Well Hyacinthists, did you remember that to-day is Saint Hyacinthos day?
Here on Crete he is the local version of Saint Valentine.

   So all people in love (including with geometry) cheer-up.

    Michael

Dear all,

In his paper on the Gergonne problem in FG Nikolaos Dergiades showed how
to find a point O in a given plane \pi such that AO+BO+CO was minimal.
Now if we take for \pi different planes parallel to the plane of ABC,
and we let O' be the orthogonal projection of O to ABC, can anything be
said about the locus of O', except of course that it passes through the
Fermat point?

Kind regards,
Sincerely,
Floor.

Dear Floor and other Hyacinthists,
Floor wrote :

> In his paper on the Gergonne problem in FG Nikolaos Dergiades
showed how
> to find a point O in a given plane \pi such that AO+BO+CO was
minimal.
> Now if we take for \pi different planes parallel to the plane of
ABC,
> and we let O' be the orthogonal projection of O to ABC, can
anything be
> said about the locus of O', except of course that it passes through
the
> Fermat point?
>

O' must be barycentric (1/OA, 1/OB, 1/OC);
As OA^2 - O'A^2 = OB^2 - O'B^2 = OC^2 - O'C^2,
a little elimination leads to the quintic
a^2.y.z.(y-z).(y.z + x^2) + circular = 0
Friendly. Jean-Pierre

Dear Jean-Pierre,

[FvL]
> > In his paper on the Gergonne problem in FG Nikolaos Dergiades
> showed how
> > to find a point O in a given plane \pi such that AO+BO+CO was
> minimal.
> > Now if we take for \pi different planes parallel to the plane of
> ABC,
> > and we let O' be the orthogonal projection of O to ABC, can
> anything be
> > said about the locus of O', except of course that it passes through
> the
> > Fermat point?

[JPE]
> O' must be barycentric (1/OA, 1/OB, 1/OC);

Probably there is a simple reason for this, but I don't immediately see
it?

[JPE]:
> As OA^2 - O'A^2 = OB^2 - O'B^2 = OC^2 - O'C^2,
> a little elimination leads to the quintic
> a^2.y.z.(y-z).(y.z + x^2) + circular = 0

Thanks!

Kind regards,
Floor.

Dear Floor,

> [JPE]
> > O' must be barycentric (1/OA, 1/OB, 1/OC);
>
> Probably there is a simple reason for this, but I don't immediately
see
> it?

Nikolaos Dergiades wrote in the FG article :
<< If O is a point on the plane /pi such as the centroidal line of
the trihedron O.ABC is perpendicular to /pi, then
OA + OB + OC <= MA + MB + MC for every point M on /pi>>
As the centroidal line is directed by
Vect(OA)/OA + Vect(OB)/OB + Vect(OC)/OC, we get, by orthogonal
projection on the plane ABC,
Vect(O'A)/OA + Vect(O'B)/OB + Vect(O'C)/OC = 0.
Friendly. Jean-Pierre

Dear Floor, Jean-Pierre,...
Jean-Pierre  wrote

>Nikolaos Dergiades wrote in the FG article :
><< If O is a point on the plane /pi such as the centroidal line of
>the trihedron O.ABC is perpendicular to /pi, then
>OA + OB + OC <= MA + MB + MC for every point M on /pi>>
>As the centroidal line is directed by
>Vect(OA)/OA + Vect(OB)/OB + Vect(OC)/OC, we get, by orthogonal
>projection on the plane ABC,
>Vect(O'A)/OA + Vect(O'B)/OB + Vect(O'C)/OC = 0.


another explanation is:
Since OO'  is the centroidal line of the trihedron O.ABC
if AO' meets BC at D then OD is bisector of triangle OBC
and hence if  (x : y : z) are the barycentrics of O'
we have  z / y = BD / DC = OB / OC   (bisector's theorem)
or   y.OB = z.OC  and similarly  x.OA = y.OB = z.OC
and the barycentrics of O' are (1/OA : 1/OB : 1/OC).
Best regards
Nikolaos Dergiades

Is there a context in which the 1st isogonic point (Fermat-Torricelli) might
also be referred to as the Steiner point?  I think I have heard it used this
way with regard to a minimum spanning tree on a graph with 4 vertices.

Chuck


____________________________________________________________________
Mr. R.S. (Chuck) Tiberio
Instructor in Mathematics
Wellesley High School
Wellesley, MA  02481

Visit me at http://www.tiac.net/users/tiberio


____________________________________________________________________
Get free email and a permanent address at http://www.amexmail.com/?A=1

Let ABC be a triangle and PaPbPc the pedal triangle of P.

                              A
                             /\
                            /  \
                           /    \
                         Pc      Pb
                         / z   y  \ t'y + t"z
              t'z + t"y /    P     \
                       /            Ab
                      /      Ma      \
                     Ac               \
                    /        x         \
                   B--------Pa----------C

Denote x = PPa, y = PPb, z = PPc (actual normals of P), and let
Ab, Ac be points on AC, AB, resp. (see the figure) such that PbAb = t'y + t"z
PcAc = t'z + t"y, and let Ma be the midpoint of AbAc.
Similarly we define Mb, Mc [= midpoints of BcBa, CaCb, resp.]

