## Re: [EMHL] Points (b,c,a) and (c,a,b)

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• Dear Steve and Francois, The discussion of this nice configuation has been very interesting. I think that there are a few things to add (and one correction)
Message 1 of 8 , Dec 7, 2006
Dear Steve and Francois,

The discussion of this nice configuation has been very interesting.
I think that there are a few things to add (and one correction)

Steve:
>1. All 6 permutations (group 1 + group 2) lie on an elliipse centers
>at G and having the same shape as the Steiner one.
>2. The midpoint of any two in group 2 is the medial (complement) of
>one of the points in group 1.
>3. The triangles formed from group 1 and group 2 have G as centroid.

My calculations suggest that 2 should be

"the mid-point of two members of the same group are the complement
of the third member of the same group"

The facts about triple perspectivity are easy from the coordinates -
for
example, u:v:w lies on cevians of 1/u:1/w:1/v, 1/v:1/u:1/w,
1/w:1/v:1/u,
as can be seen by dividing coordinates by vw,uv,uw respectively.

Notation :
a' = u:v:w b' = w:u:v c' = v:w:u
a" = u:w:v b" = v:u:w c" = w:v:u

Simple calculation then shows that
a'a", b'c", b"c' are parallel to BC,
b'b", a'c", a"c' are parallel to CA, and
c'c", a'b", a"b' are parallel to AB.

We will write x*y for the mid-point of points x and y.

a'*a", b'*c", b"*c' lie on the median through A.
b'*b", a'*c", a"*c' lie on the median through B.
c'*c", a'*b", a"*b' lie on the median through C.
These can also be deduced from the parallel properties above.

The following sets of mid-points are collinear :

a'*c', a"*b", a'*b", a"*c' and 2 similar pairs,
a'*c', b"*c", a'*c", c'*b" and 2 similar pairs,
a'*c', a'*c", a'*a", c'*c" and 2 similar pairs.
These can also be deduced from the parallel properties above.

c'*a", c'*b", c'*b" form a triangle congruent to the medial triangle
of
triangle a"b"c" - this is obvious by a homothety with centre c'.
The triangle is also the rotation of the medial triangle through pi
There are obviously 5 similar examples.

Note that all this information about parallels allows us to construct
all 6 from a single point, simply by line intersections.

Now to the matter of the common circumconic. This can be done by brute
force of course, but it is mor fun as follows :

Observe that, if we apply a homothety with centre G centroid of ABC,
then we obtain a further set of 6 points with permuted coordinates.

Suppose we apply the homothety which maps a' to a point of the Steiner
Ellipse of ABC. Then the coordinates of the other images show that
they
lie on this ellipse. The images give two triangles on the Ellipse,
each
triply perspective to ABC. This follows from the parallel results
wehave
above. It follows that we also have a common Steiner Inellipse. The
inverse transformation gives us a triangle A'B'C' inscribed in the
common
Steiner Ellipse.

Returning to the original points :
all 6 points lie on an ellipse - this is the Steiner Ellipse of both
of
the triangles a'b'c' and a"b"c".
These triangles also have a common Steiner Inellipse. This contains
the
point a'*b', and its analogues.
We can now determine the order of the points round this ellipse.

Let us suppose that we have a set on the Steiner Ellipse of ABC.
The medians of ABC give further points A',B',C' on the Steiner
Ellipse.
These divide the ellipse into 6 segments AC',C'B, BA', etc.

The knowledge about parallels to the sides of ABC, and mid-points on
the
medians allow us to deduce that segment each contains one of our 6
points.
We now CHOOSE a' as the point in the segment BA'.
Thinking about its location, the coordinates can be taken asu:v:w with
with u negative, v,W positive and v greater than w. As our point is
on the
Steiner ellipse u+v an u+w are positive. Now in our notation above,
we see
that the order is Ba'a"Cb'b"Ac'c". Note that the order is preseved by
homothety - in the general picture a' is chosen as the vertex on BC
which
is nearer to B.

