from p/d = q/a

> (aa+dd)pp + 2ad(a-2d)p + ddaa = 0

>

> Special Case :

> d = 2a and p = 2a q = a or

> d = 2a and p = 2a/5 q = a/5

The point O is (0, -d) and not (-d , 0)

Best regards

Nikolaos Dergiades