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What I don't like about the DrozFarny theorem
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Dear all,
In recent times there have been some discussions on proofs of the following
theorem by DrozFarny:
Let L1 and L2 be perpendicular and passing through H. The intersections of
Li with the sides of ABC are Ai, Bi and Ci respectively. The midpoints of
A1A2, B1B2 and C1C2 are collinear.
In some versions of this theorem the collinearity of the midpoints is
replaced by the coaxiality of the circles with diameters A1A2, B1B2 and
C1C2.
There is something I don't like about this theorem. My main objection is
that it is not necessary to take midpoints. Any points A3, B3 and C3
dividing the segments A1A2, B1B2 and C1C2 in the same ratio will do. Of
course it turns out that A1B1C1, A2B2C2 and A3B3C3 are 'similar' inscribed
degenerate triangles. In stead of these lines L1 and L2 we could have taken
any two (directly) similar inscribed triangles A1B1C1 and A2B2C2 and we
would find A3B3C3 similar to these.
Going back to the A1B1C1 and A2B2C2 perpendicular lines through H:
The more crucial fact of the DF theorem is IMHO (and here I repeat myself)
that A1B1C1 and A2B2C2 are 'similar' degenerate triangles. Or, stated in a
more classical way: the circles (AB1C1), (AB2C2), (A1BC1), (A2BC2), (A1B1C),
(A2B2C) intersect in one point on (ABC). This implies the stronger form of
DF, not only the midpoints.
Kind regards,
Floor. 0 Attachment
Dear Floor,
In Hyacinthos message #10716, you wrote:
>> Any points A3, B3 and C3 dividing the segments A1A2,
This really opened my eyes! The degenerate triangles
>> B1B2 and C1C2 in the same ratio will do. Of course
>> it turns out that A1B1C1, A2B2C2 and A3B3C3 are
>> 'similar' inscribed degenerate triangles.
A1B1C1 and A2B2C2 are similar; in other words, we have
B1C1 / C1A1 = B2C2 / C2A2. Therefore, we can find two
reals k and l with k + l = 1 such that
C1 = k A1 + l B1 and C2 = k A2 + l B2, where we
identify points with vectors from an (arbitrarily
chosen) origin. Now, if the points A3, B3, C3 divide
the segments A1A2, B1B2, C1C2 in the same ratio, we
can find two reals u and v with u + v = 1 such that
A3 = u A1 + v A2, B3 = u B1 + v B2 and
C3 = u C1 + v C2. Thus,
C3 = u C1 + v C2 = u (k A1 + l B1) + v (k A2 + l B2)
= k (u A1 + v A2) + l (u B1 + v B2)
= k A3 + l B3,
and since k + l = 1, it follows that the point C3
lies on the line A3B3, i. e. the points A3, B3, C3
are collinear. Also, it is now clear that
B1C1 : C1A1 : A1B1 = B2C2 : C2A2 : A2B2
= B3C3 : C3A3 : A3B3.
Now this explains really a lot. The DrozFarny
theorem turns out to be even simpler than I thought.
I must say I still like the DrozFarny theorem, but
now I see the most crucial part of it is the
similarity of the degenerate triangles A1B1C1 and
A2B2C2 rather than the coaxality of the circles.
Sincerely,
Darij Grinberg 0 Attachment
Dear Darij,
[DG]> Now this explains really a lot. The DrozFarny
Of course I still like the DFtheorem. I only think it needs be restated, to
> theorem turns out to be even simpler than I thought.
> I must say I still like the DrozFarny theorem, but
> now I see the most crucial part of it is the
> similarity of the degenerate triangles A1B1C1 and
> A2B2C2 rather than the coaxality of the circles.
see the complete figure, the reason why DF is true. It seems that you agree.
Your use of vectors to show this is really clear and simple.
Using Miquel theory is as simple  when M is the Miquel point, then
triangles A1MA2, B1MB2, C1MC2 are similar, and hence so are A1MA3, B1MB3 and
C1MC3, but that means that A3B3C3 is a result of pivoting as well, and thus
similar.
