## RE: [EMHL] POINT ON THE EULER LINE ?

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• Dear Antreas, [APH}: Let ABC be a triangle and L,La,Lb,Lc the Euler Lines of ABC, IBC, ICA, IAB, resp. (concurrent at point S) and L1, L2, L3 the reflections
Message 1 of 6 , Feb 6, 2013
Dear Antreas,

[APH}: Let ABC be a triangle and L,La,Lb,Lc the Euler Lines of ABC, IBC, ICA, IAB, resp. (concurrent at point S) and L1, L2, L3 the reflections of BC, CA, AB, in La, Lb, Lc, resp.

The circumcenter Q of the triangle bounded by the lines (L1,L2,L3) is on the line L.

True?

*** No.

Best regards
Sincerely
Paul
• Dear Paul This time there are too many points P on the Euler line..... ! Let ABC be a triangle, P a point, A ,B ,C the reflections of A,B,C in the
Message 2 of 6 , Feb 7, 2013
Dear Paul

This time there are too many points P on the Euler line..... !

Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C in the
perpendicular
bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian triangle of P
wrt A'B'C'.

Which is the locus of P such that ABC, A"B"C" are perspective?

APH

On Thu, Feb 7, 2013 at 1:50 AM, Paul Yiu <yiu@...> wrote:

> **
>
>
> Dear Antreas,
>
> [APH}: Let ABC be a triangle and L,La,Lb,Lc the Euler Lines of ABC, IBC,
> ICA, IAB, resp. (concurrent at point S) and L1, L2, L3 the reflections of
> BC, CA, AB, in La, Lb, Lc, resp.
>
>
> The circumcenter Q of the triangle bounded by the lines (L1,L2,L3) is on
> the line L.
>
> True?
>
> *** No.
>
> Best regards
> Sincerely
> Paul
>
> _
>

[Non-text portions of this message have been removed]
• Dear Antreas, This is wonderful. [APH]: Let ABC be a triangle, P a point, A ,B ,C the reflections of A,B,C in the perpendicular bisectors of BC, CA, AB,
Message 3 of 6 , Feb 7, 2013
Dear Antreas,

This is wonderful.

[APH]: Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C in the
perpendicular bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian triangle of P
wrt A'B'C'. Which is the locus of P such that ABC, A"B"C" are perspective?

A ' = (a^2:-(b^2-c^2):b^2-c^2),

If P = (u:v:w), then
A'' = (((b^2-c^2)u+a^2v)((b^2-c^2)u-a^2w) : b^2(v+w)((b^2-c^2)u+a^2v) : -c^2(v+w)((b^2-c^2)u-a^2w))
etc.

A''B''C'' and ABC are perspective if and only if P lies on the Euler line.
The perspector Q also lies on the Euler line.

If OP P PH = t : 1-t, then
OQ : QH = (1+t) : -8t\cos A\cos B\cos C

Here are some examples:

P Q
------------
G X(25)
O H
H X(24)
N X(3518)
L O
X(21) X (28)
X(22) G
X(23) X(468)
X(186 X(403)

Best regards
Sincerely
Paul
• Dear Paul Note that the triangle A B C , where A ,B ,C are the reflections of A,B,C in the perpendicular bisectors, is the triangle which is the antipodal
Message 4 of 6 , Feb 7, 2013
Dear Paul

Note that the triangle A'B'C', where A',B',C' are the reflections of A,B,C
in the perpendicular bisectors,
is the triangle which is the antipodal triangle of the circumcevian
triangle of H.

So we can replace H with an other fixed point T on the Euler line.

That is:

Let ABC be a triangle, T a fixed point on the Euler Line, AtBtCt the
circumcevian
triangle of T, A'B'C' the antipodal triangle of AtBtCt (ie A',B',C' are the
antipodes of At, Bt,Ct).

Let P be a variable point and A"B"C" the circumcevian triangle wrt A'B'C'.

Which is the locus of P such that ABC, A"B"C" are perspective?

I think that the locus is the Euler Line.

We can furthermore generalize it by taking T as an arbitrary fixed point.
The locus will be some line, I guess.

Greetings

Antreas

On Thu, Feb 7, 2013 at 5:14 PM, Paul Yiu <yiu@...> wrote:

> **
>
>
> Dear Antreas,
>
> This is wonderful.
>
> [APH]: Let ABC be a triangle, P a point, A',B',C' the reflections of A,B,C
> in the
>
> perpendicular bisectors of BC, CA, AB, resp. and A"B"C" the circumcevian
> triangle of P
> wrt A'B'C'. Which is the locus of P such that ABC, A"B"C" are perspective?
>
> A ' = (a^2:-(b^2-c^2):b^2-c^2),
>
> If P = (u:v:w), then
> A'' = (((b^2-c^2)u+a^2v)((b^2-c^2)u-a^2w) : b^2(v+w)((b^2-c^2)u+a^2v) :
> -c^2(v+w)((b^2-c^2)u-a^2w))
> etc.
>
> A''B''C'' and ABC are perspective if and only if P lies on the Euler line.
> The perspector Q also lies on the Euler line.
>
> If OP P PH = t : 1-t, then
> OQ : QH = (1+t) : -8t\cos A\cos B\cos C
>
> Here are some examples:
>
> P Q
> ------------
> G X(25)
> O H
> H X(24)
> N X(3518)
> L O
> X(21) X (28)
> X(22) G
> X(23) X(468)
> X(186 X(403)
>
> Best regards
> Sincerely
> Paul
>
> _
>

[Non-text portions of this message have been removed]
• Dear Antreas, You are right. [APH]: Let ABC be a triangle, T a fixed point on the Euler Line, AtBtCt the circumcevian triangle of T, A B C the antipodal
Message 5 of 6 , Feb 7, 2013
Dear Antreas,

You are right.

[APH]: Let ABC be a triangle, T a fixed point on the Euler Line, AtBtCt the
circumcevian triangle of T, A'B'C' the antipodal triangle of AtBtCt (ie A',B',C' are the
antipodes of At, Bt,Ct).

Let P be a variable point and A"B"C" the circumcevian triangle wrt A'B'C'.

Which is the locus of P such that ABC, A"B"C" are perspective?

*** The locus is the line OT, and the perspector Q lies on OT as well.

If OP : PT = t: 1-t, then
OQ : QT = R^2(1+t) : - (R^2-OT^2) t.

Best regards
Sincerely
Paul
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