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## Paracevian perspector ?

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• 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
Message 1 of 4 , Jul 23, 2004
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 ... Your S is the midpoint of Q and P-Ceva conj. of Q. Best regards Bernard [Non-text portions of this message have been removed]
Message 2 of 4 , Jul 23, 2004
Dear Eric

> [ED] 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...)

^^^^^^^^^^^^^^^^^^^^^^^+ 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 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

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

Best regards

Bernard

[Non-text portions of this message have been removed]
• Dear Eric and Bernard ... cyclic...) ... lead ... [BG] ... This leads to a construction of Q if we know P and S but I don t understand why we cannot get 4
Message 3 of 4 , Jul 23, 2004
Dear Eric and Bernard
> > [ED] 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...)
>
> ^^^^^^^^^^^^^^^^^^^^^^^+ 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
> > midpoint of Q1 and Q2.
> >
> > However I can't see why, using the formula above

[BG]
> 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
• ... Cheers, Eric, for finding such a beautiful theorem! It was indeed too beautiful for me to leave it without a proper synthetic proof, so I have been looking
Message 4 of 4 , Sep 1, 2004
In Hyacinthos message #10135, Eric Danneels wrote:

>> Consider a triangle ABC and 2 different
>> 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

Cheers, Eric, for finding such a beautiful
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
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