Some further calculation allow us to deduce that (for 6 points on
Steiner
Ellipse) :
a'*b", b'*c", c'*a" lie inside the Steiner Inellipse;
a"*b', b"*c', c"*a' lie inside ABC, but outside the Inellipse,
a'*a", b'*b", c'*c" lie outside ABC, but in the Steiner Ellipse.
Each trio form a triangle homothetic with ABC, centre G.

Of course, these properties are preserved by homothety provided that
we
replace ABC by the inverse image mentioned above.

Sorry for such a long post - I found this topic fascinating

Regards

Wilson

--- In Hyacinthos@yahoogroups.com, "Francois Rideau"
<francois.rideau@...> wrote:
>
> Dear Steve
> Anyway, this configuration is very beautiful!
> Friendly
> François
>
>
> [Non-text portions of this message have been removed]
>
• Dear Wilson Thanks for your explanations. Yes this configuration is beautiful and above all that is not euclidian but only affine geometry. I have plenty of
Message 2 of 8 , Dec 7, 2006
Dear Wilson
Yes this configuration is beautiful and above all that is not euclidian but
only affine geometry.
I have plenty of nice affine drawings in my files and I tried in the past to
interest Hyacinthists in it with no avail.
Friendly
François

On 12/7/06, Wilson Stothers < wws@...> wrote:
>
> Dear Steve and Francois,
>
> The discussion of this nice configuation has been very interesting.
> I think that there are a few things to add (and one correction)
>
> Steve:
> >1. All 6 permutations (group 1 + group 2) lie on an elliipse centers
> >at G and having the same shape as the Steiner one.
> >2. The midpoint of any two in group 2 is the medial (complement) of
> >one of the points in group 1.
> >3. The triangles formed from group 1 and group 2 have G as centroid.
>
> My calculations suggest that 2 should be
>
> "the mid-point of two members of the same group are the complement
> of the third member of the same group"
>
> The facts about triple perspectivity are easy from the coordinates -
> for
> example, u:v:w lies on cevians of 1/u:1/w:1/v, 1/v:1/u:1/w,
> 1/w:1/v:1/u,
> as can be seen by dividing coordinates by vw,uv,uw respectively.
>
> Notation :
> a' = u:v:w b' = w:u:v c' = v:w:u
> a" = u:w:v b" = v:u:w c" = w:v:u
>
> Simple calculation then shows that
> a'a", b'c", b"c' are parallel to BC,
> b'b", a'c", a"c' are parallel to CA, and
> c'c", a'b", a"b' are parallel to AB.
>
> We will write x*y for the mid-point of points x and y.
>
> a'*a", b'*c", b"*c' lie on the median through A.
> b'*b", a'*c", a"*c' lie on the median through B.
> c'*c", a'*b", a"*b' lie on the median through C.
> These can also be deduced from the parallel properties above.
>
> The following sets of mid-points are collinear :
>
> a'*c', a"*b", a'*b", a"*c' and 2 similar pairs,
> a'*c', b"*c", a'*c", c'*b" and 2 similar pairs,
> a'*c', a'*c", a'*a", c'*c" and 2 similar pairs.
> These can also be deduced from the parallel properties above.
>
> c'*a", c'*b", c'*b" form a triangle congruent to the medial triangle
> of
> triangle a"b"c" - this is obvious by a homothety with centre c'.
> The triangle is also the rotation of the medial triangle through pi
> There are obviously 5 similar examples.
>
> Note that all this information about parallels allows us to construct
> all 6 from a single point, simply by line intersections.
>
> Now to the matter of the common circumconic. This can be done by brute
> force of course, but it is mor fun as follows :
>
> Observe that, if we apply a homothety with centre G centroid of ABC,
> then we obtain a further set of 6 points with permuted coordinates.
>
> Suppose we apply the homothety which maps a' to a point of the Steiner
> Ellipse of ABC. Then the coordinates of the other images show that
> they
> lie on this ellipse. The images give two triangles on the Ellipse,
> each
> triply perspective to ABC. This follows from the parallel results
> wehave
> above. It follows that we also have a common Steiner Inellipse. The
> inverse transformation gives us a triangle A'B'C' inscribed in the
> common
> Steiner Ellipse.
>
> Returning to the original points :
> all 6 points lie on an ellipse - this is the Steiner Ellipse of both
> of
> the triangles a'b'c' and a"b"c".
> These triangles also have a common Steiner Inellipse. This contains
> the
> point a'*b', and its analogues.
> We can now determine the order of the points round this ellipse.
>
> Let us suppose that we have a set on the Steiner Ellipse of ABC.
> The medians of ABC give further points A',B',C' on the Steiner
> Ellipse.
> These divide the ellipse into 6 segments AC',C'B, BA', etc.
>
> The knowledge about parallels to the sides of ABC, and mid-points on
> the
> medians allow us to deduce that segment each contains one of our 6
> points.
> We now CHOOSE a' as the point in the segment BA'.
> Thinking about its location, the coordinates can be taken asu:v:w with
> with u negative, v,W positive and v greater than w. As our point is
> on the
> Steiner ellipse u+v an u+w are positive. Now in our notation above,
> we see
> that the order is Ba'a"Cb'b"Ac'c". Note that the order is preseved by
> homothety - in the general picture a' is chosen as the vertex on BC
> which
> is nearer to B.
>
> Some further calculation allow us to deduce that (for 6 points on
> Steiner
> Ellipse) :
> a'*b", b'*c", c'*a" lie inside the Steiner Inellipse;
> a"*b', b"*c', c"*a' lie inside ABC, but outside the Inellipse,
> a'*a", b'*b", c'*c" lie outside ABC, but in the Steiner Ellipse.
> Each trio form a triangle homothetic with ABC, centre G.
>
> Of course, these properties are preserved by homothety provided that
> we
> replace ABC by the inverse image mentioned above.
>
> Sorry for such a long post - I found this topic fascinating
>
> Regards
>
> Wilson
>
> --- In Hyacinthos@yahoogroups.com <Hyacinthos%40yahoogroups.com>,
> "Francois Rideau"
> <francois.rideau@...> wrote:
> >
> > Dear Steve
> > Anyway, this configuration is very beautiful!
> > Friendly
> > François
> >
> >
> > [Non-text portions of this message have been removed]
> >
>
>
>