Kind regards,
Floor. 0 Attachment
Dear Floor,
[FvL]: In recent times there have been some discussions on proofs of
the following theorem by DrozFarny:
Let L1 and L2 be perpendicular and passing through H. The
intersections of Li with the sides of ABC are Ai, Bi and Ci
respectively. The midpoints of A1A2, B1B2 and C1C2 are collinear.
... Tere is something I don't like about this theorem. My main
objection is that it is not necessary to take midpoints. Any points
A3, B3 and C3 dividing the segments A1A2, B1B2 and C1C2 in the same
ratio will do.
*** It is very interesting. The envelope of the line containing A3,
B3, C3, as the ratio varies, is the inscribed parabola tangent to L1
and L2.
Best regards
Sincerely
Paul 0 Attachment
Dear Floor,
[FvL]: In recent times there have been some discussions on proofs of
the following theorem by DrozFarny:
Let L1 and L2 be perpendicular and passing through H. The
intersections of Li with the sides of ABC are Ai, Bi and Ci
respectively. The midpoints of A1A2, B1B2 and C1C2 are collinear.
*** As the perpendicular lines rotate about H, the envelope of the
DrozFarny line is the inscribed ellipse with foci O and H (hence
center N).
Best regards
Sincerely
Paul 0 Attachment
Dear Floor,
[FvL]: In recent times there have been some discussions on proofs of
the following theorem by DrozFarny:
Let L1 and L2 be perpendicular and passing through H. The
intersections of Li with the sides of ABC are Ai, Bi and Ci
respectively. The midpoints of A1A2, B1B2 and C1C2 are collinear.
*** For two fixed perpendicular directions, consider the lines L1 and
L2 through a point P with these directions, intersecting the
sidelines of ABC at Ai, Bi, Ci (for i = 1, 2). The locus of P for
which the midpoints of A1A2, B1B2, C1C2 are collinear is the
rectangular circumhyperbola with asymptotes along the given
directions.
Best regards
Sincerely
Paul 0 Attachment
Dear Paul,
> [FvL]: In recent times there have been some discussions on proofs of
[PY]
> the following theorem by DrozFarny:
>
> Let L1 and L2 be perpendicular and passing through H. The
> intersections of Li with the sides of ABC are Ai, Bi and Ci
> respectively. The midpoints of A1A2, B1B2 and C1C2 are collinear.
> ... There is something I don't like about this theorem. My main
> objection is that it is not necessary to take midpoints. Any points
> A3, B3 and C3 dividing the segments A1A2, B1B2 and C1C2 in the same
> ratio will do.
> *** It is very interesting. The envelope of the line containing A3,
Let P be a point on the circumcircle and A'B'C' be its SimsonWallace line.
> B3, C3, as the ratio varies, is the inscribed parabola tangent to L1
> and L2.
Isn't for each P the collection of pivoted "triangles" A'B'C' a parabola?
Kind regards,
Floor. 0 Attachment
Dear Paul,
> [FvL]: In recent times there have been some discussions on proofs of
[PY] For two fixed perpendicular directions, consider the lines L1 and
> the following theorem by DrozFarny:
>
> Let L1 and L2 be perpendicular and passing through H. The
> intersections of Li with the sides of ABC are Ai, Bi and Ci
> respectively. The midpoints of A1A2, B1B2 and C1C2 are collinear.
> L2 through a point P with these directions, intersecting the
For two general fixed directions, the locus of P as above is the
> sidelines of ABC at Ai, Bi, Ci (for i = 1, 2). The locus of P for
> which the midpoints of A1A2, B1B2, C1C2 are collinear is the
> rectangular circumhyperbola with asymptotes along the given
> directions.
circumhyperbola with asymptotes along the given directions.
In stead of midpoints of A1A2, B1B2 and C1C2 we may again take any point
dividing these segments in the same ratio, and the locus stays the same.
