## Question on conics

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• It is known that if we have two common points of two conics and other sufficient elements of each one, we can construct the other two common points; the same
Message 1 of 15 , May 12 12:48 PM
It is known that if we have two common points of two conics and other
sufficient elements of each one, we can construct the other two common
points; the same is true for the common tangents. My question is: is it
true that if we know two common tangents (and all other elements of
each conic) we can construct the common points? And if this is true,
how can this be accomplished ?
Thanks for all illuminations

Giovanni Artico
• To construct a conic it is enough to know 5 data about it: 5 points, or 5 tangents, or combinations of these. If You know 2 tangents and 3 other points, You
Message 2 of 15 , May 12 12:52 PM
To construct a conic it is enough to know 5 data about it: 5 points, or 5 tangents, or combinations of these. If You know 2 tangents and 3 other points, You can construct the conic.

Martin

From: Giovanni Artico
Sent: Monday, May 12, 2008 9:48 PM
To: Hyacinthos@yahoogroups.com
Subject: [EMHL] Question on conics

It is known that if we have two common points of two conics and other
sufficient elements of each one, we can construct the other two common
points; the same is true for the common tangents. My question is: is it
true that if we know two common tangents (and all other elements of
each conic) we can construct the common points? And if this is true,
how can this be accomplished ?
Thanks for all illuminations

Giovanni Artico

[Non-text portions of this message have been removed]
• Dear Giovanni! I don t understand the problem. What are the given elements of conics and which means can we use for construction? Sincerely
Message 3 of 15 , May 12 11:10 PM
Dear Giovanni!
I don't understand the problem. What are the given elements of conics and which means can we use for construction?

Sincerely Alexey

It is known that if we have two common points of two conics and other
sufficient elements of each one, we can construct the other two common
points; the same is true for the common tangents. My question is: is it
true that if we know two common tangents (and all other elements of
each conic) we can construct the common points? And if this is true,
how can this be accomplished ?
Thanks for all illuminations

Giovanni Artico

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[Non-text portions of this message have been removed]
• R. Cuppens in Faire de la géométrie supérieure en jouant avec Cabri géomètre II writes: the problem of finding the common tangents to two conics is the
Message 4 of 15 , May 12 11:35 PM
R. Cuppens in 'Faire de la géométrie supérieure en jouant avec Cabri géomètre II' writes:

"the problem of finding the common tangents to two conics is the dual problem of finding the common points. If You have to construct the common points A, B, C, D of conic 1 and the porar conic of conic 2: the tangents to conic 1 by these points are the common tangents of conics 1 and 2".

I think it wold not be difficult to inverse this construction to find the common points having the common tangents.

Martin

To: Hyacinthos@yahoogroups.comFrom: zasl@...: Tue, 13 May 2008 10:10:01 +0400Subject: Re: [EMHL] Question on conics

Dear Giovanni! I don't understand the problem. What are the given elements of conics and which means can we use for construction? Sincerely Alexey It is known that if we have two common points of two conics and other sufficient elements of each one, we can construct the other two common points; the same is true for the common tangents. My question is: is it true that if we know two common tangents (and all other elements of each conic) we can construct the common points? And if this is true, how can this be accomplished ? Thanks for all illuminations Giovanni Artico Antivirus scanning: Symantec Mail Security for SMTP.[Non-text portions of this message have been removed]

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• Dear Martin and Giovanni ... There is a little problem : there are 4 conics going through 3 given points and touching 2 given lines. Of course, these 4 conics
Message 5 of 15 , May 12 11:54 PM
Dear Martin and Giovanni
> To construct a conic it is enough to know 5 data about it: 5 points,
> or 5 tangents, or combinations of these. If You know 2 tangents and 3
> other points, You can construct the conic.

There is a little problem : there are 4 conics going through 3 given
points and touching 2 given lines.
Of course, these 4 conics are constructible : if the three common
points are A,B,C and the common tangents are L1, L2 with trilinear
poles P1, P2 (wrt ABC), the perspectors of the 4 conics are the common
points of the inscribed conics with perspectors P1 and P2 and these
common points are constructible (because the degenerate conics of the
pencil generated these two inscribed conics are constructible pair of
lines intersecting respectively at the vertices of the precevian
triangle of L1 inter L2)
But now, if you want to construct the common points of two conics
knowing three points and to tangents for each conic, you have 16 pairs
of such conics and the complete construction (which is possible) will
be a bit tedious.
friendly. Jean-Pierre
• ... and which means can we use for construction? ... Example: two hyperbolae; I know the foci and vertices of each hyperbola; I know also the four common
Message 6 of 15 , May 13 12:36 AM
--- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@...>
wrote:
>
> Dear Giovanni!
> I don't understand the problem. What are the given elements of conics
and which means can we use for construction?
>