If t' = t" = 0, then AbAc coincides with PbPc, etc, and the triangle
MaMbMc is the medial triangle of PaPbPc.
In this case the locus of P such that MaMbMc, ABC are perspective is
Darboux cubic.
[In fact the locus is Darboux cubic for every MaMbMc which is the cevian
triangle of some point w.r.t PaPbPc]

In general, for arbitrary t' and t", which is the locus of P such that
MaMbMc, ABC are perspective?

The locus that I found is the isogonal cubic with pivot (in normals):

  1 + t"sinA        1 + t"sinB          1 + t"sinC
------------- - [ ------------- ] * [ ------------- ]  ::
cosA + t'sinA     cosB + t'sinB       sinC + t'sinC


t'  t"                            Pivot
__________________________________________________________________
0   0          cosA - cosBcosC :: = L (de Longchamps point)

0   1          (1 + sinA)/cosA - (1 + sinB)(1 + sinC)/cosBcosC ::

1   0          (sinA + cosA) - (sinB + cosB)(sinC + cosC) ::


Assume that t' =/= 0, and let t = t"/t'

Then PbAb = y + tz, PcAc = z + ty, etc

Pivot:

   (1 + tsinA)/(cosA + sinA) - [(1 + tsinB)/(cosB + sinB)]*
    [(1 + tsinC)/(cosC + sinC)] ::

So the locus of the pivots, as t varies, is the conic with that
parametric form (perhaps a "non-property" conic, as JPE would say !).

Greretings

Antreas


                   Geometry

I prove a theorem and the house expands:
the windows jerk free to hover near the ceiling,
the ceiling floats away with a sigh.

As the walls clear themselves of everything
but tranparency, the scent of carnations
leaves with them. I am out in the open

and above the windows have hinged into butterflies,
sunlight glinting where they've intersected.
They are going to some point true and unproven.

-- Rita Dove

R.S.Tiberio wrote:

>Is there a context in which the 1st isogonic point (Fermat-Torricelli) might
>also be referred to as the Steiner point?  I think I have heard it used this
>way with regard to a minimum spanning tree on a graph with 4 vertices.

Dear Chuck

See:

Steiner Minimal Tree Problem
Bob Bell

          http://www.css.tayloru.edu/~bbell/steiner/

Antreas

[APH]:

>> 2. Let ABC be a triangle, P a point, and A'B'C' the circumcevian triangle
>> of P.
>> Let A" = B'C /\ C'B
>>     B" = C'A /\ A'C
>>     C" = A'B /\ B'A
>>
>>Which is the locus of P such that A"B"C" and ABC are perspective?
>[...]
>>The second one, is a cubic:
>>If my calculations were correct, the equation of the locus is:
>>
>>    (1 - sinA)x(y^2sinC - z^2sinB) + cyclic = 0  (in normals)

[PY]:

>A''B''C'' being the desmic mate of the circumcevian triangle of P, is
>always perspective with ABC. Since
>
>A' = (-a^2/(b^2/v+c^2/w) : v : w)
>
>etc,
>
>A'' = (u : b^2/(c^2/w+a^2/u) : c^2/(a^2/u+b^2/v))
>
>etc, and the perspector is
>
>(a^2/(b^2/v + c^2/w) : ... : ... ),
>
>gmg(P),
>
>g = isogonal conjugate,
>m = image in medial triangle.

I had used a correct formula (see below) but my calculations were not correct.

Let's try again.


                                    A
                                    /\            B'
                         C'        /  \
                                  /1 2 \A*
                                 /      \
                                /2  P   1\
                               / 1      2 \
                              B------------C


                                     A'

                                       A"

Denote angles PAB = A1, PAC = A2, PBC = B1, PBA = B2, PCA = C1, PCB = C2

Let A*, B*, C* be the reflections of A" on BC, B" on CA, C" on CA, resp.,
and let P = (x:y:z) in actual normals.

A"B"C", A*B*C* are in perspective with ABC <==>

cotB -+ cot(B + C1)   cotC -+ cot(C + A1)    cotA -+ cot(A + B1)
------------------- * -------------------- * ------------------- = 1, resp.
cotC -+ cot(C + B2)   cotA -+ cot(A + C2)    cotB -+ cot(B + A2)

1. Let's compute the expression:

cotB - cot(B + C1)   cotC - cot(C + A1)   cotA - cot(A + B1)
------------------ * ------------------ * ------------------
cotC - cot(C + B2)   cotA - cot(A + C2)   cotB - cot(B + A2)

We have (after some calcualtions):

                             ysinC
cotB - cot(B + C1) = --------------------
                      sinB(ysinA + xsinB)


                            zsinB
cotC - cot(C + B2) = --------------------
                      sinC(zsinA + xsinC)

==>

cotB - cot(B + C1)    y     sin^2C    zsinA + xsinC
------------------ = --- * ------- * --------------
cotC - cot(C + B2)    z     sin^2B    ysinA + xsinB

Similarly:

cotC - cot(C + A1)    z     sin^2A    xsinB + ysinA
------------------ = --- * ------- * --------------
cotA - cot(A + C2)    x     sin^2C    zsinB + ysinC

cotA - cot(A + B1)    x     sin^2B    ysinC + zsinB
------------------ = --- * ------- * --------------
cotB - cot(B + A2)    y     sin^2A    xsinC + zsinA


Therefore

cotB - cot(B + C1)   cotC - cot(C + A1)   cotA - cot(A + B1)
------------------ * ------------------ * ------------------ = 1
cotC - cot(C + B2)   cotA - cot(A + C2)   cotB - cot(B + A2)

==> A"B"C", ABC are perspective.

