- View SourceDear 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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[Non-text portions of this message have been removed] - View SourceDear 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.

Your radical center is the complement of the Schiffler point and

appears as X(442) in ETC.

Friendly. Jean-Pierre - View SourceDear 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

*** You are right. Let A_b and A_c be the pedal (orthogonal projections) of

> 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 ?

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.

[Non-text portions of this message have been removed]

> Best regards,

> Khoa Lu Nguyen.

>

>

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- View SourceDear 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

In Hyacinthos message #10565, Paul Yiu wrote:

>> 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.

>> Let A_b and A_c be the pedal (orthogonal

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

>> 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),

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