## Arnold theorem

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• In message #6614, I gave a lemma for the Arnold theorem. Now I have noted that it is not really needed in the proof. Here is the ARNOLD THEOREM (for H. E.
Message 1 of 13 , Feb 28, 2003
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In message #6614, I gave a lemma for the Arnold
theorem. Now I have noted that it is not really
needed in the proof.

Here is the ARNOLD THEOREM (for H. E. Arnold, who
gave a special case of it as problem AMM 3258 in
American Mathematical Monthly):

Let ABC be a triangle with orthocenter H, and
let P be an arbitrary point. Call X, Y, Z
the traces of P on the sides of ABC (i. e.
X = AP /\ BC, Y = BP /\ CA, and Z = CP /\ AB).
The perpendicular to HX through A intersects
BC at X'; analogously, define Y' and Z'. Then,
- the points X', Y', Z' are collinear;
- the line X'Y'Z' is orthogonal to the
line HP.

More on this theorem I have written in a HTML
comment to a problem in the MathPro database. The
database is

http://problems.math.umr.edu/index.htm

Since it is not interlinked, to find my comment one
should search for the text "The perpendiculars
dropped from the vertices of a triangle upon the
lines joining the feet of the angle bisectors".

Now to the PROOF of the Arnold theorem. Let AA',
BB', CC' be the altitudes of triangle ABC. As we
know, AH * HA' = BH * HB' = CH * HC'. Let Q be a
point on the line HP, for which PH * HQ = AH * HA'
= BH * HB' = CH * HC' (with directed segments).
Since AX' is orthogonal to HX and AH is orthogonal
to XX', the point H is the orthocenter of triangle
AXX'. Thus, H lies on the perpendicular from X' to
AX. Call D the foot of this perpendicular. Then,
angle AHD = angle X'HA', and angle ADH = angle
X'A'H = 90°; therefore, triangles AHD and X'HA'
are similar, and AH : HD = X'H : HA'. Thus,
AH * HA' = X'H * HD. Together with PH * HQ = AH *
HA', this gives PH * HQ = X'H * HD, what is
equivalent to PH : HD = X'H : HQ. Together with
angle PHD = angle X'HQ, this yields that triangles
PHD and X'HQ are similar; thus, angle X'QH = angle
PDH = 90°. Therefore, X'H is perpendicular to HQ,
i. e. to HP. This shows that X' lies on the
perpendicular to HP through Q. Analogously, Y' and
Z' lies on this perpendicular. This proves the
Arnold Theorem.

EXERCISES:

1) Orthic products formula.
Prove that AH * HA' = BH * HB' = CH * HC' =
4 R² cos A cos B cos C, where R is the

2) The First Panakis theorem.
The point Q is the intersection of circles AA'P,
BB'P and CC'P. (A theorem of independent
importance.)

3) Panakis transformation as inversion.
The point Q is the inverse of P with respect to
the polar circle of triangle ABC (if the polar
circle exists, i. e. if triangle ABC is obtuse).

4) Arnold theorem and polarity.
The line X'Y'Z' in the Arnold theorem is the
polar of P with respect to the polar circle of
triangle ABC (if the polar circle exists, i. e.
if triangle ABC is obtuse).

5) Alternative proof of Arnold theorem for
obtuse triangles.
Give a proof of the Arnold theorem based on
Exercise 4) and using the properties of polars.
(This proof will be valid for obtuse triangles
ABC.)

6) The orthic axis (Altshiller-Court theorem
If P is the centroid of triangle ABC, then the
line X'Y'Z' is the orthic axis, i. e. the axis
of perspective of triangle ABC and the orthic
triangle A'B'C'.
(A quite hard exercise. I know of an ugly proof.)

Darij Grinberg
• Dear Hyacinthists, ABC and A B C are two given triangles. Do you know a way to construct the pair (M,N) of points such as M,N are isogonal conjugates wrt ABC
Message 2 of 13 , Mar 2, 2003
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Dear Hyacinthists,
ABC and A'B'C' are two given triangles.
Do you know a way to construct the pair (M,N) of points such as M,N
are isogonal conjugates wrt ABC and wrt A'B'C'?
I think that such a pair is - generally - unique and ruler and
compass constructible.
Friendly. Jean-Pierre
• Dear Jean-Pierre, ... [JP] ... the reflections of the lines AA , AB , AC about a bisector at A in ABC intersect the sidelines of A B C at three collinear
Message 3 of 13 , Mar 3, 2003
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Dear Jean-Pierre,
>
[JP]
> ABC and A'B'C' are two given triangles.
> Do you know a way to construct the pair (M,N) of points such as M,N
> are isogonal conjugates wrt ABC and wrt A'B'C'?
> I think that such a pair is - generally - unique and ruler and
> compass constructible.