                           1                            1
The Perspector we get is --- * sin^2A * ------------------------------- ::
                           x             (xsinB + ysinA)*(xsinC + zsinA)


What is that? Well... it is in barycentrics while P = (x:y:z) is in normals,
so we have to "regularize" it by taking P = (X:Y:Z) in barycentrics:

We have x = X/a, y = Y/b, z = Z/c

So, the perspector becomes:

   1                            1
----- * a^2 * -----------------------------------  :: =
(X/a)         ((X/a)b + (Y/b)a)*((X/a)c + (Z/c)a)

              a^4
= ---------------------------  ::
   X(b^2X + a^2Y)(c^2X + a^2Z)

[this must be the same with Paul's above, but is it???]


2. Let's now compute the expression:

cotB + cot(B + C1)   cotC + cot(C + A1)    cotA + cot(A + B1)
------------------ * ------------------ * -------------------
cotC + cot(C + B2)   cotA + cot(A + C2)    cotB + cot(B + A2)


We have (after some calculations):

                        1      ysin(2B+C) + xsin2B
cotB + cot(B + C1) = ----- * --------------------
                      sinB        ysinA + xsinB

                        1      zsin(2C+B) + xsin2C
cotC + cot(C + B2) = ----- * --------------------
                      sinC        zsinA + xsinC

==>

cotB + cot(B + C1)   sinC   ysin(2B+C) + xsin2B   zsinA + xsinC
------------------ = ---- * ------------------- * --------------
cotC + cot(C + B2)   sinB   zsin(2C+B) + xsin2C   ysinA + xsinB

Similarly etc

Therefore

A*B*C*, ABC are perspective <==>

cotB + cot(B + C1)   cotC + cot(C + A1)    cotA + cot(A + B1)
------------------ * ------------------ * ------------------- = 1
cotC + cot(C + B2)   cotA + cot(A + C2)    cotB + cot(B + A2)

<==>

ysin(2B+C) + xsin2B   zsin(2C+A) + ysin2C   xsin(2A+B) + zsin2A
------------------- * ------------------- * ------------------- = 1
zsin(2C+B) + xsin2C   xsin(2A+C) + ysin2A   ysin(2B+A) + zsin2B

<==>

ysin(A-B) + xsin2B   zsin(B-C) + ysin2C   xsin(C-A) + zsin2A
------------------ * ------------------ * ------------------- = 1
zsin(A-C) + xsin2C   xsin(B-A) + ysin2A   ysin(C-B) + zsin2B

<==>

  ysin(A-B) + xsin2B    zsin(B-C) + ysin2C    xsin(C-A) + zsin2A
------------------- * -------------------- * ------------------- = 1
-zsin(C-A) + xsin2C   -xsin(A-B) + ysin2A    -ysin(B-C) + zsin2B

<==>

x(sin2C[sin(C-A)sin(A-B) + sin(B-C)sin2A]y^2 - sin2B[sin(C-A)sin(A-B) -
sin(B-C)sin2A]z^2) + cyclic + 2xyzsin(A-B)sin(B-C)sin(C-A) = 0

So, if everything was correct in my calculations, the locus of P such that
A*B*C*, ABC are perspective, is the curve (cubic) with above equation.


Greetings from Athens

Antreas

Our future lawyers, clergy, and statesmen are expected at the University to
learn a good deal about curves, and angles, and numbers and proportions; not
because these subjects have the smallest relation to the needs of their lives,
but because in the very act of learning them they are likely to acquire that
habit of steadfast and accurate thinking, which is indispensible to success
in all pursuits of life.
- J.C. Fitch (1906)

Let ABC be a triangle and AA', BB', CC' its int. angle bisectors.

Consider three circles Ka(t), Kb(t), Kc(t), with common radius t
and centers Ka, Kb, Kc lying on the AA', BB', CC' (extensions or not) resp.
and touching BC, CA, AB, resp. at A", B", C" (ie A", B", C" are the
orth. proj. of Ka, Kb, Kc on BC, CA, AB, resp.)

                                    A
                                   /\
                                  /  \
                                 /    \
                                /      \
                               /        \
                              /          \
                             /            \
                            /              \
                           /                \
                          B----A"---A'-------C



                                   Ka

Compute the common radius t so that A"B"C" be in perspective with ABC.

