Antreas quoted:

> Bankoff, Leon - Erdos, Paul - Klamkin, Murray: The Asymmetric Propeller.
> Mathematics Magazine 46 (1973) 270-272.
>

Which according to ZfM, the paper seems to provide two proofs of the
theorem:

    If OAB, OCD and OEF are equilateral triangles of the same
    orientation then the midpoints of BC, DE and FA are the
    vertices of an equilateral triangle.

Regards, JGC

Julio Gonzalez Cabillon wrote:

>At 04:49 PM 28/12/1999 -0700, Richard Guy wrote:
>>
>> See also Martin Gardner's article on the Asymmetric Propellor,
>> somewhere in this year's Coll. Math. J.        R.
>>
>
>See:
>
>Gardner, Martin:
>"The Asymmetric Propeller", _College Math Journal_, vol 30 (1999),
>no 1, pp 18-22.

First Paragraph:
The late Leon Bankoff (he died in 1997) was a Beverly Hills, California,
dentist who also was a world expert on plane geometry....In 1979 he told me
about a series of fascinating discoveries he had made about what he called
the asymmetric propeller theorem. He intended to discuss them in an article,
but never got around to it. This is a summary of what he told me.

Also:
Bankoff, Leon - Erdos, Paul - Klamkin, Murray: The Asymmetric Propeller.
Mathematics Magazine 46 (1973) 270-272.

Antreas

-


At 04:49 PM 28/12/1999 -0700, Richard Guy wrote:
>
> See also Martin Gardner's article on the Asymmetric Propellor,
> somewhere in this year's Coll. Math. J.        R.
>

See:

Gardner, Martin:
"The Asymmetric Propeller", _College Math Journal_, vol 30 (1999),
no 1, pp 18-22.

Regards, JGC













-

Richard Guy wrote:

>See also Martin Gardner's article on the Asymmetric Propellor,
>somewhere in this year's Coll. Math. J.        R.

The Asymmetric Propeller
Martin Gardner

           A theorem, seventy years old at least and of unknown origin,
           says that if three congruent equilateral triangles are have
           corners meeting, the midpoints of the lines joining the other
           two vertices of the triangles are vertices of an equilateral
           triangle. The late Leon Bankoff discovered that the triangles
           don't have be congruent and don't have to meet at a point.
           Martin Gardner describes the results, and conjectures that the
           triangles don't have to be triangular--squares seem to work
           as well.
The College Mathematics Journal, January 1999.

Antreas

See also Martin Gardner's article on the Asymmetric Propellor,
somewhere in this year's Coll. Math. J.        R.

Lucien Droussent: On the Orthocentroidal Circle.
The American Math. Monthly 57 (1950) 169-171.

There is also an O. Quadrangle:

G. Casaux: Sur le quadrangle orthocentroidal.
Mathesis 54 (1945) 429-433.


Antreas

Den Roussel wrote:

>When Cevians are drawn from the vertices of a triangle ABC, they
>intersect the opposite sides at A'B'C' and the circumcircle at
>A''B''C''.
>
>Ceva's theorem shows that the product of the ratios of the segments of
>the sides is equal to 1.
>
>(AB'/B'C) (CA'/A'B) (BC'/C'A) = 1
>
>In like manner, the product of the ratios of the consecutive sides of
>the inscribed hexagon is equal to 1.
>
>(AB''/B''C) (CA''/A''B) (BC''/C''A) = 1
>
>In other words, when three concurrent lines are drawn in a circle, then
>the hexagon they form has the property that the product of three
>non-adjacent sides is equal to the product of the other three sides.
>
>I'm not sure if this is known or not.


H. S. M. Coxeter:

    Given six consecutive points A, B, C, D, E and F on a circle, prove
    that if (AB)(CD)(EF) = (BC)(DE)(FA), then AD, BE and CF are concurrent.
    _The Mathematics Student Journal_ 27:5 (1980) 3.


Antreas

At 09:46 AM 28/12/1999 -0700, Richard Guy wrote:

> Stick with `coaxal' -- `coaxial' means the same thing, but is
> normally reserved for cylinders.    R.
                            |
                            |
                            ^
                      "coaxial cable"

De gustibus non est disputandum!  I, for one, prefer "coaxial". JGC



























-

Stick with `coaxal' -- `coaxial' means the same thing, but is
normally reserved for cylinders.    R.

On Mon, 27 Dec 1999, Julio Gonzalez Cabillon wrote:

> : Re: What is the "orthocentroidal circle" ?
> :
> : It is the circle with diameter GH.
>
> There is a well-known coaxial (*) family of circles associated with a
                             ^^^^^^^
> triangle $ABC$ with centers on the Euler line of $ABC$.  The family of
> circles that each two have the same radical line includes the:
>
> o  circumcircle
> o  9-point circle
> o  orthocentroidal circle
> o  self-polar circle (if $ABC$ is an obtuse-angled triangle)
> o  de Longchamps circle (if $ABC$ is an obtuse-angled triangle)
> o  orthoptic (Monge) circle (of an ellipse that meets the midpoints of
>    the sides of $ABC$)
>
> (*) I'd rather use "coaxial" in lieu of the standard "coaxal".

>> Why "desmic", by the way?
>
> John Conway's term. Desmic means "linking."
>

The term "desmic" ("desmique", in fact) was introduced by the
mathematician C. Stephanos. Although one may suspect that the
Greek word working here is "desm/os" (= link), Stephanos based
his choice on "d/esmh", with the meaning of the French word
"faisceau" (= bundle). So the intended sense of "desmic" was
NOT "linking".

Regards, JGC

: Re: What is the "orthocentroidal circle" ?
:
: It is the circle with diameter GH.


There is a well-known coaxial (*) family of circles associated with a
triangle $ABC$ with centers on the Euler line of $ABC$.  The family of
circles that each two have the same radical line includes the:

o  circumcircle
o  9-point circle
o  orthocentroidal circle
o  self-polar circle (if $ABC$ is an obtuse-angled triangle)
o  de Longchamps circle (if $ABC$ is an obtuse-angled triangle)
o  orthoptic (Monge) circle (of an ellipse that meets the midpoints of
    the sides of $ABC$)

(*) I'd rather use "coaxial" in lieu of the standard "coaxal".

Regards, JGC

> Date: Thu, 23 Dec 1999 09:23:47 +0100
> From: Floor van Lamoen <f.v.lamoen@xxx.xxx
>
> > From: Clark Kimberling <ck6@xxxxxxxxxx.xxx>
> >
> > However, Honsberger doesn't mention (directly) a certain interesting
> > property of the Lemoine point.  For any point P, let A'B'C' denote the
> > pedal triangle of P (i.e., A' is the point in which the line through P
> > perpendicular to line BC meets line BC). Let S(P) be the vector sum
> > PA'+PB'+PC'.  Then S(P) is the zero vector if P is the Lemoine point.
>
> Isn't this property exactly equivalent to the fact that the
> Lemoine/symmedian point K is the centroid of its pedal triangle? This
> property of K is mentioned in O. Bottema's ``Hoofdstukken uit de
> Elementaire Meetkunde''.

Of course that is an equivalent property.

>
[snip]
> > I conjecture that the converse is true: that if P is a "point" (i.e.,
> > f(a,b,c) : g(a,b,c) : h(a,b,c)) such that S(P)=0, then P = a^2 : b^2 : c^2
> > (barycentric coordinates of the Lemoine point).

> This is not immediatly apparent to me.

