- Dear Hyacinthians,

I wonder if the following general perspectivity is known:

Consider a triangle ABC and 2 different points P and Q

Let A'B'C' be the cevian triangle of P wrt triangle ABC

The parallel to BC through Q intersects AP at A*

The parallel to AP through Q intersects BC at A#

Define B*, B#, C* and C# similarly

==> The triangles A*B*C* and A#B#C# are perspective

Let P = (p : q : r), Q = (u : v : w) (in barycentrics)

then the perspector S has coordinates

( p.(q + r).u.v.w + p.r.r.u.u.v + p.q.u.u.w -q.r.u.u.u : cyclic...)

I found more than 280 combinations of points P and Q in the ETC

which lead to a perspector also in the ETC

What intrigues me most about this conjugation is the following

property:

With a given point P there seem to be 2 points Q1 and Q2 that lead

to the same perspector S and moreover it seems that S is the

midpoint of Q1 and Q2.

However I can't see why, using the formula above

Greetings from Bruges

Eric Danneels - Dear Eric

> [ED] I wonder if the following general perspectivity is known:

^^^^^^^^^^^^^^^^^^^^^^^+ p r u u v (typo) + etc

>

> Consider a triangle ABC and 2 different points P and Q

> Let A'B'C' be the cevian triangle of P wrt triangle ABC

> The parallel to BC through Q intersects AP at A*

> The parallel to AP through Q intersects BC at A#

> Define B*, B#, C* and C# similarly

>

> ==> The triangles A*B*C* and A#B#C# are perspective

>

> Let P = (p : q : r), Q = (u : v : w) (in barycentrics)

> then the perspector S has coordinates

> ( p.(q + r).u.v.w + p.r.r.u.u.v + p.q.u.u.w -q.r.u.u.u : cyclic...)

> I found more than 280 combinations of points P and Q in the ETC

Your S is the midpoint of Q and P-Ceva conj. of Q.

> which lead to a perspector also in the ETC

>

> What intrigues me most about this conjugation is the following

> property:

>

> With a given point P there seem to be 2 points Q1 and Q2 that lead

> to the same perspector S and moreover it seems that S is the

> midpoint of Q1 and Q2.

>

> However I can't see why, using the formula above

Best regards

Bernard

[Non-text portions of this message have been removed] - Dear Eric and Bernard
> > [ED] I wonder if the following general perspectivity is known:

cyclic...)

> >

> > Consider a triangle ABC and 2 different points P and Q

> > Let A'B'C' be the cevian triangle of P wrt triangle ABC

> > The parallel to BC through Q intersects AP at A*

> > The parallel to AP through Q intersects BC at A#

> > Define B*, B#, C* and C# similarly

> >

> > ==> The triangles A*B*C* and A#B#C# are perspective

> >

> > Let P = (p : q : r), Q = (u : v : w) (in barycentrics)

> > then the perspector S has coordinates

> > ( p.(q + r).u.v.w + p.r.r.u.u.v + p.q.u.u.w -q.r.u.u.u :

>

lead

> ^^^^^^^^^^^^^^^^^^^^^^^+ p r u u v (typo) + etc

>

> > I found more than 280 combinations of points P and Q in the ETC

> > which lead to a perspector also in the ETC

> >

> > What intrigues me most about this conjugation is the following

> > property:

> >

> > With a given point P there seem to be 2 points Q1 and Q2 that

> > to the same perspector S and moreover it seems that S is the

[BG]

> > midpoint of Q1 and Q2.

> >

> > However I can't see why, using the formula above

> Your S is the midpoint of Q and P-Ceva conj. of Q.

This leads to a construction of Q if we know P and S but I don't

understand why we cannot get 4 solutions both by both reflected of

each other wrt S.

Note that when M moves on a line L, the Cevian quotient P/M moves on

the conic C(P,L) going through the feet of the cevians of P and the

feet of the cevians of the trilinear pole of L.

Now draw through S the lines L,L' going through the infinite points

of the circumconic going through P and S (if the conic is an

ellipse, Q cannot exist)

The solutions are the common points of L and C(P,L) and the common

points of L' and C(P,L').

Once more, I don't understand why it should be impossible to get 4

real solutions.

Friendly. Jean-Pierre - In Hyacinthos message #10135, Eric Danneels wrote:

>> Consider a triangle ABC and 2 different

Cheers, Eric, for finding such a beautiful

>> points P and Q

>> [...]

>> The parallel to BC through Q intersects

>> AP at A*

>> The parallel to AP through Q intersects

>> BC at A#

>> Define B*, B#, C* and C# similarly

>>

>> ==> The triangles A*B*C* and A#B#C# are

>> perspective

theorem! It was indeed too beautiful for me to

leave it without a proper synthetic proof, so

I have been looking for one, and finally I have

found a rather long synthetic proof with Ceva

and Menelaos. You can find it in the note

"On the paracevian perspector" on my website

http://de.geocities.com/darij_grinberg

Let me tell in brief how the proof goes: At

first, I construct the cevian triangle A'B'C'

of the point P with respect to the triangle

ABC. Then, I show that the lines through the

points A', B', C' parallel to the lines A*A#,

B*B#, C*C# concur at one point R. Finally, I

prove (and this is very easy) that the

midpoint R' of the segment QR lies on the

lines A*A#, B*B#, C*C#, and thus these lines

concur, i. e. the triangles A*B*C* and A#B#C#

are perspective. Actually, the point R is the

P-Ceva conjugate of the point Q; this fact

(which is due to Bernard Gibert, Hyacinthos

message #10136) I didn't prove in my note,

but it can be easily obtained using the

auxiliary points I constructed there.

Sincerely,

Darij Grinberg