Sorry, an error occurred while loading the content.

## Re: [EMHL] Re: excircle and circumcircle

Expand Messages
• ... Dear Darij! 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
Message 1 of 3 , Jan 27, 2004
• 0 Attachment
>
>>> 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.
>
Dear Darij!
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 triangle ABC and a point P. A1, B1 are the common point of AP and BC,
Message 2 of 3 , Jan 28, 2004
• 0 Attachment
Dear Darij!
>
>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!
>

The equivalence of 2 and 5 is a particular case 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.

Sincerely Alexey
• Dear Alexey, ... Exactly. My apologies for the typo. ... Yes, and this was exactly the same generalization I found in 2002 when I solved the problem!
Message 3 of 3 , Jan 28, 2004
• 0 Attachment
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
>> 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.

Yes, and this was exactly the same generalization
I found in 2002 when I solved the problem!

Sincerely,
Darij Grinberg
Your message has been successfully submitted and would be delivered to recipients shortly.