- Dear Jean-Pierre Ehrmann,

Thank you very much.

>> up to the factor area(ABC)/2/(ap+bq+cr)^2

I guess this means area(ABC)/2/(ax+by+cz)^2.

Sincerely,

Darij Grinberg

> > Let XYZ be the pedal triangle of a point P with

> > respect to a given triangle ABC. What is the locus of

> > the points P for which

> >

> > area(BXP) - area(CXP)

> > + area(CYP) - area(AYP)

> > + area(AZP) - area(BZP) = 0 ?

> >

> > I know that the circumcenter, the incenter and the

> > symmedian point lie on the locus. Other centers like the

> > orthocenter and the centroid don't.

> >

> > CONJECTURE: The locus is the Stammler hyperbola!

>

> Suppose that P is trlinear x:y:z; up to the factor

> area(ABC)/2/(ap+bq+cr)^2, we have

> area(BXP):area(CYP):area(AZP):area(CXP):area(AYP):area(BZP) =

> |u| : |v| : |w| : |u'| : |v'| : |w'| where

> u = x(SB.x+c.a.z), v,w cyclic and u' = x(SC.x+a.b.y), v',w' cyclic.

> You get a part of Stammler hyperbola when u,v,w,u',v',w' have the

> same sign.

> In any case, you get an union of different arcs of different conics.

> Note that a point inside ABC and not lying on Stammler hyperbola

> can

> verify your condition (for instance if one of the points X,Y,Z is

> outside the triangle ABC) - Darij Grinberg wrote:
> Dear Barry Wolk, and other Hyacinthians,

in

>

> >> Correct conjecture. It was a simple coordinate calculation.

>

> Thank you. I am trying to redo the proof online: Let P(x:y:z)

> trilinears, and P[x1|y1|z1] in actual (exact) trilinears (x1 =

PX, y1

> = PY and z1 = PZ). Then, we have to find the locus of points P

with u

> + v + w = 0, where u = area(BXP) - area(CXP), and v and w

analogously. We have

[long calculation snipped]

> But the equation of the Stammler hyperbola in trilinears is

(b�-c�)x�

> + cyclic sum = 0, as I read in an old Hyacinthos message. QED.

But

> not a simple coordinate calculation! Am I missing a good

idea??

Yes you are. If P(x:y:z) and X(0:y':z') in normalized

barycentrics, so x+y+z=1 and 0+y'+z'=1, then

y' = y + SC*x/a^2, z' = z + SB*x/a^2. Then BX = z'*BC = az',

and XP = x*(A-altitude) = x*2*(area of ABC)/a.

Dropping factors common to all 6 triangles,

area(BXP) = (1/2) BX XP ~~ z' * x = zx + SB*x^2/a^2. Similar

formulas for all 6 triangles are obtained by permuting the

variables, and this gives the Stammler equation quickly.

Of course this ignores the sign problem of negative areas,

pointed out by JP. ANd all I'm doing is filling in some of the

steps in his solution.

>

--

> Sincerely,

> Darij Grinberg

Barry Wolk

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