the reflections of the lines AA', AB', AC' about a bisector at A in ABC
intersect the sidelines of A'B'C' at three collinear points on L_A.
Let N_A be the Newton line of the quadrangle formed by L_A and the triangle
A'B'C'.
similarly, define the Newton line N_A' with triangle ABC.

the two Newton lines meet at Q center of the two conics inscribed in ABC and
A'B'C'. Their foci are your requested points.

Best regards

Bernard
• Dear Bernard ... M,N ... [Bernard] ... in ABC ... triangle ... in ABC and ... Many thanks and congratulations for this wonderful and so clever construction.
Message 4 of 13 , Mar 3, 2003
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Dear Bernard
> [JP]
> > ABC and A'B'C' are two given triangles.
> > Do you know a way to construct the pair (M,N) of points such as
M,N
> > are isogonal conjugates wrt ABC and wrt A'B'C'?
> > I think that such a pair is - generally - unique and ruler and
> > compass constructible.

[Bernard]
> the reflections of the lines AA', AB', AC' about a bisector at A
in ABC
> intersect the sidelines of A'B'C' at three collinear points on L_A.
> Let N_A be the Newton line of the quadrangle formed by L_A and the
triangle
> A'B'C'.
> similarly, define the Newton line N_A' with triangle ABC.
>
> the two Newton lines meet at Q center of the two conics inscribed
in ABC and
> A'B'C'. Their foci are your requested points.

Many thanks and congratulations for this wonderful and so clever
construction.
Friendly. Jean-Pierre
• ... I found a completely different proof of that result. We are talking about synthetic proofs, since the result is trivial computationally. Lemma: If
Message 5 of 13 , Mar 3, 2003
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Darij Grinberg wrote:
> In message #6614, I gave a lemma for the Arnold
> theorem. Now I have noted that it is not really
> needed in the proof.
>
> Here is the ARNOLD THEOREM (for H. E. Arnold, who
> gave a special case of it as problem AMM 3258 in
> American Mathematical Monthly):
>
> Let ABC be a triangle with orthocenter H, and
> let P be an arbitrary point. Call X, Y, Z
> the traces of P on the sides of ABC (i. e.
> X = AP /\ BC, Y = BP /\ CA, and Z = CP /\ AB).
> The perpendicular to HX through A intersects
> BC at X'; analogously, define Y' and Z'. Then,
> - the points X', Y', Z' are collinear;
> - the line X'Y'Z' is orthogonal to the
> line HP.

I found a completely different proof of that result. We are
talking about synthetic proofs, since the result is trivial
computationally.

Lemma: If corresponding sides of two quadrilaterals are
parallel, and if a diagonal of the first quadrilateral is
parallel to a diagonal of the second quadrilateral, then the
other diagonals are also parallel.

The interesting case is when the diagonals assumed to be
parallel are not corresponding diagonals -- if they do
correspond, then the quadrilaterals are homothetic and the
result follows trivially.

Proof: We have AB // A'B', BC // B'C', CD // C'D', DA // D'A',
AC // B'D', and must show BD // A'C'. Using an obvious
translation, dilation, and (possibly) reflection through a
point, we can reduce this to the case where B=B', C=C'.

Apply Pappus to the hexagon C A D B D' A'. Alternate vertices C
D D' are collinear, and so are A B A'. So the three
intersections
CA /\ BD', AD /\ D'A', DB /\ A'C are collinear. From the
assumptions we see that the first two intersections are points
at infinity, so the third intersection is also a point at
infinity. QED.

Corollary. If corresponding sides of two quadrilaterals are
perpendicular, and if a diagonal of the first quadrilateral is
perpendicular to a diagonal of the second quadrilateral, then
the other diagonals are also perpendicular.
Proof: Rotate one of the quadrilaterals by 90 degrees, and use
the lemma.

