## 3 nine point circles and the Euler line

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• Dear friends, I think this one is right but I haven t thought about it at all. So I think I ll call this one my conjecture : Let ABC be a triangle and I_a,
Message 1 of 4 , Oct 1, 2004
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Dear friends,
I think this one is right but I haven't thought about it at all.
So I think I 'll call this one my conjecture :
Let ABC be a triangle and I_a, I_b, I_c are excenters,
Let (W_a), (W_b), (W_c) are respectively the nine point
circles of I_aBC, I_bCA, I_cAB . Prove that the 3 radical axises of (W_a)
and (W_b), (W_b) and (W_c), (W_c) and (W_a) and the Euler line of ABC are concurent. What is its barycentric coordinate ?
Thanks.
Best regards,
Khoa Lu Nguyen.

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• Dear Hyacithists ... all. ... axises of (W_a) ... of ABC are concurent. What is its barycentric coordinate ? ... Your radical center is the complement of
Message 2 of 4 , Oct 1, 2004
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Dear Hyacithists
--- In Hyacinthos@yahoogroups.com, KHOA LU <treegoner@y...> wrote:
> Dear friends,
> I think this one is right but I haven't thought about it at
all.
> So I think I 'll call this one my conjecture :
> Let ABC be a triangle and I_a, I_b, I_c are excenters,
> Let (W_a), (W_b), (W_c) are respectively the nine point
> circles of I_aBC, I_bCA, I_cAB . Prove that the 3 radical
axises of (W_a)
> and (W_b), (W_b) and (W_c), (W_c) and (W_a) and the Euler line
of ABC are concurent. What is its barycentric coordinate ?
> Thanks.
> Best regards,
> Khoa Lu Nguyen.

appears as X(442) in ETC.
Friendly. Jean-Pierre
• Dear Khoa Lu, [KLN]: ... Let ABC be a triangle and I_a, I_b, I_c are excenters, ... *** You are right. Let A_b and A_c be the pedal (orthogonal
Message 3 of 4 , Oct 1, 2004
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Dear Khoa Lu,

[KLN]: ... Let ABC be a triangle and I_a, I_b, I_c are excenters,
> Let (W_a), (W_b), (W_c) are respectively the nine point
> circles of I_aBC, I_bCA, I_cAB . Prove that the 3 radical axises of
> (W_a)
> and (W_b), (W_b) and (W_c), (W_c) and (W_a) and the Euler line of ABC
> are concurent. What is its barycentric coordinate ?

*** You are right. Let A_b and A_c be the pedal (orthogonal projections) of
A on I_aI_b and I_aI_c respectively, then the radical axis of (W_b) and
(W_c) is the same as the Euler line
of AA_bA_c. From ETC, this line and the two analogous line interesect at
X(442), which has barycentric coordinates

((b+c)(a^2(b+c)+2abc-(b+c)(b-c)^2) : ... : ...).

This same point also appears in a recent message of Juan Carlos Salazar
(Hyacinthos 10323). See Darij's 10342.

Best regards
Sincerely
Paul

> Thanks.
> Best regards,
> Khoa Lu Nguyen.
>
>
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• Dear Khoa Lu, Jean-Pierre and Paul, ... I don t have a synthetic proof, but I have some remarks. After Theorem F2 of my Hyacinthos message #10584 (with
Message 4 of 4 , Oct 3, 2004
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Dear Khoa Lu, Jean-Pierre and Paul,

In Hyacinthos message #10563, Khoa Lu Nguyen wrote:

>> Let ABC be a triangle and I_a, I_b, I_c are
>> excenters,
>> Let (W_a), (W_b), (W_c) are respectively the
>> nine point circles of I_aBC, I_bCA, I_cAB .
>> Prove that the 3 radical axises of (W_a) and
>> (W_b), (W_b) and (W_c), (W_c) and (W_a) and
>> the Euler line of ABC are concurent.

In Hyacinthos message #10565, Paul Yiu wrote:

>> Let A_b and A_c be the pedal (orthogonal
>> projections) of A on I_aI_b and I_aI_c
>> respectively, then the radical axis of (W_b)
>> and (W_c) is the same as the Euler line of
>> AA_bA_c. From ETC, this line and the two
>> analogous line interesect at X(442),

I don't have a synthetic proof, but I have some
remarks.

After Theorem F2 of my Hyacinthos message #10584
(with modified notations), if I is the incenter of
triangle ABC, and A'b and A'c are the orthogonal
projections of the point A on the lines CI and BI,
respectively, then the Euler line of triangle
AA'bA'c is the radical axis of the nine-point
circles of triangles CIA and AIB.

After application of A-extraversion, this result
takes the following form: With Ab and Ac being the
orthogonal projections of the point A on the lines
CIa and BIa, respectively, the Euler line of
triangle AAbAc is the radical axis of the
nine-point circles of triangles CIaA and AIaB.

Meantime Paul Yiu's Hyacinthos message #10565
cited above shows that the Euler line of triangle
AAbAc is the radical axis of the nine-point
circles of triangles CIbA and AIcB.

Hence, the radical axis of the nine-point circles
of triangles CIaA and AIaB coincides with the
radical axis of the nine-point circles of
triangles CIbA and AIcB. On the other hand, one
can easily see synthetically that the radical
axis of the nine-point circles of triangles CIaA
and CIbA coincides with the radical axis of the
nine-point circles of triangles AIaB and AIcB
(in fact, both times, the radical axis is the
line joining the midpoints of the segments CA
and AB). Hence, the nine-point circles x, y, z,
w of the triangles CIaA, AIaB, AIcB and CIbA
form a quadruple of circles with the property
that the radical axis of the circles x and y
coincides with the radical axis of the circles
z and w, and that the radical axis of the
circles y and z coincides with the radical axis
of the circles w and x.

Finally, note that the Euler line of triangle
AAbAc is not only the radical axis of two
pairs of nine-point circles, but it has also,
at least, three interesting points lying on
it:

- the midpoint of the segment AIa;
- the A-Feuerbach point of triangle ABC;
- the complement of the Schiffler point of
triangle ABC (this is X(442)).

Thanks to all of you for this wonderful
problem. Note that I did not consider the
relations of this with Juan Carlos' problem in
Hyacinthos message #10342, since this would be
too much work (the problem is like a hydra:
each time one proofs one part of it, three new
conjectures turn up).

Sincerely,
Darij Grinberg
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