Example: two hyperbolae; I know the foci and vertices of each
hyperbola; I know also the four common tangents; by ruler and compass
is it possible to construct the four intersection points? If yes, how ?
Sincerely

Giovanni
• Dear Hyacinthists, I wrote ... given ... Here is a possible - and quite classical - construction of the conics going through A,B,C and touching the lines L1
Message 7 of 15 , May 13 5:08 AM
Dear Hyacinthists,
I wrote
> There is a little problem : there are 4 conics going through 3
given
> points and touching 2 given lines.
> Of course, these 4 conics are constructible.

Here is a possible - and quite classical - construction of the conics
going through A,B,C and touching the lines L1 and L2. (at most 4 real
solutions)
C1,C2 = common points of AB with L1,L2
B1,B2 = common points of AC with L1,L2
The radical axis of the circles (O) = ABC and C1C2C intersects AB at
Oc; the circle centered at Oc and orthogonal to (O) intersects AB at
P,P'.
Same way, the radical axis of the circles (O) and B1B2B intersects AC
at Ob; the circle centered at Ob and orthogonal to (O) intersects AC
at Q,Q'.
Now take L = PQ or PQ' or P'Q or P'Q'
If L intersects L1 and L2 at D1 and D2, the conic ABCD1D2 is a
solution

This due to the fact that
- P, P' are the fixed points of the involution on the line AB
swapping (A,B) and (C1,C2)
- Q, Q' are the fixed points of the involution on the line AC
swapping (A,C) and (B1,B2)
- As the line D1D2 (twice) is member of the pencil generated by the
required conic and the degenerated conic L1 union L2, the common
point of D1D2 and AB - or AC - must be a fixed point of the
involution i - ot i' -

Friendly. Jean-Pierre
• Knowing the foci and vertices You can construct as many points of the conic as You want. But I think that common tangents don t give information about common
Message 8 of 15 , May 13 5:22 AM
Knowing the foci and vertices You can construct as many points of the conic as You want. But I think that common tangents don't give information about common points. So my previous message is wrong: if You know how to construct the common points, You cant construct common tangents using the polar conic; but You can't construct common points having the common tangents, I think.

Martin

To: Hyacinthos@yahoogroups.comFrom: zeroprof@...: Tue, 13 May 2008 07:36:55 +0000Subject: Re: [EMHL] Question on conics

--- In Hyacinthos@yahoogroups.com, "Alexey.A.Zaslavsky" <zasl@...> wrote:>> Dear Giovanni!> I don't understand the problem. What are the given elements of conics and which means can we use for construction?> Example: two hyperbolae; I know the foci and vertices of each hyperbola; I know also the four common tangents; by ruler and compass is it possible to construct the four intersection points? If yes, how ?SincerelyGiovanni

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• Dear Martin ... conic as You want. But I think that common tangents don t give information about common points. The diagonal triangle of the complete
Message 9 of 15 , May 13 8:23 AM
Dear Martin

> Knowing the foci and vertices You can construct as many points of the
conic as You want. But I think that common tangents don't give

The diagonal triangle of the complete quadrilateral with sidelines the
4 common tangents = the diagonal triangle of the quadrangle with
vertices the 4 common points = the only triangle selfpolar wrt each
conic
Friendly. Jean-Pierre
• ... the ... Dear Jean-Pierre, I apologize, but to a not expert geometer as me this sounds a bit cryptic. I guess it means that the common points can be
Message 10 of 15 , May 13 1:11 PM
--- In Hyacinthos@yahoogroups.com, "jpehrmfr" <jean-pierre.ehrmann@...>
wrote:
> The diagonal triangle of the complete quadrilateral with sidelines
the
> 4 common tangents = the diagonal triangle of the quadrangle with
> vertices the 4 common points = the only triangle selfpolar wrt each
> conic
> Friendly. Jean-Pierre
>

Dear Jean-Pierre,
I apologize, but to a not expert geometer as me this sounds a bit
cryptic. I guess it means that the common points can be constructed,
and in the construction is involved a selfpolar triangle: but could you
briefly outline the principal steps of the construction? Many thanks in

Giovanni
• Dear Giovanni [JPE] ... [GA] ... you ... No, you cannot find the common points if you know only the common tangents (because there are infinitely many conics
Message 11 of 15 , May 14 12:20 AM
Dear Giovanni
[JPE]
> > The diagonal triangle of the complete quadrilateral with sidelines
> the
> > 4 common tangents = the diagonal triangle of the quadrangle with
> > vertices the 4 common points = the only triangle selfpolar wrt each
> > conic
[GA]
> I guess it means that the common points can be constructed,
> and in the construction is involved a selfpolar triangle: but could
you
> briefly outline the principal steps of the construction?

