- Let ABC be a triangle and A'B'C' the cevian

triangle of I.

If the Euler Lines of ABC, BB'C, CC'B

are concurrent, then is ABC necessarily isosceles?

Antreas - Dear Antreas,

[APH]> Let ABC be a triangle and A'B'C' the

No.

> cevian

> triangle of I.

>

> If the Euler Lines of ABC, BB'C, CC'B

> are concurrent, then is ABC necessarily isosceles?

For example construct the triangle ABC with angles

A=22.5

B=90

C= 67.5

The triangle B'BC is isosceles (BB' = BC)

and the triangles C'BC, ABC are

right angled at B.

Hence their Euler lines are concurrent at B,

while ABC is not isosceles.

It would be interesting to see if BC is fixed what is

the locus of A in order the Euler lines are concurrent.

Best regards

Nikos

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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr - Dear Nikos

[APH]> > Let ABC be a triangle and A'B'C' the

[ND]

> > cevian

> > triangle of I.

> >

> > If the Euler Lines of ABC, BB'C, CC'B

> > are concurrent, then is ABC necessarily isosceles?

> No.

So, if these Euler Lines are concurrent

> For example construct the triangle ABC with angles

> A=22.5

> B=90

> C= 67.5

>

> The triangle B'BC is isosceles (BB' = BC)

> and the triangles C'BC, ABC are

> right angled at B.

> Hence their Euler lines are concurrent at B,

> while ABC is not isosceles.

>

> It would be interesting to see if BC is fixed what is

> the locus of A in order the Euler lines are concurrent.

the triangle is either right angled or isosceles or ???

Note that we can replace Euler lines with other lines

(OK, OI etc) to have similar problems.

APH - Dear Nikos, that was a really nice example.

I get that the condition

-a^4 b^3 + 2 a^2 b^5 - b^7 + a^5 b c - a^4 b^2 c - a^3 b^3 c +

a^2 b^4 c - a^4 b c^2 - a^3 b^2 c^2 - 2 a^2 b^3 c^2 + 2 b^5 c^2 -

a^4 c^3 - a^3 b c^3 - 2 a^2 b^2 c^3 - b^4 c^3 + a^2 b c^4 -

b^3 c^4 + 2 a^2 c^5 + 2 b^2 c^5 - c^7 = 0

implies that Euler Lines of ABC, BB'C, CC'B are concurrent.

Best regards,

Francisco Javier.

--- In Hyacinthos@yahoogroups.com, "xpolakis" <anopolis72@...> wrote:

>

> Dear Nikos

>

> [APH]

> > > Let ABC be a triangle and A'B'C' the

> > > cevian

> > > triangle of I.

> > >

> > > If the Euler Lines of ABC, BB'C, CC'B

> > > are concurrent, then is ABC necessarily isosceles?

>

> [ND]

> > No.

> > For example construct the triangle ABC with angles

> > A=22.5

> > B=90

> > C= 67.5

> >

> > The triangle B'BC is isosceles (BB' = BC)

> > and the triangles C'BC, ABC are

> > right angled at B.

> > Hence their Euler lines are concurrent at B,

> > while ABC is not isosceles.

> >

> > It would be interesting to see if BC is fixed what is

> > the locus of A in order the Euler lines are concurrent.

>

> So, if these Euler Lines are concurrent

> the triangle is either right angled or isosceles or ???

>

> Note that we can replace Euler lines with other lines

> (OK, OI etc) to have similar problems.

>

> APH

>