It wasn't apparent to me either. However, the calculations are not
difficult, and this converse is true.

Write vec(X,Y) = vector from X to Y.
In homogenous barycentrics, P=(x:y:z).
In normalized barycentrics, if P=(x,y,z) with x+y+z=1, then
        vec(O,P) = x vec(O,A) + y vec(O,B) + z vec(O,C)
for any choice of origin O.

Now the B-pedal point of P(x,y,z) is P_B = (x+y SC/bb , 0 , z+y SA/bb),
and this is normalized, since SA+SC=bb.
Then vec(P,P_B) = vec(O,P_B) - vec(O,P)
       = y SC/bb vec(O,A) - y vec(O,B) + y SA/cc vec(O,C)
with similar formulas for vec(P,P_A) and vec(P,P_C)

So the equation S(P)=0 becomes
    (-x + y SC/bb + z SB/cc) vec(O,A)
  + (x SC/aa - y + z SA/cc) vec(O,B)
  + (x SB/aa + y SA/bb - z) vec(O,C) = 0

Well, one solution is obviously (x,y,z)=const*(aa,bb,cc). Any others?
Choose the origin O to be the circumcenter. Then the three basis
vectors are dependent, and we have the dependency relation
      aaSA vec(O,A) + bbSB vec(O,B) + ccSC vec(O,C) = 0
This is the only dependency relation between these three vectors.

So any other solution {x,y,z} would have to satisfy
    (-x + y SC/bb + z SB/cc) = k aaSA
    (x SC/aa - y + z SA/cc) = k bbSB
    (x SB/aa + y SA/bb - z) = k ccSC
for some non-zero constant k. However, adding these 3 equations
shows k=0, so there are no other solutions

I admit the calculations weren't as easy as I expected,
but this converse is true.
--
Barry Wolk          <wolkb@...>
Winnipeg Manitoba Canada

John Conway wrote:

>    What is the "orthocentroidal circle" ?


It is the circle with diameter GH.

Cf.:
Cercle orthocentroidal (TUCKER). Cercle  decrit sur GH pris pour diametre.
Voir _Mathesis_, 1890, p. 166, et 1893, p. 33.
(F.G. - M: Exercices de Geometrie....., p. 1264)

Antreas

Antreas - I shorten so that I can see all your points at once:

On Mon, 27 Dec 1999, Antreas P. Hatzipolakis wrote:

> >The medians: AMa of the triangle ABaCa, BMb of the triangle
> >BCbAb, and CMc of the triangle CBcAc are concurrent.

*** at  L = ( : sinB(cosB - cosCcosA): ) = (:bbSB-SCSA:)   ***

> >The symmedians: ASa of the triangle ABaCa, BSb of the triangle
> >BCbAb, and CSc of the triangle CBcAc are concurrent.

***  at  L* = (:bb/(bbSB-SCSA):)

> The feet Ta, Tb, Tc, of the exsymmedians: ATa of the triangle ABaCa, BTb
> of the triangle BCbAb, and CTc of the triangle CBcAc, are collinear.
>
> The a-Menelaus ratio is:
>
> BTa   sinC   (cosB + cosAcosC)
> --- = ---- * -----------------
> TaC   sinB   (cosC + cosAcosB)


     So this is the line  (p:q:r|0)  where

     q or  1/q   =  (cosB + cosCcosA)/sinB

   The fire alarm's just gone off in this building, so I'll have
to leave.  (It regularly does so, but one should obey!)  JHC

>
>
> Conjectures:
>
> I have drawn some figures with ISOPTIKON, and "discovered" that the
> the perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.
>
> Also, the Euler lines of the four triangles ABC, ABaCa, BCbAb, CBcAc
> are concurrent. (This point is probably an new point on the Euler line)
>
> I didn't draw figures for other lines (OK, OI, etc), but I believe
> that they concur as well.
>
>
> Happy Holidays
>
>
> Antreas
>
> >

On Sat, 25 Dec 1999, Antreas P. Hatzipolakis wrote:

> Seasonal Greetings from Athens.
> And Happy Birthday, John!

    Thanks!

> Following are some theorems / conjectures on what I call "orthial triangles".
>
> Let ABC be a triangle. We draw perpendiculars to its sides at its vertices
> A,B,C, intersecting the opposite sides at Ba, Ca; Ac, Bc; Cb, Ab :

   *** Note by JHC:  I retain only this much of the figure:

>                          /            \
>  ---Ba------------------B---Sa----Ma---C------------------Ca-----
>                        /                \

    which should be enough for me to keep the notation in mind ***
> Theorem #1
> The medians: AMa of the triangle ABaCa, BMb of the triangle
> BCbAb, and CMc of the triangle CBcAc concur in L = deLongchamps pt.
> (Ma, Mb, Mc are the midpoints of BaCa, CbAb, AcBc, respectively.)


> I had calculated trigonometrically the a-Ceva ratio:
>
> BMa    sinC   (cosC - cosBcosA)
> ---- = ----- * ----------------
> MaC    sinB   (cosB - cosCcosA)
>
> and Conway proved it using barycentrics.

***   The two methods are essentially the same, since the Cevian
ratios of the point with barycentrics  (X:Y:Z)  are  Y:Z, Z:X, X:Y.
So the barycentrics of this point are  ( : b(cosB - cosCcosA) : ),
since the sine law permits me to replace  sinB  by  b.  Alternatively,
it might be  1/that  if your Cevian ratio is the other way up - let me
check: in my notation,  SB = ca.cosB = (cc+aa-bb)/2, so multiplying
the above by  abc  we get  ( : bbSB - SCSA : ) = ( : SASB+SBSC-SCSA : ),
which is indeed the superior of the orthocenter  ( : SCSA : )  ***
> Theorem #2
> The symmedians: ASa of the triangle ABaCa, BSb of the triangle
> BCbAb, and CSc of the triangle CBcAc are concurrent.
>
>
> I calculated the a-Ceva ratio:
>
> BSa    sinC   (cosB - cosAcosC)
> ---- = ---- * -----------------
> SCa    sinB   (cosC - cosAcosB)

***   This is equivalent to barycentrics  ( : b/(cosB - cosCcosA) : ),
that is to say, the conjugal,  L*, of the deLongchamps point L  ***

> (There is no mention of a geometric property of this point in
> that entry. So, we have now one!)
>
> Conjecture:
> The perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.
> How about the Brocard Axes of the orthial triangles ABaCa, BCbAb, CBcAc?
> I would bet that they concur too!

    OK, let me work on these (about which I'm rather doubtful).  I'll
draw the triangle my usual way round:

                Cb-------C--------A-------Ab

    The vectors _|_  a,b,c  are  (-aa:SC:SB),(SC:-bb:SA),(SB:SA:-cc),
so  Cb  = (SB:0:-cc),  Ab = (-aa:0:SB),  whose coordinate-sums are
-SA  and  -SC.  I'll therefore multiply them by  -SC  and  -SA
respectively, to get a common denominator:

       Cb = (-SBSC:0:ccSC)      Ab = (aaSA:0:SASB)

so their midpoint is  (aaSA-SBSC:0:ccSC-SASB),  whence the general
point on their perpendicular is this + some multiple of (SC:-bb:SA).

This makes it clear that these perpendiculars don't concur, since there's
no way to adjust these multiples to give a symmetric point (I think - I'll
check properly in a moment).  But taking the multiplier  SB  gives us
the interesting point   (aaSA:-bbSB:ccSC),  which is one of the harmonic
associates of the circumcenter  O = (:bbSB:).  So at least we get a
theorem:  these perpendiculars pass through the preCevian or harmonic
triangle of the circumcenter.

