- Dear Nikos and Alexei,

in fact, there are 6 possible circumscribed triangles for a given

inscribed one. For each of them, as my sketch shows, the locus is a

conic, but not necessarily a hyperbola: as the initial triangle

varies, some of these conics can turn into an ellipse as well.

However, the conic built "symmetrically" must be a hyperbola, as

Alexei noted, and it can be very close to two lines.

Nice construction!

Vladimir

2013/2/21 Nikolaos Dergiades <ndergiades@...>:> Dear Alexey,

> I think that the locus is a pair of lines

> Best regards

> Nikos Dergiades

>

>

>> Dear colleagues!

>> My previous problem can be reformulated.

>> Consider all pairs of homothetic regular triangles one of

>> which is inscribed in the given triangle and the second one

>> is circumscribed around it. Which is the locus of homothety

>> centers?

>> Since the product of sides of two triangles is constant,

>> there exist two pairs of equal triangles. Thus the sought

>> locus meets the infinite line in two points and we can

>> suppose that it is a hyperbola. Is this true?

>>

>> Sincerely

>>

>> Alexey

>>

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

>>

>>

>>

>> ------------------------------------

>>

>> Yahoo! Groups Links

>>

>>

>> Hyacinthos-fullfeatured@yahoogroups.com

>>

>>

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

> - Dear Nikos and Vladimir!

Of course there are several families of pairs of triangles but I consider the simplest of them. Vertex A_1 of inscribed triangle A_1B_1C_1 lies on BC, B_1 lies on CA, C_1 lies on AB. Side A_2B_2 on circumscribed triangle A_2B_2C_2 is parallel to A_1B_1 and passes through C etc. Triangles ABC and A_1B_1C_1 have the same orientation. In this case the locus can't be an ellipse because this family contains two pairs of equal triangles. For some triangles ABC it seems to be a pair of lines but this is an optical illusion, increasing a dimension of triangle we can see that this a hyperbola with small real axis.

Sincerely Alexey

in fact, there are 6 possible circumscribed triangles for a given

inscribed one. For each of them, as my sketch shows, the locus is a

conic, but not necessarily a hyperbola: as the initial triangle

varies, some of these conics can turn into an ellipse as well.

However, the conic built "symmetrically" must be a hyperbola, as

Alexei noted, and it can be very close to two lines.

Nice construction!

Vladimir

2013/2/21 Nikolaos Dergiades ndergiades@...>:> Dear Alexey,

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

> I think that the locus is a pair of lines

> Best regards

> Nikos Dergiades

>

>

>> Dear colleagues!

>> My previous problem can be reformulated.

>> Consider all pairs of homothetic regular triangles one of

>> which is inscribed in the given triangle and the second one

>> is circumscribed around it. Which is the locus of homothety

>> centers?

>> Since the product of sides of two triangles is constant,

>> there exist two pairs of equal triangles. Thus the sought

>> locus meets the infinite line in two points and we can

>> suppose that it is a hyperbola. Is this true?

>>

>> Sincerely

>>

>> Alexey

>>

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

>>

>>

>>

>> ------------------------------------

>>

>> Yahoo! Groups Links

>>

>>

>> Hyacinthos-fullfeatured@yahoogroups.com

>>

>>

>

>

> ------------------------------------

>

> Yahoo! Groups Links

>

>

>