Variant: The three circles touch the circumcircle (int. or externally)
at A",B",C" instead of the triangle sides.


Antreas

Recently, the Encyclopedia of Triangle Centers - ETC acquired 200+ more
centers.  Several originated in Hyacinthos discussions. To find your
contributions, click on

http://cedar.evansville.edu/~ck6/encyclopedia/ETC.html .

and search for your surname. Please send me corrections and suggestions.

If you can't download the above large file, let me know and I'll
rerereconsider updating the "Parts" that are accessible from the Front
Cover, which has address

http://cedar.evansville.edu/~ck6/encyclopedia/ .

The support-page SEARCH has been updated.  Other support-pages (LINKS,
GLOSSARY, MORE, etc.) are ripe for updates and repairs, probably in August.
(The support-pages are all clickable from the first mentioned address.  To
reach them from the second address, first select "All" or any "Part".)

Clark Kimberling

Dear Antreas.

[APH]: Let ABC be a triangle and AA', BB', CC' its int. angle
bisectors.  Consider three circles Ka(t), Kb(t), Kc(t), with common
radius t  and centers Ka, Kb, Kc lying on the AA', BB', CC'
(extensions or not) resp.  and touching BC, CA, AB, resp. at A", B",
C" (ie A", B", C" are the  orth. proj. of Ka, Kb, Kc on BC, CA, AB,
resp.)
>
>                                    A
>                                   /\
>                                  /  \
>                                 /    \
>                                /      \
>                               /        \
>                              /          \
>                             /            \
>                            /              \
>                           /                \
>                          B----A"---A'-------C
>
>
>
>                                   Ka

Compute the common radius t so that A"B"C" be in perspective with ABC.

***
The incircle is a trivial case.

Another case occurs when the points of tangency are the traces of the
isogonal conjugate of the Clawson point, i.e. the point X(63) with
homogeneous barycentric coordinates (a.SA : b.SB : c.SC). The
perpendiculars to the side lines at these traces intersect the
respective angle bisectors at points at a common distance 2R from the
sides. Thus these equal circles has radius equal to the
circumdiameter of ABC.

These two are the only possibilities.

Best regards
Sincerely
Paul



> Variant: The three circles touch the circumcircle (int. or
externally)
> at A",B",C" instead of the triangle sides.
>
>
> Antreas

Dear Paul

[APH]
>>Let ABC be a triangle and AA', BB', CC' its int. angle
>>bisectors.  Consider three circles Ka(t), Kb(t), Kc(t), with common
>>radius t  and centers Ka, Kb, Kc lying on the AA', BB', CC'
>>(extensions or not) resp.  and touching BC, CA, AB, resp. at A", B",
>>C" (ie A", B", C" are the  orth. proj. of Ka, Kb, Kc on BC, CA, AB,
>>resp.)
>>
>>                                    A
>>                                   /\
>>                                  /  \
>>                                 /    \
>>                                /      \
>>                               /        \
>>                              /          \
>>                             /            \
>>                            /              \
>>                           /                \
>>                          B----A"---A'-------C
>>
>>
>>
>>                                   Ka
>>Compute the common radius t so that A"B"C" be in perspective with ABC.

[PY]:
>The incircle is a trivial case.

I would bet that there should be an other case which is
more trivial than that ! :-)

[PY]:
>Another case occurs when the points of tangency are the traces of the
>isogonal conjugate of the Clawson point, i.e. the point X(63) with
>homogeneous barycentric coordinates (a.SA : b.SB : c.SC). The
>perpendiculars to the side lines at these traces intersect the
>respective angle bisectors at points at a common distance 2R from the
>sides. Thus these equal circles has radius equal to the
>circumdiameter of ABC.
>
>These two are the only possibilities.

Well.... The problem can be solved by the Ceva theorem.
By this theorem we get something like this:

f(t)/f'(t) * g(t)/g'(t)  * h(t)/h'(t) = 1

where f(t) etc are first degree polynomials of t.

So we get an equation of third degree (cubic).

You already found two real roots of this equation.
So the third one must be REAL too! Which is it?

A circle is a real circle even if its radius is zero! :-)

PS: I think that we can formulate an analogous problem based on Menelaus:
That is, the centers Ka, Kb, Kc lie on the ext. angle bisectors,
and the points of tangency A",B",C" be collinear.

Greetings

Antreas

The point about zero is that we do not need to use it in the operations
of daily life. No one goes out to buy zero fish. It is in a way the most
civilized of all cardinals, and its use is only forced on us by the needs
of cultivated modes of thought.
- Alfred North Whitehead.

Dear Antreas
you wrote


>Let ABC be a triangle and AA', BB', CC' its int. angle bisectors.
>
>Consider three circles Ka(t), Kb(t), Kc(t), with common radius t
>and centers Ka, Kb, Kc lying on the AA', BB', CC' (extensions or not) resp.
>and touching BC, CA, AB, resp. at A", B", C" (ie A", B", C" are the
>orth. proj. of Ka, Kb, Kc on BC, CA, AB, resp.)