     But I'm getting a thought which makes me think that perhaps they DO
concur, at a very interesting point.  No - I think it's the wrong way
round.  What I'm thinking of is the fact that the contact points of
the Steiner deltoid with the edges have the property that they are
the Cevian feet of one point and the pedal feet of another.  But in
fact they're the pedal feet of  L  and the Cevian feet of the point I
used to call  M  but think I now call it the Retrocenter  R.

     I tried doing that check online, but it began to look a bit complicated,
so I'll postpone it.

    JHC

On Sun, 26 Dec 1999, Richard Guy wrote:

> Many happy returns of the day, John.
>
> When will you next survive three consecutive orders of simple groups?   R.

    That's almost as bad as telling Erdos he was duplicating the cube
for the last time!

     Looking forward to your next biquadrate,  John C.

PS.  I never saw that "Caian".  What did it say?   JHC

I wrote:

>Following are some theorems / conjectures on what I call "orthial triangles".
>
>Let ABC be a triangle. We draw perpendiculars to its sides at its vertices
>A,B,C, intersecting the opposite sides at Ba, Ca; Ac, Bc; Cb, Ab :
>
>                        \              /
>                        Ab            Ac
>                          \          /
>                           \        /
>                            \      /
>                             \    /
>                              \  /
>                                A
>                               /\
>                              /  \
>                            Sc    Mb
>                            /      \
>                          Mc        Sb
>                          /          \
>                         /            \
> ---Ba------------------B---Sa----Ma---C------------------Ca-----
>                       /                \
>                      /                  \
>                     /                    \
>                    /                      \
>                   Bc                       Cb
>                  /                          \
>
>In the figure:  BaA _|_ AC, CaA _|_ AC; AcC _|_ CB, BcC _|_ CB;
>
>CbB _|_ BA , AbB _|_ BA  (The symbol _|_ means "perpendicular to")
>
>I call the triangles ABaCa, BCbAb, CBcAc as "Orthial Triangles" of ABC
>(for the lack of a better name!).
>
>Theorem #1
>The medians: AMa of the triangle ABaCa, BMb of the triangle
>BCbAb, and CMc of the triangle CBcAc are concurrent.
>(Ma, Mb, Mc are the midpoints of BaCa, CbAb, AcBc, respectively.)
>
>Note: John Conway and I have discussed it in geometry-college.
>I had calculated trigonometrically the a-Ceva ratio:
>
>BMa    sinC   (cosC - cosBcosA)
>---- = ----- * ----------------
>MaC    sinB   (cosB - cosCcosA)
>
>and Conway proved it using barycentrics.
>
>See the thread:
>       http://forum.swarthmore.edu/epigone/geom.college/styterben/
>
>The point of concurrency is the de Longchamps Point of ABC.
>(X_20 in Clark Kimberling's _TCCT_).
>
>Theorem #2
>The symmedians: ASa of the triangle ABaCa, BSb of the triangle
>BCbAb, and CSc of the triangle CBcAc are concurrent.
>
>Note:
>I calculated the a-Ceva ratio:
>
>BSa    sinC   (cosB - cosAcosC)
>---- = ---- * -----------------
>SCa    sinB   (cosC - cosAcosB)
>
>The point of concurrency is X_64 in _TCCT_.
>(There is no mention of a geometric property of this point in
>that entry. So, we have now one!)

Theorem #3:

Theorem #3
The feet Ta, Tb, Tc, of the exsymmedians: ATa of the triangle ABaCa, BTb
of the triangle BCbAb, and CTc of the triangle CBcAc, are collinear.

I proved it by Menelaus theorem.

The a-Menelaus ratio is:

BTa   sinC   (cosB + cosAcosC)
--- = ---- * -----------------
TaC   sinB   (cosC + cosAcosB)


Conjectures:

I have drawn some figures with ISOPTIKON, and "discovered" that the
the perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.

Also, the Euler lines of the four triangles ABC, ABaCa, BCbAb, CBcAc
are concurrent. (This point is probably an new point on the Euler line)

I didn't draw figures for other lines (OK, OI, etc), but I believe
that they concur as well.


Happy Holidays


Antreas

<quote>
                  A visual homage to Alberto P. Calderon

           Miguel de Guzman Garcia-Monge - Miguel de Guzman Ozamiz

Madrid, December 1998

A solution to the Apollonius problem proposed by Alberto Calderon.
Alberto Calderon proposed a very elegant way to solve the classical
Apollonius problem: Given three circles in a plane a, find another one
tangent to them.

Calderon's idea can be easily explained. Let us assume, in order to set
the context, that the three given circles are external to each other and that
we are looking for a circle which is externally tangent to them.

For each one of the given circles we construct a straight circular cone
with a base on that circle, with its vertex located below the plane awhich
contains the circles and having a 90 deg. angle at the vertex. These three
cones intersect at two points in space, one above the plane a, M, and the
other below it, N. We now take the straight circular cone with vertex at M,
90&ordm; angle at M and whose axis is perpendicular to a.The circle it
intersects on the plane a is externally tangent to the three given circles
and so is the solution to the Apollonius problem we wanted to solve. The
proof is very simple and will be left to the reader.

(Analogously, one can proceed in order to obtain the other solutions to
the Apollonius problem. If one is interested, for example, in obtaining a
circle such that two of the given circles are internally tangent to it and the
other one is externally tangent, one constructs auxiliary cones with vertices
above or below the plane of the circles according to the required conditions.)

The Feuerbach circle theorem in light of Calderon's solution.
Equipped with the above idea we are going to interpret, in a similar
way and in three dimensions, one of the most beautiful results in the
geometry of the triangle, the one related to the Feuerbach circle, i.e. the
circle which passes through the three midpoints of the sides of a triangle,
which is also called the nine-point circle. Here is the statement of the
theorem:
The Feuerbach circle of a triangle is tangent to the circle I inscribed
in the triangle and to its three exinscribed circles I(a), I(b), I(c).

(I(a), for instance, is the circle internally tangent to sides b and c and
externally tangent to side a.)

The strinkingly beautiful result of this way to look at the Feuerbach
theorem will now be visually presented through a sequence of images. All
the facts they show are simple consequences of the ideas above and there
remains nothing further to prove.

Figure 1.
Let ABC be an arbitrary triangle on the plane  p. Let  I be its inscribed
circle and  I(a), I(b), I(c), its three exinscribed circles.

Figure 2.
With base on I we construct a straight circular cone K with its vertex
above the plane  p and such that its angle at the vertex is 90 deg.
With base at I(a) we contruct another straight circular cone K(a)
with its vertex below the plane p and such that its angle at the vertex is
90 deg. The intersection of K and K(a) is a hyperbola H(a).

Figure 3.
In an analogous way we construct the cones  K(b) and  K(c), also with
their vertices below the plane p,and thus we obtain the hyperbolas  H(b),H(c).

Figure 4.
We thus have the three hyperbolas  H(a), H(b), H(c) in space.

Figure 5.
The intersection of the cone K(a) with K(b) is another hyperbola
H(a,b). In a similar way we obtain  H(b,c), H(c,a), another three
hyperbolas in space.

Figure 6.
The six hyperbolas thus obtained meet at one point F!
We also construct the straight circular cone  K(F) with vertex at F,
with its axis perpendicular to p and a 90 deg. angle at F.