Kind regards,
Floor. 0 Attachment
Dear Floor and Paul,
> [FvL] In recent times there have been some discussions on proofs of
[PY] For two fixed perpendicular directions, consider the lines L1 and
> the following theorem by DrozFarny:
>
> Let L1 and L2 be perpendicular and passing through H. The
> intersections of Li with the sides of ABC are Ai, Bi and Ci
> respectively. The midpoints of A1A2, B1B2 and C1C2 are collinear.
> L2 through a point P with these directions, intersecting the
For any point P different from orthocenter H there is unique
> sidelines of ABC at Ai, Bi, Ci (for i = 1, 2). The locus of P for
> which the midpoints of A1A2, B1B2, C1C2 are collinear is the
> rectangular circumhyperbola with asymptotes along the given
> directions.
> [FvL]
>For two general fixed directions, the locus of P as above is the
>circumhyperbola with asymptotes along the given directions.
pair of two perpendiculars through P with DrozFarny
property(midpoints of A1A2,B1B2,C1C2 are collinear).
I have mentioned this result earlier when we have tried to
find good proof of DrozFarny theorem..
I have two questions.
1.What is the reason that for orthocenter any pair of
two perpendiculars through H has DF property,but
for any other point there is only one pair of perpendiculars
through it with DF property?
2.What is the meaning of this unique pair of perpendiculars
through point P different form orthocenter H,with DF property?
Are they asymptotes of some hyperbola?
Best regards
Milorad R.Stevanovic
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Dear Floor, Milorad and Paul,
> [FvL] In recent times there have been some discussions on proofs of
The locus of P for which the midpoints of A1A2, B1B2, C1C2 form a
> > the following theorem by DrozFarny:
> >
> > Let L1 and L2 be perpendicular and passing through H. The
> > intersections of Li with the sides of ABC are Ai, Bi and Ci
> > respectively. The midpoints of A1A2, B1B2 and C1C2 are collinear.
>
> [PY] For two fixed perpendicular directions, consider the lines L1 and
> > L2 through a point P with these directions, intersecting the
> > sidelines of ABC at Ai, Bi, Ci (for i = 1, 2). The locus of P for
> > which the midpoints of A1A2, B1B2, C1C2 are collinear is the
> > rectangular circumhyperbola with asymptotes along the given
> > directions.
>
> > [FvL] For two general fixed directions, the locus of P as above is
> the
> >circumhyperbola with asymptotes along the given directions.
triangle perspective to ABC is a central pK
 whose pole is the perspector of the circumhyperbola above,
 whose center is the center of the circumhyperbola,
 whose pivot is the anticomplement of the isoconjugate of the center.
This gives another characterization of the central pKs as seen in
Special Isocubics §3.1.3.
Best regards
Bernard
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Dear Milorad,
[MS]:> For any point P different from orthocenter H there is unique
I think the following is also true:
> pair of two perpendiculars through P with DrozFarny
> property(midpoints of A1A2,B1B2,C1C2 are collinear).
> I have mentioned this result earlier when we have tried to
> find good proof of DrozFarny theorem..
When we consider pairs of lines making a fixed acute angle, then for any
point P different from H,A,B,C there are two pairs with the DFproperty. For
H there are none.
Can someone confirm?
Kind regards,
Floor. 0 Attachment
Dear Floor and Milorad> [MS]:
Two lines L, L' have the DrozFarny property if and only if they
> > For any point P different from orthocenter H there is unique
> > pair of two perpendiculars through P with DrozFarny
> > property(midpoints of A1A2,B1B2,C1C2 are collinear).
> > I have mentioned this result earlier when we have tried to
> > find good proof of DrozFarny theorem..
touch the same inscribed parabola (in which case the line joining
the midpoints touch the parabola too)
Hence, if P<>H, there is an unique pair of perpendiculars through P
with DrozFarny property : the tangents from P to the inscribed
parabola with directrix HP (this gives an easy construction of these
lines)
[FvL]> I think the following is also true:
for any
> When we consider pairs of lines making a fixed acute angle, then
> point P different from H,A,B,C there are two pairs with the DF
property. For
> H there are none.