>Compute the common radius t so that A"B"C" be in perspective with ABC.
>
>Variant: The three circles touch the circumcircle (int. or externally)
>at A",B",C" instead of the triangle sides.


If my calculations are correct
then for the first case
we have
t = 0
t = r  (inradius)
t = -2R   (R = circumradius )
Best regards
Nikolaos Dergiades

Dear Hyacinthists,

If the three angles in degrees a, b, c,
are rational numbers such that
a < b < c < 90    and     2.sina.sinb = sinc,
can we prove or disprove that
a + b = 90    and    c = 2a ?

Best regards
Nikolaos Dergiades

[APH]:
>Let ABC be a triangle and AA', BB', CC' its int. angle bisectors.
>
>Consider three circles Ka(t), Kb(t), Kc(t), with common radius t
>and centers Ka, Kb, Kc lying on the AA', BB', CC' (extensions or not) resp.
>and touching BC, CA, AB, resp. at A", B", C" (ie A", B", C" are the
>orth. proj. of Ka, Kb, Kc on BC, CA, AB, resp.)
>
>                                   A
>                                  /\
>                                 /  \
>                                /    \
>                               /      \
>                              /        \
>                             /          \
>                            /            \
>                           /              \
>                          /                \
>                         B----A"---A'-------C
>
>
>
>                                  Ka
>
>Compute the common radius t so that A"B"C" be in perspective with ABC.

Variant:

Let A*B*C* be the medial triangle of ABC.

The circles Ka(t), Kb(t), Kc(t) touch at A", B", C" the sides B*C*, C*A*, A*B*
(resp.) of the medial triangle [instead of the sides BC, CA, AB of ABC]

If I remember correctly, one of the possible cases, namely:
t = common inradius of the triangles AB*C*, BC*A*, CA*B*, was proposed
as a problem by Paul in the _Crux M/m_

APH

The following paper has been published in Forum Geometricorum. It can be
viewed at

http://forumgeom.fau.edu/FG2001volume1/FG200114index.html

The Editors
Forum Geometricorum

--------------
Clark Kimberling, Multiplying and Dividing Curves by Points,

Forum Geometricorum, 1 (2001) 99 -- 105.

Abstract: Pointwise products and quotients, defined in terms of barycentric
and trilinear coordinates, are extended to products $P\cdot\Gamma $ and
quotients $\Gamma/P$, where $P$ is a point and $\Gamma $ is a curve. In
trilinears, for example, if $\Gamma_0$ denotes the circumcircle, then
$P\cdot\Gamma_0$ is a parabola if and only if $P$ lies on the Steiner
inscribed ellipse.  Barycentric division by the triangle center $X_{110}$
carries $\Gamma_0$ onto the Kiepert hyperbola $\Gamma'$; if $P$ is on
$\Gamma_0$, then the point $P'=P/X_{110}$ is the point, other than the
Tarry point, $X_{98}$, in which the line $PX_{98}$ meets $\Gamma'$, and if
$\Omega_1$ and $\Omega_2$ denote the Brocard points, then
$|P'\Omega_1|/|P'\Omega_2|=|P\Omega_1|/|P\Omega_2|$; that is, $P'$ and $P$
lie on the same Apollonian circle with respect to $\Omega_1$ and $\Omega_2$.

Let ABC be a triangle, and PaPbPc the pedal triangle of point P.
The perp. bisectors of PPc, PPb meet at A', and intersect BC at Ab, Ac, resp.

                               A
                              /\
                             /  \
                            /    \
                           /  A'  \
                          Pc       Pb
                         /    P     \
                        /  B'    C'  \
                       B-Ab-------Ac--C

Similarly we define the points B', C' and Bc,Ba on CA and Ca, Cb on AB.
The triangle A'B'C' is the Midway triangle of P.

For which point P we have that: AbAc = BcBa = CaCb ?

Generalization:
Let P be a point and A', B', C' points on PA, PB, PC, resp., such that
PA'/PA = PB'/PB = PC'/PC = t
[A'B'C' = t-way triangle of P]

Let Ab = A'B' /\ BC, Ac = A'C' /\ BC
Similarly we define the points Bc,Ba; Ca,Cb.

Which is the locus of P such that AbAc = BcBa = CaCb, as t varies?

APH

Dear Antreas,

[APH]:
> Let ABC be a triangle, and PaPbPc the pedal triangle of point P.
> The perp. bisectors of PPc, PPb meet at A', and intersect BC at Ab, Ac, resp.
>
>                               A
>                              /\
>                             /  \
>                            /    \
>                           /  A'  \
>                          Pc       Pb
>                         /    P     \
>                        /  B'    C'  \
>                       B-Ab-------Ac--C
>
> Similarly we define the points B', C' and Bc,Ba on CA and Ca, Cb on AB.
> The triangle A'B'C' is the Midway triangle of P.

We can simply take Ab to be A'B' /\ BC and Ac to be A'C'/\BC (etc.).

[APH]:
> For which point P we have that: AbAc = BcBa = CaCb ?

This is the point in barycentrics (-3bc+ac+ab : bc-3ac+ab : -3ab+ab+ac).
I have not checked whether it is in ETC, but you probably have, Antreas?