Figure 7.
The intersection of  K(F) with p is Feuerbach's circle!


Bibliography.

Alberto P. Calderon, Reflexiones sobre el aprendizaje y ensenanza de la
matematica, La Gaceta de la Real Sociedad Matematica Espanola 1 (1998),
80-88. [Reproduccion del texto de la Conferencia Rey Pastor en la XXXVI
Reunion Anual de la Union Matematica Argentina y IX Reunion de Educacion
Matematica, Santa Fe y Parana, 1986]

Laura Guggenbuhl, Karl Wilhelm Feuerbach, Mathematician, The Scientific
Monthly, 81 (1955), 71-76. [Reprinted as an Appendix in: Dan Pedoe,
Circles. A Mathematical View (The Mathematical Association of America,
Washington, 1995), 89-100.]

</quote>

   http://www.mat.ucm.es/deptos/am/guzman/homcalderon/homapceng/oohomcal.htm

----------------------------------------------------------------------------


Antreas

on 12/26/99 6:37 PM Antreas P. Hatzipolakis wrote

>From: xpolakis@... (Antreas P. Hatzipolakis)
>
>X_76 [in Clark Kimberling's _TCCT_]:
>
>[Name]: 3rd Brocard Point.
>
>[Trilinears]: (:b^(-3):)
>
>[Geometrical Properties]: The Brocard midpoint (X_39) of the
>anticomplementary triangle.
>
>
>Another Property:
>
>The intersection point of WW_2, W'W_1, where W, W'; W_1, W_2, are the
>Brocard points of a triangle and its medial triangle, respectively.
>(R. Tucker)
>
>In Tucker's wording: WW_2, W'W_1 intersect in
>a^3 * alpha = b^3 * beta = c^3 * gamma.
>


An important point.

Conway calls this the Junction point, the isotomic conjugate of the
symmedian point.
If W and W' are the Brocard points of Triangle ABC. Then WW'J is triply
in perspective with the first Brocard triangle.

It is on the line that connects H, isoH, Hminor, one of the sides of the
small conway triangle. It's subordinate is the Brocard midpoint of ABC
and is on the OK line, which I believe is equivalent to something Antreas
said. Note H-minor is the orthocenter of the first Brocard triangle.

Steve

X_76 [in Clark Kimberling's _TCCT_]:

[Name]: 3rd Brocard Point.

[Trilinears]: (:b^(-3):)

[Geometrical Properties]: The Brocard midpoint (X_39) of the
anticomplementary triangle.


Another Property:

The intersection point of WW_2, W'W_1, where W, W'; W_1, W_2, are the
Brocard points of a triangle and its medial triangle, respectively.
(R. Tucker)

In Tucker's wording: WW_2, W'W_1 intersect in
a^3 * alpha = b^3 * beta = c^3 * gamma.


Happy Holidays

Antreas

Many happy returns of the day, John.

When will you next survive three consecutive orders of simple groups?   R.

ABC right-angled triangle at A <==> sin^2(A) = sin^2(B) + sin^2(C)
(Pythagorean Theorem)

How about to call a triangle ABC as "pseudo-right-angled" if holds the
equality: cos^2(A) = cos^2(B) + cos^2(C) ?

Which might be a geometric property of such a triangle ?

Well... This one:
The bases of its "orthial triangles" form a right-angled triangle.

I wrote:

>Let ABC be a triangle. We draw perpendiculars to its sides at its vertices
>A,B,C, intersecting the opposite sides at Ba, Ca; Ac, Bc; Cb, Ab :
>
>                        \              /
>                        Ab            Ac
>                          \          /
>                           \        /
>                            \      /
>                             \    /
>                              \  /
>                                A
>                               /\
>                              /  \
>                            Sc    Mb
>                            /      \
>                          Mc        Sb
>                          /          \
>                         /            \
> ---Ba------------------B---Sa----Ma---C------------------Ca-----
>                       /                \
>                      /                  \
>                     /                    \
>                    /                      \
>                   Bc                       Cb
>                  /                          \
>
>In the figure:  BaA _|_ AC, CaA _|_ AC; AcC _|_ CB, BcC _|_ CB;
>
>CbB _|_ BA , AbB _|_ BA  (The symbol _|_ means "perpendicular to")
>
>I call the triangles ABaCa, BCbAb, CBcAc as "Orthial Triangles" of ABC
>(for the lack of a better name!).

The "bases" BaCa, CbAb, BcAc of these triangles are equal to:

atanBtanC, btanCtanA, ctanAtanB.

Now, if the triangle formed with these "bases" is right-angled (with the angle
opposite to BaCa equal to 90 d.) then:

(BaCa)^2 = (CbAb)^2 + (BcAc)^2 <==> (atanBtanC)^2 = (btanCtanA)^2+(ctanAtanB)^2
<==> cos^2(A) = cos^2(B) + cos^2(C)


Happy Holidays

Antreas

Dear Friends,

Seasonal Greetings from Athens.
And Happy Birthday, John!


Following are some theorems / conjectures on what I call "orthial triangles".

Let ABC be a triangle. We draw perpendiculars to its sides at its vertices
A,B,C, intersecting the opposite sides at Ba, Ca; Ac, Bc; Cb, Ab :

                         \              /
                         Ab            Ac
                           \          /
                            \        /
                             \      /
                              \    /
                               \  /
                                 A
                                /\
                               /  \
                             Sc    Mb
                             /      \
                           Mc        Sb
                           /          \
                          /            \
  ---Ba------------------B---Sa----Ma---C------------------Ca-----
                        /                \
                       /                  \
                      /                    \
                     /                      \
                    Bc                       Cb
                   /                          \

In the figure:  BaA _|_ AC, CaA _|_ AC; AcC _|_ CB, BcC _|_ CB;

CbB _|_ BA , AbB _|_ BA  (The symbol _|_ means "perpendicular to")

I call the triangles ABaCa, BCbAb, CBcAc as "Orthial Triangles" of ABC
(for the lack of a better name!).

Theorem #1
The medians: AMa of the triangle ABaCa, BMb of the triangle
BCbAb, and CMc of the triangle CBcAc are concurrent.
(Ma, Mb, Mc are the midpoints of BaCa, CbAb, AcBc, respectively.)

Note: John Conway and I have discussed it in geometry-college.
I had calculated trigonometrically the a-Ceva ratio:

BMa    sinC   (cosC - cosBcosA)
---- = ----- * ----------------
MaC    sinB   (cosB - cosCcosA)

and Conway proved it using barycentrics.

See the thread:
        http://forum.swarthmore.edu/epigone/geom.college/styterben/

The point of concurrency is the de Longchamps Point of ABC.
(X_20 in Clark Kimberling's _TCCT_).

Theorem #2
The symmedians: ASa of the triangle ABaCa, BSb of the triangle
BCbAb, and CSc of the triangle CBcAc are concurrent.

Note:
I calculated the a-Ceva ratio:

BSa    sinC   (cosB - cosAcosC)
---- = ---- * -----------------
SCa    sinB   (cosC - cosAcosB)

The point of concurrency is X_64 in _TCCT_.
(There is no mention of a geometric property of this point in
that entry. So, we have now one!)

Conjecture:
The perpendicular bisectors of BaCa, AcBc, CbAb are concurrent.
How about the Brocard Axes of the orthial triangles ABaCa, BCbAb, CBcAc?
I would bet that they concur too!