I don't think so. If t is the measure in [Pi/2,Pi/2] of the
required line angle, I think that there exist two pairs of solutions
if and only if OP*>R cos(t) where P* = isogonal conjugate of P.
Of course, if P* is outside the circumcircle, there are two pairs of
solutions for any t.
If Ta is the part of the plane limited by the halflines AB, AC and
the segment BC and opposite to A, P* is outside the circumcircle if
and only if P lies in Ta union Tb union Tc.
Friendly. JeanPierre
Friendly. JeanPierre 0 Attachment
Dear JeanPierre and Milorad,
> > [MS]:
[JPE]
> > > For any point P different from orthocenter H there is unique
> > > pair of two perpendiculars through P with DrozFarny
> > > property(midpoints of A1A2,B1B2,C1C2 are collinear).
> > > I have mentioned this result earlier when we have tried to
> > > find good proof of DrozFarny theorem..
> Two lines L, L' have the DrozFarny property if and only if they
Nice!
> touch the same inscribed parabola (in which case the line joining
> the midpoints touch the parabola too)
> Hence, if P<>H, there is an unique pair of perpendiculars through P
> with DrozFarny property : the tangents from P to the inscribed
> parabola with directrix HP (this gives an easy construction of these
> lines)
In accordance with H for the perpendicular pairs of lines, for any P the
pairs of Pperpendicular through P lines have the DFproperty. For each P
there is one pair of directions that is as well perpendicular as
Pperpendicular. These are the directions of asymptotes of the
circumhyperbola through H and P.
> [FvL]
[JPE]:
> > I think the following is also true:
> > When we consider pairs of lines making a fixed acute angle, then
> for any
> > point P different from H,A,B,C there are two pairs with the DF
> property. For
> > H there are none.
> I don't think so. If t is the measure in [Pi/2,Pi/2] of the
I think you are right, and I have mistaken here. Thanks!
> required line angle, I think that there exist two pairs of solutions
> if and only if OP*>R cos(t) where P* = isogonal conjugate of P.
> Of course, if P* is outside the circumcircle, there are two pairs of
> solutions for any t.
> If Ta is the part of the plane limited by the halflines AB, AC and
> the segment BC and opposite to A, P* is outside the circumcircle if
> and only if P lies in Ta union Tb union Tc.
> Friendly. JeanPierre
Kind regards,
Floor. 0 Attachment
Dear Floor and Milorad> [JPE]
through P
> > Two lines L, L' have the DrozFarny property if and only if they
> > touch the same inscribed parabola (in which case the line joining
> > the midpoints touch the parabola too)
> > Hence, if P<>H, there is an unique pair of perpendiculars
> > with DrozFarny property : the tangents from P to the inscribed
these
> > parabola with directrix HP (this gives an easy construction of
> > lines)
P the
> [FvL]
> In accordance with H for the perpendicular pairs of lines, for any
> pairs of Pperpendicular through P lines have the DFproperty. For
each P
> there is one pair of directions that is as well perpendicular as
Of course, we have another characterization :
> Pperpendicular. These are the directions of asymptotes of the
> circumhyperbola through H and P.
A pair of lines L, L' intersecting at P have the DFproperty if and
only if the circumhyperbola going through the infinite points of L
and L'goes through P.
Friendly. JeanPierre
>
then
> > [FvL]
> > > I think the following is also true:
> > > When we consider pairs of lines making a fixed acute angle,
> > for any
solutions
> > > point P different from H,A,B,C there are two pairs with the DF
> > property. For
> > > H there are none.
>
> [JPE]:
> > I don't think so. If t is the measure in [Pi/2,Pi/2] of the
> > required line angle, I think that there exist two pairs of
> > if and only if OP*>R cos(t) where P* = isogonal conjugate of P.
pairs of
> > Of course, if P* is outside the circumcircle, there are two
> > solutions for any t.
and
> > If Ta is the part of the plane limited by the halflines AB, AC
> > the segment BC and opposite to A, P* is outside the circumcircle
if
> > and only if P lies in Ta union Tb union Tc.