[Non-text portions of this message have been removed]
• ... Wilson et al., Thanks for the correction and the nice additions to this schema. I too find this a fascinating topic. Conway does not so I have had a hard
Message 3 of 8 , Dec 9, 2006
On Dec 7, 2006, at 10:25 AM, Wilson Stothers wrote:

> The discussion of this nice configuation has been very interesting.
> I think that there are a few things to add (and one correction)
>
> Steve:
> >1. All 6 permutations (group 1 + group 2) lie on an elliipse centers
> >at G and having the same shape as the Steiner one.
> >2. The midpoint of any two in group 2 is the medial (complement) of
> >one of the points in group 1.
> >3. The triangles formed from group 1 and group 2 have G as centroid.
>
> My calculations suggest that 2 should be
>
> "the mid-point of two members of the same group are the complement
> of the third member of the same group"
>
>
>
> Sorry for such a long post - I found this topic fascinating

Wilson et al.,

Thanks for the correction and the nice additions to this schema. I
too find this a fascinating topic. Conway does not so I have had a
hard time convincing him to put it into our book. He is coming
around, though, as more and more of these types of points are being
discovered in normal geometrical contexts.

The tripolars and duals of the permuted points are also interesting.
If P> and P< are the cyclic permutations of P, then the duals of P>
and P< meet on the GP line at the inverse of P in the Steiner
ellipse. The triplars meet at the Steiner inverse of tP (I think).

Steve

Notation:
20web/notation.html

Triangle web page:
http://paideiaschool.org/TeacherPages/Steve_Sigur/geometryIndex.htm

Other math:
http://paideiaschool.org/TeacherPages/Steve_Sigur/interesting2.htm

[Non-text portions of this message have been removed]
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