Antreas

Dear Steve and friends,

I have found a quite different approach to the Fermats,etc based on
projective geometry :

Let's call (F) the pencil (faisceau in French) of conics with basis the two
Fermats Fn & Fs and the two Napoleons Nn & Ns. Kiepert hyperbola is
obviously one of the conics from (F).

the triangle KON (K = Lemoine, O = circumcenter, N = 9pts center i.e.
midpoint of [HO] ) is the autopolar triangle of (F) : each vertex is the
polar point of the opposite side w.r.t. any conic of (F). In particular, the
polar lines of K, O ,N w.r.t Kiepert hyperbola are ON (Euler line), KN (name
?), KO (Brocard line) but this is true for any conic.

We know that (F) leads to a Desarguesean involution on the Brocard line KO
(and more generally on any line) with fixed points O and K obtained with the
two degenerate conics consisting of the pairs of lines concurrent in O and
K.

Then, (g) being a conic of (F) intersecting the line KO in m and m', those
two points are in involution and therefore are inverse w.r.t the circle with
diameter OK : one is inside, the other outside unless they are on the circle
itself i.e. m = O or m = K.

Let's call now :
     m* and m'* the isogonal conjugates of m and m' (they are on Kiepert
hyperbola),
     n = Hm inter Om* and n' = Hm' inter Om'*   (they are on Kiepert
hyperbola as well) (H = orthocenter),
     n* and n'* the isogonal conjugates of n and n' (they are on KO and
inverse w.r.t the circle with diameter OK).

We have 8 points [not always distincts, see below]: 4 on KO and 4 on Kiepert
which are all simply related to a whole bunch of properties :

1). nn' and m*m'* concur in K
     mm'*, m*m', nn'*, n*n' concur in G
     mn*, n'm'*, m*n concur in O
     mn, m'n', m*n*, m'*n'* concur in H
     m*n' and m'*n concur in N
2). two points on KO (m & m' or n & n') and the 4 on Kiepert are on a same
conic. (Pascal's theorem reciprocal)
3). m & m' are harmonic conjugates w.r.t O & K and there are many other
harmonic conjugates (it is a consequence of the polarity above)

So, any point M chosen on KO leads to a group of 8 points (including M) and
if M is one of the two isodynamics, the 8 points will be the Fx, Nx, Ix, Nx*
with all the properties above.

****************************************

One particular case is very interesting :

we know that any polar line of a point m on KO w.r.t. any conic of (F) goes
through N so, if we take m = O (circumcenter), this polar line will meet
Kiepert in two points V1 & V2 and the tangent lines to Kiepert are OV1 &
OV2. In a previous message, I have called "Vecten points" those two points
(centers of concurrence of lines going through one vertex and the centre of
the square built externally or internally on the opposite side)

then if we call Y1 = GV1 inter HV2 and Y2 = GV2 inter HV1, those two points
are on KO and are inverse w.r.t the circle with diameter OK.
In this situation, the 8 points are in 4 pairs of 2 merged points : V1, V2,
Y1, Y2 vertices of a quadrangle the diagonal triangle of which is HGK, and
this implies a lot of harmonic conjugacy between all those points and N & O.

QUESTION : are those points Y1 & Y2 simply related to ABC ? are they known
under a specific name ?

****************************************

Another thing to study is based on the fact that the Kiepert hyperbola is
the isotomic conjugate of the line KG (G = centroid) therefore it is
possible to adapt all that has been said for the involution on KO to another
involution on KG but the situation is less familiar : new points have to be
identified and a new work has to be done. Maybe, someone will be
interested...

regards and Joyeux Noël to everybody

bernard gibert

Hello friends and season's greetings,

This note is to comment on the connection between the Kiepert hyperbola,
the Euler line, the Brocard line, and the four pairwise families of
points Fx, Nx, Ix, Nx* (where x = n or s, and where the * indicates the
isogonic conjugate).

Since I am putting this on the whole newsgroup, I will restate the
notation. These points are weak points that occur in pairs. The n
indicates the normal version of the point. The s is the second or
switched version. John Conway uses the term "fissile" to describe points
that have two parts. Since I need an adjective for this, I will use that.
What I plan to show is that points on the Kiepert hyperbola are defined
in pairs and that lines through the points of two of these pairs
intersect on the Euler line. Points on the Brocard line share this
property in that if one fissile pair in on the Brocard line and one on
the Kiepert hyperbola, then lines between them intersect on the Euler
line.

There is an appendix on notation and formulas.

Four pairs of fissile points:

Fermat points: Fn and Fs
isodynamic points: In and Is
Napoleon points: Nn and Ns
The conjugate Napoleons Nn* and Ns*


We have shown in the last few days that these 6 points are on a conic. I
am hoping that someone has a name for this conic.

Of these points, two pairs are on the Kiepert hyperbola, a well known
hyperbola that goes through many named points including the centroid, the
orthocenter, the Fermats, the Napoleons, and the vertices of ABC.

Any point Pn on the Kiepert hyperbola has barycentric coordinates of the
form

         Pn =  :1/(SB + u): =:= : (SA+u)(SC+u) :
                             = : SA SC + (SA+SC)u + uu :
                             = : SA SC + bb u + uu :

where u is symmetric in the coordinates. Here we used the identity SA+SC
= bb. SA and SC are defined below in a notation and formula appendix. bb
is email for b^2. The notation : y : is shorthand for (x:y:z) and is used
when, as in our cases, the algebraic expressions for each coordinate are
symmetrical. In the same way, we only do our computations for the y
coordinate.

The point Ps = :1/(SB - u): =:=  : SA SC - bb u + uu : is the other
version of the pair Pn, Ps.

Using these formulas we can immediately show a number of things about
these 2 points. First we know that the points  Fn + Fs     Fn     Fn - Fs
      Fs  are harmonic conjugate pairs. From the above coordinates we have
that Fn-Fs = : 2u bb :, which is K, the symmedian point. Fn+Fs = 2 :SA
SC: + 2 uu :1: , which is on the Euler line since the first term is H,
the orthocenter, and the second term is G, the centroid.

The isogonic conjugate Pn* = : bb(SB + u) : . We immediately prove
similar properties for Pn* and Ps*. As before the four points Pn*+Ps*
Pn*   Pn*-Ps*  Ps* are harmonic conjugate pairs. Pn*+Ps* = : bb(SB + u) +
bb(SB -u): = 2 :bbSB:, the circumcenter. Pn*-Ps* = : bb(SB + u) - bb(SB
-u): = 2 u : bb :, the symmedian point. Hence these points are on the
Brocard line OK.

Hence the isogonic conjugate of the Kiepert hyperbola is the Brocard
line.

Since In = Fn* and Is = Fs* and the Fermats are on the Kiepert hyperbola,
the isodynamics are on the Brocard line. Likewise the conjugates of the
Napoleon points, Nn* and Ns*, are on the Brocard line.

I now want to demonstrate the connection to the Euler line.

If Pn,s = : (SA +- u)(SC +- u) : and Qn,s : (SA +- v)(SC +- v) : are
fissile points on the Kiepert hyperbola, then the intersection of the
lines PnQn and PsQs is on the Euler line. Likewise the intersection of
lines PnQs and PsQn is on the Euler line.

This can easily be shown in coordinates, where as usual we will only do
the computation for the y coordinate.

  Consider the points

      v Pn,s - u Qn,s  =  :v (SA +- u)(SC +- u) - u (SA +- v)(SC +- v):
                          = (v-u) :SA SC: - (v-u) uv :1:
                                     H                G

Hence lines PnQn and PsQs go through the same point on the Euler line. By
similar methods (look at v Pn,s + u Qs,n) it can be shown the PnQs and
PsQn intersect on the Euler line.