[APH]:
> Generalization:
> Let P be a point and A', B', C' points on PA, PB, PC, resp., such that
> PA'/PA = PB'/PB = PC'/PC = t
> [A'B'C' = t-way triangle of P]
>
> Let Ab = A'B' /\ BC, Ac = A'C' /\ BC
> Similarly we define the points Bc,Ba; Ca,Cb.
>
> Which is the locus of P such that AbAc = BcBa = CaCb, as t varies?

I get the point ( (-2-t)bc+ab+ac : ... : ... ) but here t= AA'/PA'.

This shows that the locus is the line through the centroid and the
isotomic conjugate of the incenter.

Kind regards,
Sincerely,
Floor.

Dear Floor,

[APH]:
>> Let ABC be a triangle, and PaPbPc the pedal triangle of point P.
>> The perp. bisectors of PPc, PPb meet at A', and intersect BC at Ab, Ac, resp.
>>
>>                               A
>>                              /\
>>                             /  \
>>                            /    \
>>                           /  A'  \
>>                          Pc       Pb
>>                         /    P     \
>>                        /  B'    C'  \
>>                       B-Ab-------Ac--C
>>
>> Similarly we define the points B', C' and Bc,Ba on CA and Ca, Cb on AB.
>> The triangle A'B'C' is the Midway triangle of P.

[FVL]:
>We can simply take Ab to be A'B' /\ BC and Ac to be A'C'/\BC (etc.).

[APH]:
>> For which point P we have that: AbAc = BcBa = CaCb ?

[FvL]:
>This is the point in barycentrics (-3bc+ac+ab : bc-3ac+ab : -3ab+ab+ac).
>I have not checked whether it is in ETC, but you probably have, Antreas?

No, I have not.

[APH]:
>> Generalization:
>> Let P be a point and A', B', C' points on PA, PB, PC, resp., such that
>> PA'/PA = PB'/PB = PC'/PC = t
>> [A'B'C' = t-way triangle of P]
>>
>> Let Ab = A'B' /\ BC, Ac = A'C' /\ BC
>> Similarly we define the points Bc,Ba; Ca,Cb.
>>
>> Which is the locus of P such that AbAc = BcBa = CaCb, as t varies?

[FvL]:
>I get the point ( (-2-t)bc+ab+ac : ... : ... ) but here t= AA'/PA'.
>
>This shows that the locus is the line through the centroid and the
>isotomic conjugate of the incenter.

I don't remember what I got, so I have to solve it again!

Let PPa = x, PPb = y, PPc = z (actual normals of P)

                               A
                              /\
                             /  \
                            /    \
                           /  A'  \
                          Pc       Pb
                         /   A2  A3 \
                        /      P     \
                       /   B'      C' \
                      B-Ab-----Pa--Ac--C


Let A2 = A'B' /\ PPc, A3 = A'C' /\ PPb

PA'/PA = PA2 / PPc = t ==> PA2 = tz
PC'/PC = PA3 / PPb = t ==> PA3 = ty

        (PPa * cos(A3AcPa)) + PA3
AbPa = -------------------------
                sin(A3AcPa)

==>

        xcosC + ty                       xcosB + tz
AbPa = ----------, and similarly PaAc = -----------
           sinC                             sinB

==>
                       xsinA + tysinB + tzsinC    ax + bty + ctz
AbAc = AbPa + PaAc = ------------------------- = --------------
                                sinBsinC                bc

Similarly we get cyclical expressions of BcBa,CaCb.

Now,
AbAc = BcBa = CaCb

==>

   (a^2)x + taby + tcaz =
= tabx + (b^2)y + tbcz =
= tcax + tbcy + (c^2)z

==>

         x                    y                     z
------------------ = ------------------ = ------------------
| 1   tab   tca  |   | a^2   1   tca  |   | a^2   tab   1  |
|                |   |                |   |                |
| 1   b^2   tbc  |   | tab   1   tbc  |   | tab   b^2   1  |
|                |   |                |   |                |
| 1   tbc   c^2  |   | tca   1   c^2  |   | tca   tbc   1  |


==>

                     x                           y               z
---------------------------------------  = -------------- = -----------
(1 - t) * bc * [-bc + t(-bc + ca + ab)]     (1-t)* cycl.    (1-t)*cycl.

And since 1 - t =/= 0 [otherwise ABC is necessarily equilateral]

==>

P = (x:y:z) = (bc[-bc + t(-bc + ca + ab)] :: ) in Normals

Let's now see if our results are the same:

You found that:

P = ((-2-T)bc+ca+ab ::) in Barycentrics, where T = AA'/PA' = ... = ...

We have:

  1     PA     PA' + A'A
--- = ---- = ---------- =  1 + T  ==> T = (1/t) - 1 = (1-t)/t
  t     PA'       PA'

P = ((-2-T)bc+ca+ab ::) = ((-2 - [(1-t)/t])bc + ca + ab ::) =

= (-bc + t(bc + ca + ab) ::) in barycentrics =

= (bc[-bc + t(-bc + ca + ab)] ::) in normals.

