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## One point, One Line Conjugate

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• Dear All My Friends, I suddenly see that we can generalize the fact which I discovered in Jean-Luis Ayme message as follow: Given triangle ABC, any point P,
Message 1 of 6 , Jun 25, 2006
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Dear All My Friends,
I suddenly see that we can generalize the fact which I discovered in Jean-Luis Ayme message as follow:
Given triangle ABC, any point P, any line L. The line L cuts three sidelines BC, CA, AB at A', B', C' respectively. One line La passes through A' and perpendicular to line AP. Similarly define Lb, Lc. Three lines La, Lb, Lc bound one triangle A1B1C1.
Results:
1. A1B1C1 and ABC are perspective with perspector Q.
With each point P and each lines L we can map to one Q. We denote this map as M(P, L). So we can write Q = M(P, L). This map also generates one triangle A1B1C1, we denote the triangle map as T(P, L). So we can write: A1B1C1 = T(P, L).
2. It is interesting that if Q = M(P, L) then M(Q, L) = P. We can say M is one conjugate.
3. If Q = M(P, L) and A1B1C1 = T(P, L) and A2B2C2 = T(Q, L) then A1B1C1 and A2B2C2 are perspective at one point D.
Dear All My Friends,
What is the type of this conjugate? What is coordinates of Q, D? There are some thing other interesting?
Please kindly help me to know about it!
Best regards,
Bui Quang Tuan

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• Dear All My Friends, I see some facts of this conjugate: 1. P, Q, D are collinear on one line L* perpendicular to L. 2. Triangle A2B2C2 and ABC are perspective
Message 2 of 6 , Jun 25, 2006
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Dear All My Friends,
I see some facts of this conjugate:
1. P, Q, D are collinear on one line L* perpendicular to L.
2. Triangle A2B2C2 and ABC are perspective at P (so it generates conjugate feature).
3. Sidelines of A1B1C1 and A2B2C2, by pair, intersect on line L. As I know L is their perspective axis or perspectrix.
These facts may be useful for the identification of this type of conjugate.
Best regards,
Bui Quang Tuan

Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
I suddenly see that we can generalize the fact which I discovered in Jean-Luis Ayme message as follow:
Given triangle ABC, any point P, any line L. The line L cuts three sidelines BC, CA, AB at A', B', C' respectively. One line La passes through A' and perpendicular to line AP. Similarly define Lb, Lc. Three lines La, Lb, Lc bound one triangle A1B1C1.
Results:
1. A1B1C1 and ABC are perspective with perspector Q.
With each point P and each lines L we can map to one Q. We denote this map as M(P, L). So we can write Q = M(P, L). This map also generates one triangle A1B1C1, we denote the triangle map as T(P, L). So we can write: A1B1C1 = T(P, L).
2. It is interesting that if Q = M(P, L) then M(Q, L) = P. We can say M is one conjugate.
3. If Q = M(P, L) and A1B1C1 = T(P, L) and A2B2C2 = T(Q, L) then A1B1C1 and A2B2C2 are perspective at one point D.
Dear All My Friends,
What is the type of this conjugate? What is coordinates of Q, D? There are some thing other interesting?
Please kindly help me to know about it!
Best regards,
Bui Quang Tuan

---------------------------------
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• Dear All My Friends, Point D, the perspector of two triangle A1B1C1 and A2B2C2 is alway on one special line: the orthocentric line of complete quadrilateral
Message 3 of 6 , Jun 26, 2006
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Dear All My Friends,
Point D, the perspector of two triangle A1B1C1 and A2B2C2 is alway on one special line: the orthocentric line of complete quadrilateral bounded by ABC and line L. (The collinear line of four orthocenters of four triangles of this complete quadrilateral). Therefor D can be easy determined by intersection of this orthocentric line and the line from P perpendicular to L.
Best regards,
Bui Quang Tuan

Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
I see some facts of this conjugate:
1. P, Q, D are collinear on one line L* perpendicular to L.
2. Triangle A2B2C2 and ABC are perspective at P (so it generates conjugate feature).
3. Sidelines of A1B1C1 and A2B2C2, by pair, intersect on line L. As I know L is their perspective axis or perspectrix.
These facts may be useful for the identification of this type of conjugate.
Best regards,
Bui Quang Tuan

