- Dear Francisco,

I can follow your reasoning and it looks like you are quite right.

So that's why I constructed your way and my way together.

After doing this I saw immediately the origin of the appararent

contradiction.

When P is on NPC, the line Q.R is parallel to DP', etc. indeed.

However it still doesn't pass through P.

Actually there is no ground to go through P.

However this line can be constructed.

0. As a matter of fact when P is on NPC the points Q and R are

infinitypoints and QR intersects The BrocardAxis in a special point T.

This point T is derived by a mapping from NPC on the BrocardAxis.

1. P is a point on NPC.

2. S1 and S2 are different points on any radius of NPC.

3. Construct QR-lines as introduced by you before for S1 and S2.

These lines intersect in the same point T (on the BrocardAxis) for

all different S1 and S2 (unequal to X5, P, antipode of P).

So you could say that for every line through X5 there is a mapping on

a point on the BrocardAxis. Because of continuity-considerations this

is also valid for the intersectionpoint P of this line with NPC.

4. So draw the line through T parallel to DP'.

This is the line QR when Q and R are infinitypoints.

This was confirmed by my drawings in Cabri.

I hope it still is understandable.

Best regards,

Chris van Tienhoven

--- In Hyacinthos@yahoogroups.com, "garciacapitan"

<garciacapitan@...> wrote:>

we

> Dear Chris, I guess that you make some error in your construction,

> please do the following:

>

> ABC is a triangle, DEF its orthic triangle, LMN its medial triangle.

> P any point on the nine point circle of ABC.

> We have that altitude AD is the internal bisector of angle EDF. So

> call P' the reflection of P with respect to line AD.

bisector

> Call P'' the reflection of P with respect the internal angle

> of angle MLN.

and

>

> You can see that (isogonal) lines DP' and LP'' are parallel. If you

> construct the isogonals with respect the other vertices of the two

> triangles you we'll see that these lines are also parallel to DP'

> LP'', meaning that both isogonal conjugates of P with respect the

two

> triangles are the same infinity point.

we

>

> Hence if P belongs to NPC, the three points P, Q, R are in fact only

> two points, always collinear.

>

> --- In Hyacinthos@yahoogroups.com, "chris.vantienhoven"

> <van10hoven@> wrote:

> >

> >

> > Dear Francisco,

> >

> > --- In Hyacinthos@yahoogroups.com, "garciacapitan"

> > <garciacapitan@> wrote:

> > > If P is on the NPC then Q and R are the same infinity point, so

> > can

have

> > > say that P, Q and R are collinear (if P is an infinity point we

> > > also that Q=R, hence P, Q and R can be considered collinear in

this

> > > case).

construct

> >

> > Take any point P on any line through Center X5 from NPC and

> > Q and R as described earlier.

QR

> > Draw the line QR.

> > Let P run over this line. Then you can see that P is only on line

> > when it crosses the KiepertHyperbola.

collineate

> > At least this can be done in Cabri.

> > When it crosses NPC you can see also where the line from one

> > InfinityPoint to another InfinityPoint tends to.

> > This is a line not through P.

> > So as far as I can see it, when P is on NPC it does not

> > with Q and R.

> > Best regards,

> >

> > Chris van Tienhoven

> >

> - Dear Francisco.

I understand what you are saying.

It is an interesting question.

Everything is allright for me. Except for your statement "The line

PQR here is the parallel to DP', LP", etc. through P." That is just

the point you started with. Under what conditions is there a line PQR?

For a better understanding let us name:

Line L = the line through Q and R

Pi = Infinity Point, being the IntersectionPoint of parallel

lines DP', LP'', etc.

We started with the question under what conditions P,Q,R are

collinear?

You noticed that when P is a point on the Kiepert Hyperbola that in

that case P,Q,R are collinear.

This is easy to be confirmed with drawings.

Now you state that when P is on NPC, then Q and R coincide with

Infintypoint Pi.

Then you state (in other words): so P.Q.R = P.Pi.Pi = P.Pi, ergo they

are collinear.

Is that right?

However the question is not to draw the line P.Pi. The original

question is to draw a line through Pi.Pi and then look if this line

coincides with P.

When you draw a line through Pi.Pi you actually draw a line through

Pi (evident) but you don't know if it comes from the right or the

left side of the plane.

It can be any line parallel to lines DP', LP'', etc. There is an

endless amount of these lines possible.

As soon as we found that very special line that is inherent to the

problem then we have to review if P is on this line.

Of course this should be confirmed with drawings (as in real life we

should try to confirm all things that are said to be true).

When I make drawings in Cabri and take a variable point P and move it

to any point on NPC I see what happens to QR in these drawings:

1. It is moving to a parallel line to DP', LP'', etc.

2. It is going through the point T on the BrocardAxis that I

described in earlier mail.

3. It is NOT going through P !

So this is a confirmation of the contrary of your statement. For me

it is clear that P,Q,R are not collinear when P is on NPC, except for

the situation when it is an intersectionpoint of NPC with the Kiepert

Hyperbola.

Best regards,

Chris van Tienhoven

p.s. There is a comparable situation when you take 2 points Q and R

on a circle and any point P outside the circle and you want to know

when P,Q,R are collinear. As soon as Q and R coincide you notice that

in this case (inherent to the problem) QR is the tangent to the

circle and P is not on the tangent.