> > Friendly. JeanPierre
>
> I think you are right, and I have mistaken here. Thanks!
>
> Kind regards,
> Floor. 0 Attachment
If P<> H, there is an unique point Q on the
circle through ABC so that HP is the Steinerline of Q wrt ABC. Then
the DrozFarny line of P is the perpendicular bissector of PQ.
If P=H, then we can take any point Q on the ABCcircle and so one.....
Is that true?
F.Rideau 0 Attachment
Dear Francois, Floor and Milorad> If P<> H, there is an unique point Q on the
Then
> circle through ABC so that HP is the Steinerline of Q wrt ABC.
> the DrozFarny line of P is the perpendicular bissector of PQ.
Yes.
> Is that true?
Consider the points U1, U2 on the line HP such as PU1 = PU2 = PQ;
the perpendicular bisector L1  or L2  of QU1  or QU2  intersects
the sidelines of ABC at A1,B1,C1  or A2, B2, C2 
Then the midpoints of A1A2, B1B2, C1C2 lie on the perpendicular
bisector of PQ and, of course, L1 and L2 are perpendicular.
This is related to the fact that the inscribed parabola with
directrix HP is the envelope of the perpendicular bisectors of QM,
where M lies on HP.
Now come back to Floor's problem : given a point P and an angle T,
find the pairs of lines (L,L') with the DFproperty, intersecting at
P and such as <L,L' = T.
IF MM' is a choird of the circumcircle, if d is the distance from O
to MM', the cosinus of the angle of two lines with infinite points
the isogonal conjugates of M and M' is d/R.
As the circumhyperbola through the infinite points of L and L' must
go through P, the isogonal conjugate P* of P must lie on the line
MM', where M and M' are the isogonal conjugates of the infinite
points of L and L'.
Hence the construction : draw the tangents from P* to the circle (O,
R cos(T)); if one of these tangents intersect the circumcircle at M
and M', we have the solution (PM*, PM'*)
Obviously, we have two distinct solutions iff OP*>R cos(T) and no
solution iff OP* < R cos(T)
Friendly. JeanPierre 0 Attachment
My dear JeanPierre
You know what? I am happy. It's the first time I get an answer. I was
looking at Hyacinthos Forum
since a few week without understanding all the discussions. It seems
to me most of the questions
are too Euclidian oriented. There is also some interesting problems in
projective and affine geometry.
The DrozFarny problems raises some new questions I don't know the answers.
For example, each point P<>H provides 2 orthogonal lines, what are the
associated flows? What about these flows around H?
Here another similar problem in affine geometry: Let ABC a triangle in
an affine real plane.
We know that through each point P there his 2 isotomic lines. What are
the associated flows? Besides it's a good problem to draw these 2
isotomic lines through P, using Cabri or another software but I am
interested in the flows.
It's relatively easy to write the differential equations providing these flows
but it's another matter to draw them! Maybe with Maple or Mathematica?
Friendly
François 0 Attachment
Dear Francois,
> We know that through each point P there his 2 isotomic lines. What are
what do you exactly mean by "flow" ?
> the associated flows?
> Besides it's a good problem to draw these 2
I suppose you call "isotomic lines" 2 lines meeting each sideline of
> isotomic lines through P, using Cabri or another software but I am
> interested in the flows.
ABC at two points points symmetric in the midpoint of the side. If so,
these lines are the asymptotes of the circumconic with center P.
Obviously they are not always real.
this is related to the cubics I call "Allardice cubics".
Best regards
Bernard
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My dear Bernard
I apologyze for "flow". It's not the best word. I want to suggest:
flow of a vector field
but what I mean is to search the plane curves so that in each point P
of the curve
the direction of the tangent is known, for example in the DF case,
the direction of the 2 orthogonal lines through P or in the isotomic
case the direction of the 2 isotomic lines through P.
Yes that what I mean by isotomic lines . They are not always real. I remember
the medial triangleis a separator?
It's very easy to draw the 2 isotomic lines through P when they are real.
I want to draw the associated integral curves.
Best regards
Francois
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