So, our results agree.


Antreas

Let ABC be a triangle and PaPbPc the pedal triangle of point P.

Let (Ab),(Ac) be the NPCs of the (right-anglred) triangles PaBP, PaCP, resp.
     (Bc),(Ba)                                              PbCP, PbAP
     (Ca),(Ba)                                              PcAP, PcBP

Let La be the common chord (rad. axis) of the circles (Ba),(Ca)
     Lb                                                (Cb),(Ab)
     Lc                                                (Ac),(Bc).

Then the lines La, Lb, Lc concur.

The point of concurrence is the circumcenter of the triangle PaPbPc (center
of pedal circle).

This point is the midpoint of PP', where P' is the isog. conj. of P.
So, the cases of P = {O or H} are interesting: In both, the point of
concurrence is the point N [= Nine Point Circle Center] of ABC.

How about the position of the six centers Ab, Ac, Bc, Ba, Ca, Cb ?
Are they lying on some remarkable curve? (conic?)

[If I remember correctly, Paul and I have discussed the case of P = H]

Antreas

I wrote:

>                      xsinA + tysinB + tzsinC    ax + bty + ctz
>AbAc = AbPa + PaAc = ------------------------- = --------------- *(2R)
>                               sinBsinC                bc

I added the "*(2R)"

Antreas

Let ABC be a triangle and consider an arbitrary triangle AB'C' such that:
AB'C' ~ [=similar to] ABC [the vertex A is in common].
Find the locus of the intersection of the lines BB' and CC'.
(I. Ioannides: Mathematics of D' of Gymnasium. Vol II.
OEDB: Athens, 1968, p. 52, #66)


Antreas


Time out:
"OEDB" are the initials of the Greek governemental Publ. house which
publishes didactical books for primary and secondary Greek schools.
[It means: ORGANISMOS      EKDOSEWS    DIDAKTIKWN BIBLIWN
            Organization of Edition of  Didactical Books]

It is written on the books with this form:  O E
                                             D B

Last week my 8 y.o. son showed me the logo above (in one of his books)
and asked me:
- Do you know what this means ?
- Yes. ORGANISMOS etc
- Not quite! It means:

        O [=The]             EINAI  [= is]

        DASKALOS [=teacher]  BLAKAS [= stupid]

[read it vertically]

APH

Nikolaos Dergiades wrote:
> Dear Hyacinthists,
>
> If the three angles in degrees a, b, c,
> are rational numbers such that
> a < b < c < 90    and     2.sina.sinb = sinc,
> can we prove or disprove that
> a + b = 90    and    c = 2a ?

A check of all cases where a,b,c are integers didn't find
any new solutions.

The identity  2 * sin a * sin b = cos(a-b) - cos(a+b)
can be used to transform the problem into that of
describing the solutions to sin x + sin y = sin z,
where x,y,z are rational in degrees. This seems to be
a more natural question than the other version: perhaps
something has been published about this. With
0<x<y<z<90, the known solutions are y=60-x, z=60+x.
--
Barry Wolk

Dear Antreas and Floor,

[APH]:
Let ABC be a triangle, and PaPbPc the pedal triangle of point P.
The perp. bisectors of PPc, PPb meet at A', and intersect BC at Ab,
Ac, resp.

> >>                               A
> >>                              /\
> >>                             /  \
> >>                            /    \
> >>                           /  A'  \
> >>                          Pc       Pb
> >>                         /    P     \
> >>                        /  B'    C'  \
> >>                       B-Ab-------Ac--C

Similarly we define the points B', C' and Bc,Ba on CA and Ca, Cb on
AB. The triangle A'B'C' is the Midway triangle of P.

[FVL]:
We can simply take Ab to be A'B' /\ BC and Ac to be A'C'/\BC (etc.).

[APH]:
For which point P we have that: AbAc = BcBa = CaCb ?

[FvL]:
This is the point in barycentrics (-3bc+ac+ab : bc-3ac+ab : -
3ab+ab+ac).

***
This is P = 4t(I) - 3G, where t(I) is the isotomic conjugate of the
incenter. It is closely related to the equal-parallelian point Q, the
parallel intercepts through which are equal in length. this point Q
is X(192) in [ETC]:

Q = 3G - 2t(I)

(1) P and Q are symmetric with respect to t(I). Equivalently, P is
the superior of Q.

(2) AbAc = BcBa = CaCb = the equal parallelians = 2abc/(ab+bc+ca).

Best regards
Sincerely
Paul

Let ABC be a triangle, and PaPbPc the pedal traingle of point P.


                               A
                              /\
                             /  \
                            /    \
                           /      \
                          Pc       Pb
                         /          \
                        /            \
                       /      P       \
                      /                \
                     /                  \
                    /                    \
                   B-------Ab-------Ac----C


The perp. bisector of PbC intersects BC at Ab
         "             PcB      "     BC    Ac

Similarly we define the points Bc,Ba on CA and Ca,Cb on AB.

For which point P we have that: AbAc = BcBa = CaCb ?