Quang Tuan Bui wrote: Dear All My Friends,
I suddenly see that we can generalize the fact which I discovered in Jean-Luis Ayme message as follow:
Given triangle ABC, any point P, any line L. The line L cuts three sidelines BC, CA, AB at A', B', C' respectively. One line La passes through A' and perpendicular to line AP. Similarly define Lb, Lc. Three lines La, Lb, Lc bound one triangle A1B1C1.
Results:
1. A1B1C1 and ABC are perspective with perspector Q.
With each point P and each lines L we can map to one Q. We denote this map as M(P, L). So we can write Q = M(P, L). This map also generates one triangle A1B1C1, we denote the triangle map as T(P, L). So we can write: A1B1C1 = T(P, L).
2. It is interesting that if Q = M(P, L) then M(Q, L) = P. We can say M is one conjugate.
3. If Q = M(P, L) and A1B1C1 = T(P, L) and A2B2C2 = T(Q, L) then A1B1C1 and A2B2C2 are perspective at one point D.
Dear All My Friends,
What is the type of this conjugate? What is coordinates of Q, D? There are some thing other interesting?
Please kindly help me to know about it!
Best regards,
Bui Quang Tuan

---------------------------------
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• Dear All My Friends, Two points P and Q (conjugate of P) are always on one rectangular circumhyperbola of ABC. Therefore if given P and L, we can construct Q
Message 4 of 6 , Jun 26, 2006
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Dear All My Friends,
Two points P and Q (conjugate of P) are always on one rectangular circumhyperbola of ABC. Therefore if given P and L, we can construct Q as following:
- Construct conic (Co) passing through A, B, C, P and H, orthocenter of ABC
- Construct L* passing through P and perpendicular with L
- L* cuts (Co) at two points: one is P, another is Q.
Please note that:
- The conic (Co) is independent of L. It is determined completely only by P.
- If L is changed but always parallel with one fixed direction so Q is not changed. Here L is only important by its direction.
- We can see this conjugate as direction conjugate based on rectangular circumhyperbola.
Based on these analyses we can construct similar conjugate with arbitrary conic as follow:
Given triangle ABC, point T, line L. With any point P we can construct Q = M(P, T, L) as follow:
- Construct conic (Co) passing through A, B, C, T and P.
- Construct line L* passing through P and perpendicular with L.
- L* cuts (Co) at two points: one is P, another is Q.
Easy to show that if Q = M(P, T, L) then M(Q, T, L) = P
Remarks:
- The conic (Co) depend on point P only. I can be circle, ellipse, parabola or hyperbola.
- Line L is used for direction only.
- What we can do with circumconic so we can do the same with inconic.
- We can choose another one to one map f instead of L direction. The map f determined on points of conic and should be "conjugate" i.e. f(f(X)) = X
- If T is one special point we can have some special case of this map. (Before is one example when T = H).
- Some interesting features can be seen when T is some special points (so conic is some special conics) and P is also some special points.
- I don't know if in general case Q = M(P, T, L) we can construct by compass and ruler (without conic using) as in the case when T = H?
- I don't know if we can choose another curve instead of conic or not?
Dear All My Friends,
That all what I can understand of this strange conjugate type.
If any one have good idea or nice results so please kindly post here. I am very happy to read any advices!
Best regards,
Bui Quang Tuan

Quang Tuan Bui <bqtuan1962@...> wrote: Dear All My Friends,
Point D, the perspector of two triangle A1B1C1 and A2B2C2 is alway on one special line: the orthocentric line of complete quadrilateral bounded by ABC and line L. (The collinear line of four orthocenters of four triangles of this complete quadrilateral). Therefor D can be easy determined by intersection of this orthocentric line and the line from P perpendicular to L.
Best regards,
Bui Quang Tuan

