>

Dear Darij!

>>> Let AX and BY are the bisectrix of triangle ABC,

>>> O, R - the center and radius of its circumcircle,

>>> I1, r1 - the center and radius of excircle

>>> touching the sideline AB, A', B', C' - the

>>> touching points of excircle with BC, CA, AB.

>>> Then next conditions are equivalent:

>>> 1. R=r1.

>>> 2. O is in line XY.

>>> 3. O is the orthocenter of A'B'C' (I suppose

>>> also that A'B'C' is autopolar with respect to

>>> circumcicle of ABC).

>>> 4. cosA+cosB=cosC.

>

>You can add the following equivalent condition:

>5. If I is the incenter of triangle ABC, and Hb

>and Hc are the feet of the altitudes from B and C,

>then I lies on HbHc.

>

I think that I is in HaHb. In this case the equivalence of 4 and 5 may be

proved by trygonomethric transformations. The triangles ABC and HaHbC are

similar with coefficient cosC, so CL=CC1cosC, where CL is the bisectrix of

HaHbC, CC1 - the bisectrix of ABC. But CI/CC1=(sinA+sinB)/(sinA+sinB+sinC)

and by 4 I=L.

Sincerely Alexey- Dear Darij!
>

The equivalence of 2 and 5 is a particular case of next fact. Let given a

>The equivalence of the conditions 2. and 5. was the

>subject of Problem 4 in the Bundeswettbewerb

>Mathematik (German National Mathematics Olympiad)

>2002, 2 round. In fact, it was a very hard problem;

>I was lucky enough to know the method of trilinear

>coordinates to solve it!

>

triangle ABC and a point P. A1, B1 are the common point of AP and BC, BP and

AC. Q - a point in A1B1, Q' is isogonally conjugated to Q, A2, B2 are the

common points of AQ' and BC, BQ' and AC. Then the point P' isogonally

conjugated to P is in A2B2. This can be easily proved by trilinear

coordinates.

Sincerely Alexey - Dear Alexey,

In Hyacinthos message #9144, you wrote:

>> I think that I is in HaHb.

Exactly. My apologies for the typo.

In Hyacinthos message #9145, you wrote:

>> The equivalence of 2 and 5 is a particular case

Yes, and this was exactly the same generalization

>> of next fact. Let given a triangle ABC and a

>> point P. A1, B1 are the common point of AP and

>> BC, BP and AC. Q - a point in A1B1, Q' is

>> isogonally conjugated to Q, A2, B2 are the

>> common points of AQ' and BC, BQ' and AC. Then

>> the point P' isogonally conjugated to P is in

>> A2B2. This can be easily proved by trilinear

>> coordinates.

I found in 2002 when I solved the problem!

Sincerely,

Darij Grinberg