Generalization:

PaPbPc = pedal triangle of P

                               A
                              /\
                             /  \
                            /    \
                           /      \
                          Pc       Pb
                         /          \
                      A'c            A'b
                       /      P       \
                      /                \
                     /         A*       \
                    /                    \
                   B-------Ab-------Ac----C


Let A'b be a point on AC, and A'c a point on AB, such that:

PcA'c / PcB = PbA'b / PbC = t

The perpendicular to AC at A'b intersects BC at Ab
          "           AB    A'c     "      BC    Ac

Similarly we define the points Bc,Ba on CA and Ca,Cb on AB.

Which is the locus of P such that AbAc = BcBa = CaCb, as t varies?

PS: If A'bAb /\ A'cAc = A*, and similarly B* = ..., C* = ...
then which is the locus of P such that A*B*C*, ABC are perspective?

PPS: Special case: t = 0. In this case, A'b = Pb, A'c = Pc and
Ab = PPb /\ BC, Ac = PPc /\ BC
The point P such that AbAc = BcBa = CaCb is:
           (1/(tanB + tanC) ::) = (csc2A ::) in normals
IIRC, I have posted this some time ago.


Antreas

Barry Wolk wrote:


>Nikolaos Dergiades wrote:
>> Dear Hyacinthists,
>>
>> If the three angles in degrees a, b, c,
>> are rational numbers such that
>> a < b < c < 90    and     2.sina.sinb = sinc,
>> can we prove or disprove that
>> a + b = 90    and    c = 2a ?
>
>A check of all cases where a,b,c are integers didn't find
>any new solutions.
>
>The identity  2 * sin a * sin b = cos(a-b) - cos(a+b)
>can be used to transform the problem into that of
>describing the solutions to sin x + sin y = sin z,
>where x,y,z are rational in degrees. This seems to be
>a more natural question than the other version: perhaps
>something has been published about this. With
>0<x<y<z<90, the known solutions are y=60-x, z=60+x.


A further check of all cases a = a'/n   b = b'/n  c = c' /n
where n = 2,3 permits to conjecture that all the
rational solutions in degrees are of the form
a = (30n + k)/n      b = (60n - k)/n   c = (60n + 2k)/n
where n is an arbitrary natural number and
k = 1 , 2 , . . . , 15n - 1.

Best regards
Sincerely
Nikolaos Dergiades

Dear  Antreas
you wrote:

>Let ABC be a triangle and consider an arbitrary triangle
AB'C' such that:
>AB'C' ~ [=similar to] ABC [the vertex A is in common].
>Find the locus of the intersection of the lines BB' and
CC'.
>(I. Ioannides: Mathematics of D' of Gymnasium. Vol II.
>OEDB: Athens, 1968, p. 52, #66)


At first glance, if the order of AB, AC is the
same as the order of AB', AC',
we can say that if  s = AC/AB then
a rotation around A by angle BAC moves B to B''
and the homothetic of B'' with ratio s is the point C.
The same synthesis of tranformations moves B' to C'
and hence BB'  and CC' form an angle equal to angle BAC.
So the locus of the intersection of BB' and CC' is a
circular arc (and the symmetric about BC) every point
of which sees BC with a constant angle BAC.

But if the order of AB, AC is not the
same as the order of AB', AC', then the problem
(if my calculations were correct) I think is complicated.
In the special case where the angle A = 90 and
AB = AB' , AC = AC'  then there is no intersection
of BB' and CC', since BB' is parallel to CC'.

Best regards
Nikolaos Dergiades

Dear Nikos

[Antreas]:
>>Let ABC be a triangle and consider an arbitrary triangle AB'C' such that:
>>AB'C' ~ [=similar to] ABC [the vertex A is in common].
>>Find the locus of the intersection of the lines BB' and CC'.
>>(I. Ioannides: Mathematics of D' of Gymnasium. Vol II.
>>OEDB: Athens, 1968, p. 52, #66)

[Nikos]:
>At first glance, if the order of AB, AC is the
>same as the order of AB', AC',

Yes, it's the same.

[Nikos]:
>we can say that if  s = AC/AB then
>a rotation around A by angle BAC moves B to B''
>and the homothetic of B'' with ratio s is the point C.
>The same synthesis of tranformations moves B' to C'
>and hence BB' and CC' form an angle equal to angle BAC.
>So the locus of the intersection of BB' and CC' is a
>circular arc (and the symmetric about BC) every point
>of which sees BC with a constant angle BAC.

That is, the circumcircle of ABC.

                               C'


                   A           P
                           B'



             B                C


Let P = BB' /\ CC'

trABC ~ trAB'C' ==>

angBAC = angB'AC' ==> angBAB' = angCAC'     (1)

and

AB / AC = AB' / AC' or AB / AB' = AC / AC'  (2)

(1) and (2) ==> trABB' ~ trACC' ==> angAB'B = angAC'C ==>
quadAB'PC' is cyclic ==> angBPC = angB'AC' = angBAC ==>
The locus of P is the circumcircle of ABC.


Greetings

Antreas