No, you cannot find the common points if you know only the common
tangents (because there are infinitely many conics with 4 given
tangents).
I just wanted to say that there exists is a very close relation between
the complete quadrilateral with sidelines the 4 common tangents, the
quadrangle with vertices the 4 common points and the triangle selfpolar
wrt both conics.
Of course, if you know one common point M and the 4 common tangents,
the remark above gives an immediate construction (with ruler only) of
the 3 other common points : they are the vertices of the precevian
triangle of M wrt to the diagonal triangle of the complete quadrileral
with sidelines the 4 common tangents.
Same way, if you know the 4 common point and one common tangent, you
can construct the 3 other common tangents.
Friendly. Jean-Pierre
• ... The question is a bit different: if I know four common tangents AND another point on each conic, can I construct the four intersection points? (in this
Message 12 of 15 , May 14 9:15 AM
> No, you cannot find the common points if you know only the common
> tangents (because there are infinitely many conics with 4 given
> tangents).

The question is a bit different: if I know four common tangents AND
another point on each conic, can I construct the four intersection
points? (in this case there are not infinitely many conics).
The problem could be restated : if I know five points of each conic AND
four of them are common, can I costruct the four common tangents?
I don't have so far an answer to this question.
Many thanks for your help Jean-Pierre

Giovanni
• Dear Giovanni ... AND ... Yes, of course. Here is a possible solution (there is probably a better solution): 1)Find the diagonal triangle A B C of the
Message 13 of 15 , May 15 8:35 AM
Dear Giovanni

> The question is a bit different: if I know four common tangents AND
> another point on each conic, can I construct the four intersection
> points? (in this case there are not infinitely many conics).
> The problem could be restated : if I know five points of each conic
AND
> four of them are common, can I costruct the four common tangents?
> I don't have so far an answer to this question.
> Many thanks for your help Jean-Pierre

Yes, of course.
Here is a possible solution (there is probably a better solution):
1)Find the diagonal triangle A'B'C' of the quadrangle with vertices
the 4 common points
2) Find the centers W,W' of the two conics
3) If the line WW' intersects B'C' at S, find the points U1, U2 on
the line B'C' with midpoint S and harmonic conjugates wrt (B',C')
These points are opposite vertices of the complete quadrilateral with
sidelines the 4 common tangents
Do the same with the lines WW' and C'A' to get the 4 common tangents.

Friendly. Jean-Pierre
• Merci beaucoup Jean-Pierre Giovanni
Message 14 of 15 , May 15 12:03 PM
Merci beaucoup Jean-Pierre

Giovanni
• Dear Giovanni [Giovanni] ... AND ... intersection ... conic ... [JP] ... with ... tangents. I ve just used two simple facts : - U1,U2 are harmonic conjugates
Message 15 of 15 , May 15 2:34 PM
Dear Giovanni
[Giovanni]
> > The question is a bit different: if I know four common tangents
AND
> > another point on each conic, can I construct the four
intersection
> > points? (in this case there are not infinitely many conics).
> > The problem could be restated : if I know five points of each
conic
> AND
> > four of them are common, can I costruct the four common tangents?
> > I don't have so far an answer to this question.
[JP]
> Yes, of course.
> Here is a possible solution (there is probably a better solution):
> 1)Find the diagonal triangle A'B'C' of the quadrangle with vertices
> the 4 common points
> 2) Find the centers W,W' of the two conics
> 3) If the line WW' intersects B'C' at S, find the points U1, U2 on
> the line B'C' with midpoint S and harmonic conjugates wrt (B',C')
> These points are opposite vertices of the complete quadrilateral
with
> sidelines the 4 common tangents
> Do the same with the lines WW' and C'A' to get the 4 common
tangents.

I've just used two simple facts :
- U1,U2 are harmonic conjugates wrt B', C'
- their midpoint lies on the Newton line of the complete
quadrilateral with vertices the four common tangents and this Newton
line is the line WW'
In fact, there is probably something better if, instead of the
centers of the conics, we use the poles of a well choosen line wrt
the conics.
Friendly. Jean-Pierre
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