Now for the Arnold result. Note that X' is the orthocenter of
AHX, and similarly for Y' and Z'. Use the two quadrilaterals
C X' H Y' C and H A P B H. Corresponding sides are
perpendicular, and the diagonals CH and AB are also
perpendicular.
So X'Y' _|_ HP by the corollary. Similarly Y'Z' _|_ HP.

--
Barry Wolk

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• Dear Barry Wolk, Many thanks for the proof. It is indeed simpler. By the way, the Corollary is equivalent to the theorem on Orthologic Triangles, isn t it?
Message 6 of 13 , Mar 3, 2003
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Dear Barry Wolk,

Many thanks for the proof. It is indeed simpler.

By the way, the Corollary is equivalent to the theorem on Orthologic
Triangles, isn't it?

Sincerely,
Darij Grinberg

--- In "Hyacinthos", Barry Wolk wrote:

> Darij Grinberg wrote:
> > In message #6614, I gave a lemma for the Arnold
> > theorem. Now I have noted that it is not really
> > needed in the proof.
> >
> > Here is the ARNOLD THEOREM (for H. E. Arnold, who
> > gave a special case of it as problem AMM 3258 in
> > American Mathematical Monthly):
> >
> > Let ABC be a triangle with orthocenter H, and
> > let P be an arbitrary point. Call X, Y, Z
> > the traces of P on the sides of ABC (i. e.
> > X = AP /\ BC, Y = BP /\ CA, and Z = CP /\ AB).
> > The perpendicular to HX through A intersects
> > BC at X'; analogously, define Y' and Z'. Then,
> > - the points X', Y', Z' are collinear;
> > - the line X'Y'Z' is orthogonal to the
> > line HP.
>
> I found a completely different proof of that result. We are
> talking about synthetic proofs, since the result is trivial
> computationally.
>
> Lemma: If corresponding sides of two quadrilaterals are
> parallel, and if a diagonal of the first quadrilateral is
> parallel to a diagonal of the second quadrilateral, then the
> other diagonals are also parallel.
>
> The interesting case is when the diagonals assumed to be
> parallel are not corresponding diagonals -- if they do
> correspond, then the quadrilaterals are homothetic and the
> result follows trivially.
>
> Proof: We have AB // A'B', BC // B'C', CD // C'D', DA // D'A',
> AC // B'D', and must show BD // A'C'. Using an obvious
> translation, dilation, and (possibly) reflection through a
> point, we can reduce this to the case where B=B', C=C'.
>
> Apply Pappus to the hexagon C A D B D' A'. Alternate vertices C
> D D' are collinear, and so are A B A'. So the three
> intersections
> CA /\ BD', AD /\ D'A', DB /\ A'C are collinear. From the
> assumptions we see that the first two intersections are points
> at infinity, so the third intersection is also a point at
> infinity. QED.
>
> Corollary. If corresponding sides of two quadrilaterals are
> perpendicular, and if a diagonal of the first quadrilateral is
> perpendicular to a diagonal of the second quadrilateral, then
> the other diagonals are also perpendicular.
> Proof: Rotate one of the quadrilaterals by 90 degrees, and use
> the lemma.
>
> Now for the Arnold result. Note that X' is the orthocenter of
> AHX, and similarly for Y' and Z'. Use the two quadrilaterals
> C X' H Y' C and H A P B H. Corresponding sides are
> perpendicular, and the diagonals CH and AB are also
> perpendicular.
> So X'Y' _|_ HP by the corollary. Similarly Y'Z' _|_ HP.
>
> --
> Barry Wolk
• Darij has recently discussed an Arnold theorem (for a given point P the perpendicular from A to HPa meets BC at X, etc, then X,Y,Z are collinear). I am left
Message 7 of 13 , Mar 4, 2003
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Darij has recently discussed an Arnold theorem (for a given point P the
perpendicular from A to HPa meets BC at X, etc, then X,Y,Z are collinear).
I am left with a number of questions ....

The theorem is one of those results that leave me curious to know how they
arose in the first place. Would it have been after some casual doodling
with a drawing program like Cabri? In which case we would already "know"
the result. Proof then becomes a verification.

As Barry Wolk pointed out, an algebraic verification is trivial. Why do we
then seek a "synthetic" proof? I recognise the pleasure that this gives me
when I can hit upon one - and I enjoyed reading Barry's. But is my pleasure
very different from the brief "high" after solving a crossword clue? What
is the investment here? Why isn't the trivial algebraic calculation
satisfying enough? Why, for that matter, isn't the Cabri verification (with
its hidden algebra)?

