- Dear all,

xpolakis@... wrote:>

This particular example has been discussed in the thread "Don Wallace

> Let P be a point and P', P" its isogonal, isotomic conjugates, resp.

> We have six lines: PP', PP", P'P", tripolars of P,P',P", which can form

> C(6,3) = 20 triads of lines.

>

> For each one of the triads we can ask for the locus of P such that

> the members of the triad concur.

>

> For example:

> Which is the locus of P such that the tripolars of P,P',P" concur?

> I found that the locus is a conic (hyperbola) passing through the four

> Ix and G. [equation: a^2(b^2-c^2)x^2 + cycl. = 0]

Locus".

Kind regards,

Floor. >> For example:

Floor is not mentioning that, before Hyacinthos, he also posted (and

>> Which is the locus of P such that the tripolars of P,P',P" concur?

>> I found that the locus is a conic (hyperbola) passing through the four

>> Ix and G. [equation: a^2(b^2-c^2)x^2 + cycl. = 0]

>

> This particular example has been discussed in the thread "Don Wallace

> Locus".

>

> Kind regards,

> Floor.

solved) this in the geometry forum.

Steve- [APH]:
>> Let P be a point and P', P" its isogonal, isotomic conjugates, resp.

[FvL]:

>> We have six lines: PP', PP", P'P", tripolars of P,P',P", which can form

>> C(6,3) = 20 triads of lines.

>>

>> For each one of the triads we can ask for the locus of P such that

>> the members of the triad concur.

>>

>> For example:

>> Which is the locus of P such that the tripolars of P,P',P" concur?

>> I found that the locus is a conic (hyperbola) passing through the four

>> Ix and G. [equation: a^2(b^2-c^2)x^2 + cycl. = 0]

>This particular example has been discussed in the thread "Don Wallace

Yes, the Don Wallace locus, "The locus of P such that P, gP [= isog. conj.

>Locus".

of P], tP [= isog. conj. of P] are collinear" (AMM 43(1936) 500), is the

hyperbola a^2(b^2 - c^2)x^2 + b^2(c^2 - a^2)y^2 + c^2(a^2 - b^2)z^2 = 0

But is the locus of P, such that the tripolars of P, gP, tP are concurrent,

the Wallace hyperbola, as I wrote ?

According to my new calculations:

The locus of P such that the tripolars of :

- P, gP, tP concur, is the quartic:

(b^2 - c^2)/x^2 + (c^2 - a^2)/y^2 + (a^2 - b^2)/z^2 = 0,

- P, gP gtP concur, is the conic (Wallace hyperbola):

a^2(b^2 - c^2)x^2 + b^2(c^2 - a^2)y^2 + c^2(a^2 - b^2)z^2 = 0

The reader may find the loci of P such that the tripolars of

- P, tP, tgP

- P, tgP, gtP

- P, gtgP, tgtP

etc, are concurrent.

Antreas - on 5/1/01 1:13 PM, xpolakis@... wrote:

> Yes, the Don Wallace locus, "The locus of P such that P, gP [= isog. conj.

This equation is in trilinears, I assume. When I do this in barycentrics, I

> of P], tP [= isog. conj. of P] are collinear" (AMM 43(1936) 500), is the

> hyperbola a^2(b^2 - c^2)x^2 + b^2(c^2 - a^2)y^2 + c^2(a^2 - b^2)z^2 = 0

get

(b^2 - c^2)x^2 + (c^2 - a^2)y^2 + (a^2 - b^2)z^2 = 0

which is better named as the Kiepert hyperbola of the medial triangle.

ABC is self polar with respect to it. It has many point on it which may be

found by applying (: x-y+z:) to any point (:y:) on the Kipert hyperbola. In

particular the 4 incenters lie on it, as well as H and L.

Apologies if this has already been pointed out.

The Triangle Book nears completion, when I look forward to being a day to

day Hyacinthian again.

Friendly from Princeton,

Steve - [APH]:
>> Yes, the Don Wallace locus, "The locus of P such that P, gP [= isog. conj.

isot.

>> of P], tP [= isog. conj. of P] are collinear" (AMM 43(1936) 500), is the

>> hyperbola a^2(b^2 - c^2)x^2 + b^2(c^2 - a^2)y^2 + c^2(a^2 - b^2)z^2 = 0

[SS]:

>This equation is in trilinears, I assume. When I do this in barycentrics, I

Yes in trilinears, or better in normals.

>get

^^^^^^

>

>(b^2 - c^2)x^2 + (c^2 - a^2)y^2 + (a^2 - b^2)z^2 = 0

>

>which is better named as the Kiepert hyperbola of the medial triangle.

antimedial

>ABC is self polar with respect to it. It has many point on it which may be

^

>found by applying (: x-y+z:) to any point (:y:) on the Kipert hyperbola. In

>particular the 4 incenters lie on it, as well as H and L.

G

L lies on it, since L(ABC) = H(antimedial), and every rect. hyperbola

passes through the orthocenter of the triangle.

>Apologies if this has already been pointed out.

APH

>

>The Triangle Book nears completion, when I look forward to being a day to

>day Hyacinthian again.