The Kiepert hyperbola is loaded with fissile points and this statement is
true for any two pairs.
____________________
Summary: points on the Kiepert hyperbola divide into pairs. The 4 lines
drawn between any two pairs intersect in pairs on the Euler line.
____________________

Now to include the Brocard line.
___________________
Given fissile points on the Kiepert hyperbola and fissile points on the
Brocard line. The 4 lines drawn between the two pairs intersect in pairs
on the Euler line.
-------------------

v Pn,s - u Qn,s  =  : v (SA SC +- bb u + uu)  - u (bb SB +- v bb) :
                     =  : v SA SC - u bb SB + v uu :
                     =  : v SA SC - u (4 kk - SA SC) + v uu :
                     =  (v+u) : SA SC : + u ( uv - 4kk) :1:
                                 H                       G

v Pn,s + u Qs,n  =  : v (SA SC +- bb u + uu)  + u (bb SB -+ v bb) :

                     =  : v SA SC + u bb SB + v uu :
                     =  : v SA SC + u (4 kk - SA SC) + v uu :
                     =  (v-u) : SA SC : + u ( uv + 4kk) :1:
                                 H                       G

So that lines  PnQn and PsQs intersect on the Euler line. Also lines PnQs
and PsQn intersect at a different point on the Euler line.

I think these are very interesting results.

Steve Sigur


_______________________________

Notation and formula APPEMDIX (this is repeated from a previous post).

       We will use Conway's shorthand notation, which makes computation
very easy. In this notation identities divide into parallel weak and
strong identities,

      s = (a + b + c)/2         S = ( aa + bb + cc)/2
      sa = s - a                SA = S - aa
      sb = s - b                SB = S - bb
      sc = s - c                SC = S - cc

When these expressions are used with barycentric coordinates, geometric
properties are intimately related to algebraic ones. Symmetric
expressions often represent the centroid, while other important points
are given by expressions with small breaks in the symmetry. The structure
of algebraic identites in a,b,c are mirrored in the geometric structure
of the triangle.

Some identities we will use are the following

      SA + SC = bb
      bb SB + SA SC =    4kk    where k is the area of the triangle.
      bb SB + 2 SA SC =  y coordinate of nine point center.
      bb SB - 2 SA SC =  y coordinate of Euler pt at infinity.
      3 bb SB + 2 SA SC =  y coordinate of midpt between 9pt center and O.
      SA SC + 4 kk    =  y coordinate of N, the nine pt center
      SA SC - 4 kk    =  y coordinate of O, the circumcenter

Using this notation, we can write the barycentric coordinates of some
important points

     Circumcenter O = ( aa SA : bb SB : cc SC )
     Orthocenter  H = ( SB SC : SC SA : SA SB )
     Centroid     G = ( 1 : 1 : 1)
     Symmedian    K = ( aa : bb : cc )

Since the coordinates are symmetric, we often use a shorthand giving only
the y coordinate.  In this shorthand K = : bb : . In a similar vein we
only do computations for the y coordinate assuming that the other two can
be easily filled in.

The Fermat points have the coordinates

    Fn,s   =     : 1/(Cot B +- Cot pi/3) : = : 1/(SB +- u) :
            =:=  : (SA +- u)(SC +- u) :
            =    : SA SC +- bb u + uu :

with u = 2K / root3. In this we used the identity   2K cot B = SB. These
coordinates are computed by using isosceles Napoleons.

The isodynamic points are the conjugals of the Fermat points

    In,s   =   : bb(SB +- u) :  =  : bb SB + bb u :

The Napoleons are defined in a similar fashion to the Fermats

    Nn,s   =   : 1/(Cot B +- Cot pi/6) : = : 1/(SB +- v) :
            =:= : (SA +- v)(SC +- v) :
            = : SA SC +- bb v + vv :

with v = 2K  root3.

Note that v = 3u and uv = 4 kk.

As an interesting final posting about the Fermats, I will compute the
points where the lines FnFs and NnNs cross the Euler line. The notation
is explained in an appendix.

The coordinates of the two Fermat points are

Fn,s ->  SA SC +- u bb + uu

From this we see the following harmonic conjugate pairs

                     Fn + Fs     Fn     Fn - Fs      Fs

Using the y coordinates Fn-Fs = 2u bb, the symmedian point. The other
point is on the Euler line which we can compute as follows.

Fn+Fs  =   2 SA SC + 2 uu  . Since for the Fermat points uu = 4/3 kk
where k is the area of ABC.

Hence Fn+Fs   =:= 3 SA SC + 4 kk  =:=  4 SA SC + bb SB

which is 1/3 of the way from H to O. [G is 1/3 of the way from O to H].

--------

The Napoleons are done similarly

  Nn+Ns   =:=  SA SC +  vv

  For the Napoleon points vv = 12 kk where k is the area of ABC. Hence

  Nn+Ns   =:=  SA SC + 12 kk = 4 SA SC + 3 bb SB .

  which is 2/5 of the way from O to H.

  Steve

  ------------------------------















  APPENDIX on notation and identities

  We will use Conway's shorthand notation, which makes computation very
easy. In this notation identities divide into parallel weak and strong
identities,

      s = (a + b + c)/2         S = ( aa + bb + cc)/2
      sa = s - a                SA = S - aa
      sb = s - b                SB = S - bb
      sc = s - c                SC = S - cc

When these expressions are used with barycentric coordinates, geometric
properties are intimately related to algebraic ones. Symmetric
expressions often represent the centroid, while other important points
are given by expressions with small breaks in the symmetry. The structure
of algebraic identites in a,b,c are mirrored in the geometric structure
of the triangle.

Some identities we will use are the following

      SA + SC = bb
      bb SB + SA SC =    4kk    where k is the area of the triangle.
      bb SB + 2 SA SC =  y coordinate of nine point center.
      bb SB - 2 SA SC =  y coordinate of Euler pt at infinity.
      3 bb SB + 2 SA SC =  y coordinate of midpt between 9pt center and O.
      SA SC + 4 kk    =  y coordinate of N, the nine pt center
      SA SC - 4 kk    =  y coordinate of O, the circumcenter

Using this notation, we can write the barycentric coordinates of some
important points

     Circumcenter O = ( aa SA : bb SB : cc SC )
     Orthocenter  H = ( SB SC : SC SA : SA SB )
     Centroid     G = ( 1 : 1 : 1)
     Symmedian    K = ( aa : bb : cc )

Since the coordinates are symmetric, we often use a shorthand giving only
the y coordinate.  In this shorthand K = : bb : . In a similar vein we
only do computations for the y coordinate assuming that the other two can
be easily filled in.

The Fermat points have the coordinates

    Fn,s   =     : 1/(Cot B +- Cot pi/3) : = : 1/(SB +- u) :
            =:=  : (SA +- u)(SC +- u) :
            =    : SA SC +- bb u + uu :

with u = 2K / root3. In this we used the identity   2K cot B = SB. These
coordinates are computed by using isosceles Napoleons.

The isodynamic points are the conjugals of the Fermat points

    In,s   =   : bb(SB +- u) :  =  : bb SB + bb u :

The Napoleons are defined in a similar fashion to the Fermats

    Nn,s   =   : 1/(Cot B +- Cot pi/6) : = : 1/(SB +- v) :
            =:= : (SA +- v)(SC +- v) :
            = : SA SC +- bb v + vv :

with v = 2K  root3.