Quang Tuan Bui wrote: Dear All My Friends,
I see some facts of this conjugate:
1. P, Q, D are collinear on one line L* perpendicular to L.
2. Triangle A2B2C2 and ABC are perspective at P (so it generates conjugate feature).
3. Sidelines of A1B1C1 and A2B2C2, by pair, intersect on line L. As I know L is their perspective axis or perspectrix.
These facts may be useful for the identification of this type of conjugate.
Best regards,
Bui Quang Tuan

Quang Tuan Bui wrote: Dear All My Friends,
I suddenly see that we can generalize the fact which I discovered in Jean-Luis Ayme message as follow:
Given triangle ABC, any point P, any line L. The line L cuts three sidelines BC, CA, AB at A', B', C' respectively. One line La passes through A' and perpendicular to line AP. Similarly define Lb, Lc. Three lines La, Lb, Lc bound one triangle A1B1C1.
Results:
1. A1B1C1 and ABC are perspective with perspector Q.
With each point P and each lines L we can map to one Q. We denote this map as M(P, L). So we can write Q = M(P, L). This map also generates one triangle A1B1C1, we denote the triangle map as T(P, L). So we can write: A1B1C1 = T(P, L).
2. It is interesting that if Q = M(P, L) then M(Q, L) = P. We can say M is one conjugate.
3. If Q = M(P, L) and A1B1C1 = T(P, L) and A2B2C2 = T(Q, L) then A1B1C1 and A2B2C2 are perspective at one point D.
Dear All My Friends,
What is the type of this conjugate? What is coordinates of Q, D? There are some thing other interesting?
Please kindly help me to know about it!
Best regards,
Bui Quang Tuan

---------------------------------
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• Dear Tuan, A very quick thought .. As your conjugate is direction based, we can consider its meet with infinity, say X. Q will now be on the line P, oX, where
Message 5 of 6 , Jun 27, 2006
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Dear Tuan,

A very quick thought ..

As your conjugate is direction based, we can consider its meet with infinity, say X.
Q will now be on the line P, oX, where oX is the orthopoint of X.

For oX{x,y,z} and some P{p,q,r}, your conjugate is ..

(q SB - r SC) / (y r - z q) ::

For example, some line parallel to the OI line will meet infinity at X=X(517), the orthopoint, oX = X(513).
If P = X(7), then Q = X(885).

For whatever it is worth, (q SB - r SC) is an orthopoint operation and (y r - z q) is the crossdifference.

Best regards,
Peter.

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• Dear Peter, Thank you for your deeply analysis. The formula is not very complicated. I am thinking now about two problems: 1. I see in some special cases: A
Message 6 of 6 , Jun 27, 2006
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Dear Peter,
Thank you for your deeply analysis. The formula is not very complicated. I am thinking now about two problems:
1. I see in some special cases:
''A little result'' when L=OI and P=I
your example when L=OI and P=X(7)
the circumcircle of A1B1C1 is always touching circumcircle of ABC at one point and circumcircle of A2B2C2 passes through this point also.
2. If we choose one degenereted conic (two lines) so we have one trivial conjugate map. Its construction is very easy:
Given one point T and one line L (direction). We can map any point P to point Q by following way:
- Draw line L1 passes through P and T
- Draw line L2 passes through T and perpendicular to L1. (rectangular map)
- Draw line L3 passes through P and perpendicular to L. (for direction only)
- Q is intersection of L3 and L2
(The angle between L2 and L1 can be any other fixed direction angle. In this case we will have non-rectangular map).
Best regards,
Bui Quang Tuan

"Moses, Peter J. C." <mows@...> wrote: Dear Tuan,

A very quick thought ..

As your conjugate is direction based, we can consider its meet with infinity, say X.
Q will now be on the line P, oX, where oX is the orthopoint of X.

For oX{x,y,z} and some P{p,q,r}, your conjugate is ..

(q SB - r SC) / (y r - z q) ::

For example, some line parallel to the OI line will meet infinity at X=X(517), the orthopoint, oX = X(513).
If P = X(7), then Q = X(885).

For whatever it is worth, (q SB - r SC) is an orthopoint operation and (y r - z q) is the crossdifference.

Best regards,
Peter.

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