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• ## RE: [EMHL] IMO training 2004 homework problem sets I, II, III

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• 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
Message 1 of 12 , Apr 11, 2004
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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
• 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
Message 1 of 12 , Apr 11, 2004
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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, ... By a curious coincidence, I have found this paper yesterday before receiving your mail! It was indeed nice to see my conjecture verified that
Message 1 of 12 , Apr 12, 2004
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Dear Alexey,

In Hyacinthos message #9659, you wrote:

>> 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).

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.

Sincerely,
Darij Grinberg
• Dear Darij! ... You are right, for this points there are a synthetical proof. But without coordinates we could not prove the most interesting our result: the
Message 1 of 12 , Apr 12, 2004
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Dear Darij!
>
>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.
>
You are right, for this points there are a synthetical proof. But without
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 ... ******** The point O is (0, -d) and not (-d , 0) Best regards Nikolaos Dergiades
Message 1 of 12 , Apr 12, 2004
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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