And back inside the problem, what was so special about H? Was the original
doodle in terms of the locus of such points? Which the "trivial algebra"
would show was a cubic particular to the given point P - but always passing
through H. So the underlying theorem might have been: for a given point P
the perpendicular from A to QPa meet BC at X, etc, then X,Y,Z are collinear)
when Q lies on a cubic through H. But this still leaves me with my
curiosity about the original context of such explorations? And my own
inability to judge the significance of such seemingly isolated results.

Dick Tahta
• ... Orthologic ... [snip] ... is ... then ... Yes it is equivalent. I untangled the notation last night. Orthologic theorem : If perpendiculars from A to B C ,
Message 8 of 13 , Mar 4, 2003
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Darij Grinberg wrote:
> Dear Barry Wolk,
>
> Many thanks for the proof. It is indeed simpler.
>
> By the way, the Corollary is equivalent to the theorem on
Orthologic
> Triangles, isn't it?
>
> Sincerely,
> Darij Grinberg
>
> --- In "Hyacinthos", Barry Wolk wrote:

[snip]

> > Corollary. If corresponding sides of two quadrilaterals are
> > perpendicular, and if a diagonal of the first quadrilateral
is
> > perpendicular to a diagonal of the second quadrilateral,
then
> > the other diagonals are also perpendicular.

Yes it is equivalent. I untangled the notation last night.

Orthologic theorem : If perpendiculars from A to B'C', from B to
C'A', from C to A'B' are concurrent, then the perpendiculars
from A' to BC, from B' to CA, from C' to AB are also concurrent
(assuming A'B'C' not collinear).

Take AP_|_B'C', BP_|_C'A', CP_|_A'B', A'Q_|_BC, B'Q_|_CA, and we
must show C'Q_|_AB. Use the corresponding quadrilaterals
C A P B C and
Q B' C' A' Q
Corresponding sides are perpendicular, and also CP_|_B'A' for
two non-corresponding diagonals. So my corollary is just this
orthologic result.

This means that the proof I gave (rotate by 90 degrees,
transform and use Pappus) becomes an unusual way of proving the
orthologic theorem. How is it proved in textbooks?

The Arnold result has now been reduced to a one-liner, namely
that the triangles X' Y' C and B A P are orthologic, with both
orthologic centers being H.

> If two orthologic triangles are perspective,
> then the perspector is collinear with the
> two orthologic centers, and the join of the
> three points is orthogonal to their
> perspectrix!

Calculation has verified both these conjectures.
--
Barry Wolk

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• Dear Dick Tahta, Barry Wolk, Hyacinthos members, Many thanks for the notes on synthetic and analytic proofs. At first, many of you already know that I try to
Message 9 of 13 , Mar 5, 2003
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Dear Dick Tahta, Barry Wolk, Hyacinthos members,

Many thanks for the notes on synthetic and analytic proofs. At first,
many of you already know that I try to find synthetic proofs wherever
possible. Of course, I am also satisfied with an analytic proof if I
understand it (for example, the proofs of the Sharygin points
trilinears - quite long but I have made all transformations myself).

I don't know of which computational verification is Barry Wolk
speaking, but I think he uses barycentric coordinates. Perhaps a
theorem is that I am still too lazy to learn representing
perpendicular lines in barycentrics. Another reason is that, if
a "trivial" barycentric proof of Arnold's theorem is possible, it
uses a very nontrivial lemma, namely the orthogonality condition, and
the proof together with the proof of the orthogonality condition gets
quite long.

By the way, there are theorems which, as far as I know, nobody has
proved synthetically yet, actually quite well-known theorems like:
The centers of similtude of circumcircle and incircle are the
isogonal conjugates of the Gergonne and Nagel points.

>> The theorem is one of those results that leave me
>> curious to know how they arose in the first place.
>> Would it have been after some casual doodling
>> with a drawing program like Cabri?

The problem was published in 1927... But the idea is correct that
most of the interesting geometry theorems discovered recently were
found by computer drawing -- for example, Floor van Lamoen discovered
his circle in 2000 by computer drawing, I discovered it in 2002
(independently) by computer drawing... These dynamic geometry
utilities strongly accelerate the finding of theorems.

By the way, most of us use Sketchpad or Cabri. I use the nice (also
relatively primitive) German program "Euklid DynaGeo".