Note that v = 3u and uv = 4 kk.

Hello all,

This is my fist post to the new Hayacinthos Geometry mailing list. I
thank Antreas for establishing it and hope it will prosper.

The following is the mathematics behind the properties of the Fermat,
Isodynamic, and Napoleon points that we have been discussing. This is a
long post.

PRELIMINARIES

We will use Conway's shorthand notation, which makes computation very
easy. In this notation identities divide into parallel weak and strong
identities,

      s = (a + b + c)/2         S = ( aa + bb + cc)/2
      sa = s - a                SA = S - aa
      sb = s - b                SB = S - bb
      sc = s - c                SC = S - cc

When these expressions are used with barycentric coordinates, geometric
properties are intimately related to algebraic ones. Symmetric
expressions often represent the centroid, while other important points
are given by expressions with small breaks in the symmetry. The structure
of algebraic identites in a,b,c are mirrored in the geometric structure
of the triangle.

Some identities we will use are the following

      SA + SC = bb
      bb SB + SA SC =    4kk    where k is the area of the triangle.
      bb SB + 2 SA SC =  y coordinate of nine point center.
      bb SB - 2 SA SC =  y coordinate of Euler pt at infinity.
      3 bb SB + 2 SA SC =  y coordinate of midpt between 9pt center and O.
      SA SC + 4 kk    =  y coordinate of N, the nine pt center
      SA SC - 4 kk    =  y coordinate of O, the circumcenter

Using this notation, we can write the barycentric coordinates of some
important points

     Circumcenter O = ( aa SA : bb SB : cc SC )
     Orthocenter  H = ( SB SC : SC SA : SA SB )
     Centroid     G = ( 1 : 1 : 1)
     Symmedian    K = ( aa : bb : cc )

Since the coordinates are symmetric, we often use a shorthand giving only
the y coordinate.  In this shorthand K = : bb : . In a similar vein we
only do computations for the y coordinate assuming that the other two can
be easily filled in.

The Fermat points have the coordinates

    Fn,s   =     : 1/(Cot B +- Cot pi/3) : = : 1/(SB +- u) :
            =:=  : (SA +- u)(SC +- u) :
            =    : SA SC +- bb u + uu :

with u = 2K / root3. In this we used the identity   2K cot B = SB. These
coordinates are computed by using isosceles Napoleons.

The isodynamic points are the conjugals of the Fermat points

    In,s   =   : bb(SB +- u) :  =  : bb SB + bb u :

The Napoleons are defined in a similar fashion to the Fermats

    Nn,s   =   : 1/(Cot B +- Cot pi/6) : = : 1/(SB +- v) :
            =:= : (SA +- v)(SC +- v) :
            = : SA SC +- bb v + vv :

with v = 2K  root3.

Note that v = 3u and uv = 4 kk.



COLINEARITIES

We can immediately generate results. Here we will use the principle that
if the coordinates of two points add to a third, the three are colinear.
Also the four points a, a+b, b, a-b are harmonic.

Each pair of points is colinear with K, the symmedian point. We are only
giving the computation for the y coordinate

The Fermats

           Fn - Fs  ->   (SA SC + bb u + uu) - (SA SC - bb u + uu) = 2u bb
.

Since 2 u is a constant, we conclude that Fn and Fs are colinear with K.

The Napoleons work out the same way:

          Nn - Ns  ->   (SA SC + bb v + vv) - (SA SC - bb v + vv) = 2v bb .

The isodynamics

          In - Is ->   (bb SB + v bb) - (bb SB - v bb) = 2v bb

Hence the lines joining the Fermats, the isodynamics, and the Napoleons
all go through K.

A consequence of this is that a triangle formed by choosing one Fermat
point, one isodynamic, and one Napoleon point, will be in perspective
with the triangle formed by the remaining three. Another consequence is
that, working Brianchon's theorem backwards, these 6 points lie on a
conic (Thanks to Barry Wolk for this observation).

The harmonic conjugates are all on the Euler line as can be see from

   Fn + Fs  ->   (SA SC + bb u + uu) +  = 2 SA SC + 2 uu .

On the right hand side, SA SC are the coordinates of H and 2 uu = 2uu 1,
the centroid, hence this point is at the intersection of the line that
joins the two Fermat points and the line that joins H and G, the Euler
line. Hence it is on the Euler line.

   In + Is -> (bb SB + v bb) - (bb SB - v bb) = 2v bb SB.

hence the isodynamics go through the circumcenter and are on the OK
Brocard Meridian line.

   Nn + Ns  ->    + (SA SC - bb v + vv) = 2 SA SC + 2 vv .

Which is again on the Euler line.


GENERAL THEOREM

It is no coincidence that the Euler line and the point K keep showing up
as the following theorems show.

For any points of the form 1/(SB+-u), the line between the two point will
go through K. This should be obvious from the above.

For any two pairs of points of the form Xn,s = 1/(SB +- u) and Yn,s =
1/(SB +- v), the lines XnYs and XsYn intersect on the Euler line. For
consider this y coordinate

     v Xn + u Ys = v (SA SC + bb u + uu) + u (SA SC - bb v + vv)
                 = (u+v) SA SC + (u+v) uv

Similarly v Xs + u Yn = (u+v) SA SC + (u+v) uv .

The first term in this result represents H; the second represents G so
that this point is on the Euler line and XnYs and XsYn intersect there.


OTHER INTERSECTIONS

FnIs and FsIn intersect at G
----------------------------

   Fn + Is   ->  (SA SC + bb u + uu) + (bb SB - bb u) = (bb SB + SA SC) +
uu
                                                      = 4 kk + uu = 16/3 kk

Since k is symmetric in a, b,c  we conclude the line through Fn and Is
goes through G. Similarly the FsIn line goes through G.


FnIn and FsIs intersect at Euler point at infinity
--------------------------------------------------

   Fn - In  -> (SA SC + bb u + uu) - (bb SB + bb u) = (-bb SB + SA SC) + uu
                                   = (-bb SB + SA SC) + 1/3 (bb SB + SA SC)
                                    = 2/3(-bb SB + 2 SA SC)

which is the Euler point at infinity. The line FsIs also goes through
this point. Here we used uu = 4kk/3 = 1/3 (bb SB + SA SC).


NnFs and NsFn intersect at 9 pt center
-------------------------------------

   u Nn + v Fs = u (SA SC + bb v + vv) + v (SA SC - bb u + uu)
               = (u+v) SA SC + (u+v) uv  =:= SA SC + 4 kk

so that the line NnFs goes through the nine pt center.
Similarly for NsFn


NnFn and NsFs intersect at the circumcenter
___________________________________________

u Nn - v Fn = u (SA SC + bb v + vv) - v (SA SC + bb u + uu)
               = (u-v) SA SC - (u-v) uv  =:= SA SC - 4 kk

which is the circumcenter. Here we used uv = 4kk. Similarly NsFs go
through O.


NnIs and NsIn intersect at the midpoint between N and O
-------------------------------------------------------

   u Nn + v Is = u (SA SC + bb v + vv) + v (bb SB - u bb)
               = u SA SC + v bb SB + uvv
               = u ( (SA SC + 3 bb SB) + 3 (SA SC + bb SB))
               =:= 2 SA SC + 3 bb SB

which is the midpoint between the 9pt center and the circumcenter. NsIn
also goes through this point. Here we used v = 3u and uv = 4kk = SA SC +
bb SB.