>> In which case we would already "know" the
>> result. Proof then becomes a verification.

Now, this is very often so. But also watch out for "merely"
concurrent lines, "merely" collinear points and "merely" similar
triangles. For instance, regard the "Locus?" discussion #6375, #6377,
#6379, #6381, #6382, #6384, #6386, #6391, #6392, #6393.

In message #6375, Lev Emelyanov asked:

>> Let A1B1C1 be P-cevian triangle of ABC (A1 is
>> on BC etc.). Let c_A, c_B and c_C be incircles
>> of AB1C1, BC1A1 and CA1B1 respectively, ab be
>> common extangent (not sideline of ABC) of c_A
>> and c_B. Lines bc and ca we define similarly.
>> What is the locus of P such that ab, bc and
>> ca concur?

I drew a picture and was immediately sure that they always concur,
and that the locus is the whole plane. This caused some trouble until
finally Jean-Pierre Ehrmann found out that the locus is not the whole
plane, and that the lines usually form a very little triangle.

>> Why, for that matter, isn't the Cabri
>> verification (with its hidden algebra)?

Forgive a question from a Cabri non-specialist: Does Cabri test the
conjectures by calculation?

>> And back inside the problem, what was so

The special property of H is that H is the center of the polar circle
of triangle ABC. A simple proof of the Arnold theorem is possible if
we accept using polarity with respect to a non-always real circle:
Regard the polar p of P in the polar circle of triangle ABC. (Of
course, p is perpendicular to HP.) Then,

* the polar of A is BC (property of polar circle);
* the polar of P is p;
* the polar of X is AX', because AX' is orthogonal to HX and AX'
passes through the pole A of BC (but X lies on BC).

Thus, from the collinearity of A, P and X we get the concurrence of
BC, p and AX', i. e. the point X' lies on p. Analogously, Y' and Z'
lie on p, qed.

>> Which the "trivial algebra" would show was a
>> cubic particular to the given point P - but
>> always passing through H.

Thanks - a nice locus.

Many thanks to Barry Wolk for his elegant proof of the orthologic
triangles theorem. Perhaps this proof is somewhere in one of the
books that have been written on the subject of reciprocal figures
(Pedoe, Geometry - a comprehensive course, p. 36), but I have not
seen it before.

This has got much longer than I thought, and I will continue with
answering Barry's mail in the next message.

Sincerely,
Darij Grinberg
• Dear Barry Wolk, Hyacinthos members, ... Thank you for verification. Let me describe the context of the conjecture and my search for a further generalization
Message 10 of 13 , Mar 5, 2003
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Dear Barry Wolk, Hyacinthos members,

Barry Wolk wrote in message #6652:

>> In another thread, Darij wrote
>> > If two orthologic triangles are perspective,
>> > then the perspector is collinear with the
>> > two orthologic centers, and the join of the
>> > three points is orthogonal to their
>> > perspectrix!
>>
>> Calculation has verified both these conjectures.

Thank you for verification. Let me describe the context of the
conjecture and my search for a further generalization of the
Longchamps point.

I wanted to generalize my conjecture about pedal-cevian points
(proved by Floor van Lamoen) (thread "Darboux and a line with 5
points"). In fact, the Darboux-Lamoen theorem states:

If a triangle ABC is given, and A'B'C' is a pedal-cevian
triangle, i. e. the pedal triangle of a point P and
the cevian triangle of a point Q, and if P' is the
isogonal conjugate of P, and if R is the circumcenter
of triangle A'B'C', and if L is the Longchamps point of
triangle ABC, then

(a) P, P' and L are collinear. (Darboux)
(b) P, P' and Q are collinear. (Lamoen and me)
(c) P, Q and R are collinear. (Proof attempt)
(d) P, P', Q, L and R are collinear. (Corollary)

Now I asked myself which of these four theorems could be generalized
to arbitrary orthologic and perspective triangles. (Two orthologic
and perspective triangles, one of them being inscribed in the other,
are a less symmetrical configuration than to arbitrary orthologic and
perspective triangles.) I immediately saw that if A'B'C' should be an
arbitrary triangle orthologic and perspective to ABC, then P would be
the intersection of the perpendiculars from A' to BC, from B' to CA
and from C' to AB, and P' would be the intersection of the
perpendiculars from A to B'C', from B to C'A' and from C to A'B'. (If
A'B'C' is inscribed to ABC, than this intersection is really the
isogonal conjugate of P - easy to prove!) So P and P' are the two
orthologic centers of triangles ABC and A'B'C'.

I easily found that (b) remains true for arbitrary orthologic and
perspective triangles. I still haven't a proof, but I hope that there
will be a synthetic one...

Now I asked myself what we could do with (a): If A'B'C' is inscribed
in ABC, then P, P' and L are collinear. I call this Darboux miracle,
because at first appearance, L has nothing to do with triangle
A'B'C'. So I thought that in the general case, P and P' must be
collinear with a point depending on both triangles ABC and A'B'C',
but giving L in the special case.

Unfortunately, I haven't found a suitable L generalization yet! (...
what makes the Darboux miracle more miraculous.) I think that we
should begin with the fact that L is the orthocenter of the
anticomplementary triangle of ABC, i. e. the generalization of L will
be the intersection of some perpendiculars. But of which ones?

Sincerely,
Darij Grinberg
• ... call them ABC and A B C ... ... and searched for another point collinear with the two orthologic centers P and P and the perspector Q. If triangle A B C
Message 11 of 13 , Mar 5, 2003
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I wrote:

>> If two orthologic triangles
call them ABC and A'B'C'
>> are perspective, then the perspector is
>> collinear with the two orthologic centers,
>> and the join of the three points is
>> orthogonal to their perspectrix!

... and searched for another point collinear with the two orthologic
centers P and P' and the perspector Q.

If triangle A'B'C' is inscribed in triangle ABC, then the Longchamps
point L of triangle ABC is collinear with P, P' and Q.

If triangle A'B'C' is inscribed in the circumcircle of triangle ABC,
then (I conjecture that) the circumcenter O is collinear with P, P'
and Q.

THERE MUST BE A GENERALIZATION!
----

Sincerely,
Darij Grinberg
• ... Correct. ... I call a calculation trivial if I can do it without needing a computer to handle the polynomials. In many problems the polynomials tend to
Message 12 of 13 , Mar 5, 2003
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Darij Grinberg wrote:

> I don't know of which computational verification is Barry Wolk

> speaking, but I think he uses barycentric coordinates.

Correct.

> Perhaps a reason why
> theorem is that I am still too lazy to learn representing
> perpendicular lines in barycentrics.

I call a calculation "trivial" if I can do it without needing a
computer to handle the polynomials. In many problems the
polynomials tend to get ridiculously large.

It is very easy to handle perpendicular lines. Take two points
at infinity, P(x:y:z) and P'(x':y':z'), so x+y+z = x'+y'+z' = 0.
Then lines through P are perpendicular to lines through P' iff
x x' SA + y y' SB + z z' SC = 0

So given P, we can find P' by letting x' = y SB - z SC, y' and
z' similarly.

And the line through H and P' has coordinates
[x SA : y SB : z SC]
--
Barry Wolk

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• Dear Barry Wolk, ... Many thanks... the next step would be (perhaps) generalizing to Lamoen s Pl-perpendicularity, but I think this is simply replacing SA, SB,
Message 13 of 13 , Mar 5, 2003
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Dear Barry Wolk,

In Hyacinthos message #6661, you wrote:

>> Darij Grinberg wrote:
>>
>> > I don't know of which computational verification is Barry Wolk
>> > speaking, but I think he uses barycentric coordinates.
>>
>> Correct.
>>
>> > Perhaps a reason why
>> > theorem is that I am still too lazy to learn representing
>> > perpendicular lines in barycentrics.
>>
>> I call a calculation "trivial" if I can do it without needing a
>> computer to handle the polynomials. In many problems the
>> polynomials tend to get ridiculously large.
>>
>> It is very easy to handle perpendicular lines. Take two points
>> at infinity, P(x:y:z) and P'(x':y':z'), so x+y+z = x'+y'+z' = 0.
>> Then lines through P are perpendicular to lines through P' iff
>> x x' SA + y y' SB + z z' SC = 0
>>
>> So given P, we can find P' by letting x' = y SB - z SC, y' and
>> z' similarly.
>>
>> And the line through H and P' has coordinates
>> [x SA : y SB : z SC]
>> --
>> Barry Wolk

Many thanks... the next step would be (perhaps) generalizing to
Lamoen's Pl-perpendicularity, but I think this is simply replacing
SA, SB, SC by some constants.

Sincerely,
Darij Grinberg
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