NnIn and NsIs intersect at H, the orthocenter
---------------------------------------------

   u Nn - v In = u (SA SC + bb v + vv) - v (bb SB + u bb)
               = u SA SC - v bb SB + uvv
               =:=   SA SC - 3 bb SB + 12kk
               =     SA SC - 3 bb SB + 3 (SA SC + bb SB)
               = 4 SA SC

Hence the line NnIn goes through the orthocenter. Here we used v = 3u and
vv = 12 kk. Similarly NsIs goes through the orthocenter.

The result of all this is that these points form a desmic triple of
quadrangles as shown in this (corrected) chart.

               o     a     b      b

       I       U1    Fn    In     Nn
       II      U2    Fs    Is     Ns
       III     K     M     N      G

Where M is the midpoint of ON and U1 and U2 are hitherto unknown pairwise
points. My next posting will be about U1 and U2.

Enough for now.

Happy Holidays,

Steve











----------------- End Forwarded Message -----------------

Dear Steve,

Steve Sigur wrote on the Fermat et al. points:
<...>
> GENERAL THEOREM
>
> It is no coincidence that the Euler line and the point K keep showing up
> as the following theorems show.
>
<...>
>
> For any two pairs of points of the form Xn,s = 1/(SB +- u) and Yn,s =
> 1/(SB +- v), the lines XnYs and XsYn intersect on the Euler line. For
> consider this y coordinate
>
>     v Xn + u Ys = v (SA SC + bb u + uu) + u (SA SC - bb v + vv)
>                 = (u+v) SA SC + (u+v) uv
>
> Similarly v Xs + u Yn = (u+v) SA SC + (u+v) uv .
>
> The first term in this result represents H; the second represents G so
> that this point is on the Euler line and XnYs and XsYn intersect there.
>
> OTHER INTERSECTIONS
>
> FnIs and FsIn intersect at G
> ----------------------------
<...>
>
> NnIs and NsIn intersect at the midpoint between N and O
> -------------------------------------------------------
>
<...>
>
> NnIn and NsIs intersect at H, the orthocenter
> ---------------------------------------------
>
<...>

It is no coincidence that points on the Euler line are found, when we
use In,s. This follows from the following theorem:

Let Xn,s and Yn,s be as above. Then the lines XnY*n and and XsY*s (where
* denotes isogonal conjugate) intersect in a point on the Euler line.

Proof:
We have

   Y*n = (:vbb + bbSB:) and Y*s = (:-vbb + bbSB:)

So, Y*n and Y*s are harmonic conjugates w.r.t. O and K.

Note that Xn = (:uu + ubb + SA SC:)

So we have:

     Y*n = (:vbb + bbSB:)         Y*s = (:-vbb + bbSB:)
     Xn  = (:uu + ubb + SA SC:)   Xs = (:uu - ubb + SA SC:)

v Xn - u Y*n = (:v uu + v SA SC - u bb SB:)
v Xs - u Y*s = (:v uu + v SA SC - u bb SB:)

This point, clearly the intersection of Y*nXn and Y*sXs, can be
rewritten as uu(:1:) + v(:SA SC:) + u(:bb SB:).

Since (:1:) represents the centroid, (:SA SC:) the orthocenter and
(:bb SB:) the circumcenter, this point is on the Euler line.

Kind regards, and happy holidays,
Floor.

Dear all,

Let me start in a similar way as Clark did, and thank Antreas for
coining the idea to start a discussion list and then finding a good way
to really start one. Thanks a lot, Antreas! I was a bit surprised by the
name Hyacinthos, which I first associated with bulbs in my garden, and
second with ``Hyacinth Bucket'' (say Bouquet) from the British comedy
series ``Keeping up appearances''... ;)
Attributing this list to E. Lemoine is excellent.

> From: Clark Kimberling <ck6@...>
>
> Dear triangle geometers,
>
> I'm sure I speak for many who thank Antreas for establishing
> Hyacinthos@onelist.
>
> Antreas mentioned that that name Hyacinthos honors E.  Lemoine, of whose
> full name Hyacinthe is a part. The Lemoine point is often called the
> symmedian point. In Ross Honsberger's Episodes in Nineteenth and Twentieth
> Century Euclidean Geometry (Mathematical Association of America, 1995), a
> whole chapter is devoled to this point.
>
> However, Honsberger doesn't mention (directly) a certain interesting
> property of the Lemoine point.  For any point P, let A'B'C' denote the
> pedal triangle of P (i.e., A' is the point in which the line through P
> perpendicular to line BC meets line BC). Let S(P) be the vector sum
> PA'+PB'+PC'.  Then S(P) is the zero vector if P is the Lemoine point.

Isn't this property exactly equivalent to the fact that the
Lemoine/symmedian point K is the centroid of its pedal triangle? This
property of K is mentioned in O. Bottema's ``Hoofdstukken uit de
Elementaire Meetkunde''.

We can generalize this in the sence of P-trace (P-pedal) triangles: Let
X be a point, the inscribed triangle A'B'C' in ABC is called the P-trace
triangle of X if XA'//AP, XB'//BP and XC'//CP.

If P = f:g:h (barycentrics) then the P-version K_P of the Lemoine point
is f(g+h):g(f+h):h(f+g).

The centroid of the P-trace triangle of X = x:y:z is found as:

  2x + f/(f+h) y + f/(f+g) z :
  2y + g/(g+h) x + g/(g+f) z :
  2z + h/(h+g) x + h/(h+f) y

It is easy to verify that in case that X=K_P then indeed this centroid
is equal to X.

> I conjecture that the converse is true: that if P is a "point" (i.e.,
> f(a,b,c) : g(a,b,c) : h(a,b,c)) such that S(P)=0, then P = a^2 : b^2 : c^2
> (barycentric coordinates of the Lemoine point).

This is not immediatly apparent to me.

> By the way, many other vector sums involving triangle centers will be
> included in ETC (Encyclopedia of Triangle Centers), which should appear
> sometime before March 1, 2000.

Excellent news!

Another property of the Lemoine point: It is the center of the conic
inscribed in ABC, tangent to the sides of ABC in the orthic triangle.
(Problem 525 in Nieuw Archief voor Wiskunde, by O. Bottema).

Kind regards,
Floor.

Dear triangle geometers,

I'm sure I speak for many who thank Antreas for establishing
Hyacinthos@onelist.

Antreas mentioned that that name Hyacinthos honors E.  Lemoine, of whose
full name Hyacinthe is a part. The Lemoine point is often called the
symmedian point. In Ross Honsberger's Episodes in Nineteenth and Twentieth
Century Euclidean Geometry (Mathematical Association of America, 1995), a
whole chapter is devoled to this point.

However, Honsberger doesn't mention (directly) a certain interesting
property of the Lemoine point.  For any point P, let A'B'C' denote the
pedal triangle of P (i.e., A' is the point in which the line through P
perpendicular to line BC meets line BC). Let S(P) be the vector sum
PA'+PB'+PC'.  Then S(P) is the zero vector if P is the Lemoine point.

I conjecture that the converse is true: that if P is a "point" (i.e.,
f(a,b,c) : g(a,b,c) : h(a,b,c)) such that S(P)=0, then P = a^2 : b^2 : c^2
(barycentric coordinates of the Lemoine point).

By the way, many other vector sums involving triangle centers will be
included in ETC (Encyclopedia of Triangle Centers), which should appear
sometime before March 1, 2000.

Best holiday regards to all.

Clark Kimberling