## Re: [EMHL] Triangle Perimeter Trisector Points

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• The incenter I as areal center and the symmetric J of I wrt the centroid G as equicenter have this meaning: 1° J is the barycenter of points D with mass a =
Message 1 of 9 , Sep 17, 2010
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The incenter I as areal center and the symmetric J of I wrt the centroid G
as equicenter have this meaning:
1� J is the barycenter of points D with mass a = BC, E with mass b = CA F
with mass c = AB
2� The signed area S(IEAF), S(IFBD), S(IDCE) are equal
Friendly
Francois

On Fri, Sep 17, 2010 at 11:50 PM, Francois Rideau <francois.rideau@...
> wrote:

> hello, bbbblow
> If you look at what I have said on chowchow a few years ago on this site,
> you will see that points D,E, F are on affine correspondence on lines BC,
> CA, AB* with incenter I as areal center* and the symmetric of the
> incenter Iwrt the centroid G as equicenter.
> Friendly
> Francois
>
>
> On Thu, Sep 16, 2010 at 5:08 PM, bbblow <bbbbbblow@...> wrote:
>
>>
>>
>> For sidelines BC, CA, AB choose D, E, F such that
>> D, E, F trisect the perimeter (i.e. EA+AF=FB+BD=DC+CE with signed length.
>> direction AB, BC, CA are taken to be positive)
>>
>> Using GSP I found that the trace of centroids of DEF is a line, and the
>> trace of circumcenters is a conic (which seems to be always ellipse)
>>
>> also, the velocity of centroid when one of the trisectors was moved with
>> constant velocity seemed to be constant too.
>>
>> The centroid proposition seems to be solvable using vector algebra; we
>> parametize the trisectors, find their barycentrics and add their vectors to
>> parametize the centroid too.
>>
>> However circumcenter doesn't seem to be easily solvable using any of usual
>> barycentric coordinates nor vector things. Anyone with suggestion for its
>> solution?
>>
>> A possible extension would be to trace out 'anticevian' trisectors, i.e.
>> triangles such that their sides include A, B, C, while ABC are the perimeter
>> trisectors of the triangle.
>>
>> Since the notion of centroid is easily defined using vector algebra, it
>> seems possible to generalize the 'centroid collinearity' property to
>> tetrahedrons and the higher dimensional analogues too.
>>
>>
>>
>
>

[Non-text portions of this message have been removed]
• I correct some typos: The incenter I as areal center and the symmetric J of I wrt the centroid G as equicenter have this meaning: 1° J is the barycenter of
Message 2 of 9 , Sep 17, 2010
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I correct some typos:
The incenter I as areal center and the symmetric J of I wrt the centroid G
as equicenter have this meaning:
1� J is the barycenter of points D with mass a = BC, E with mass b = CA, F
with mass c = AB
2� The signed areas S(IEAF), S(IFBD), S(IDCE) are equal
Friendly
Francois
- Show quoted text -

O

>
> On Fri, Sep 17, 2010 at 11:50 PM, Francois Rideau <
> francois.rideau@...> wrote:
>
>> hello, bbbblow
>> If you look at what I have said on chowchow a few years ago on this site,
>> you will see that points D,E, F are on affine correspondence on lines BC,
>> CA, AB* with incenter I as areal center* and the symmetric of the
>> incenter Iwrt the centroid G as equicenter.
>> Friendly
>> Francois
>>
>>
>> On Thu, Sep 16, 2010 at 5:08 PM, bbblow <bbbbbblow@...> wrote:
>>
>>>
>>>
>>> For sidelines BC, CA, AB choose D, E, F such that
>>> D, E, F trisect the perimeter (i.e. EA+AF=FB+BD=DC+CE with signed length.
>>> direction AB, BC, CA are taken to be positive)
>>>
>>> Using GSP I found that the trace of centroids of DEF is a line, and the
>>> trace of circumcenters is a conic (which seems to be always ellipse)
>>>
>>> also, the velocity of centroid when one of the trisectors was moved with
>>> constant velocity seemed to be constant too.
>>>
>>> The centroid proposition seems to be solvable using vector algebra; we
>>> parametize the trisectors, find their barycentrics and add their vectors to
>>> parametize the centroid too.
>>>
>>> However circumcenter doesn't seem to be easily solvable using any of
>>> usual barycentric coordinates nor vector things. Anyone with suggestion for
>>> its solution?
>>>
>>> A possible extension would be to trace out 'anticevian' trisectors, i.e.
>>> triangles such that their sides include A, B, C, while ABC are the perimeter
>>> trisectors of the triangle.
>>>
>>> Since the notion of centroid is easily defined using vector algebra, it
>>> seems possible to generalize the 'centroid collinearity' property to
>>> tetrahedrons and the higher dimensional analogues too.
>>>
>>>
>>>
>>
>>
>

[Non-text portions of this message have been removed]
• Actually the property does not only hold for points trisecting a triangle perimeter, but also works in the following generalized form: On three given lines,
Message 3 of 9 , Sep 18, 2010
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Actually the property does not only hold for points trisecting a triangle perimeter, but also works in the following generalized form:

On three given lines, three points move in equal speed. Then the circumcenters of the three points trace out a conic.

An important point to be taken is that if the three points are moving in different speeds then the locus is not a conic. (All these were verified using GSP)

Possible further studies would be:
1. If we put n lines in n dimension, and animate n points on each line, what would the hyper-circumcenter trace out?
2. What determines the type of the conic?
3. How do we 'standardize' this property to barycentric coordinates? (Three lines mean three intersection points, and hence triangle geometry would be relevant)
4. Why circumcenter? How do we prove this property? (The most important question, actually)

And for Francois, I'm sorry but I didn't really get what you meant there. If possible can you provide me with the 'chowchow' that you referred to?
• Dear bbblow All I can say is you trying to search on all my past posts in Hyacinthos. Three points moving with equal speed on three given lines is just what I
Message 4 of 9 , Sep 18, 2010
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Dear bbblow
All I can say is you trying to search on all my past posts in Hyacinthos.
Three points moving with equal speed on three given lines is just what I
have called chowchow, remembering the noise of our old steam trains rolling
on their tracks.
A chowchow have in general 2 centers: the areal center and the equicenter.
It is Neuberg who found the equicenter at the turn of the 20th century.
In the other way, given the areal center and the equicenter, you get a
unique chowchow, up to an affine change of the law of time.

So a chowchow, your trisector configuration have its two centers: the areal
center is I the incenter and the equicenter is J the symmetric of I wrt the
centroid G.

Now , I give you a funny construction of your triples (D, E, F).
In fact, I construct a special triple (Do, Eo, Fo) in this way.
Let O be the circumcenter and O' be the point of the line IO such that the
affine ratio: OO'/OI = -1/3
Then the triangle DoEoFo is the pedal triangle of O' wrt the triangle of
reference ABC.

Now you get any other triple (D, E, F) since the signed segments DoD = EoE =
FoF are equal.
Friendly
Francois
For a general chowchow, the trace of the circumcenter of the triple (D, E,
F) is no longer a conic!

On Sat, Sep 18, 2010 at 2:25 PM, bbblow <bbbbbblow@...> wrote:

>
>
> Actually the property does not only hold for points trisecting a triangle
> perimeter, but also works in the following generalized form:
>
> On three given lines, three points move in equal speed. Then the
> circumcenters of the three points trace out a conic.
>
> An important point to be taken is that if the three points are moving in
> different speeds then the locus is not a conic. (All these were verified
> using GSP)
>
> Possible further studies would be:
> 1. If we put n lines in n dimension, and animate n points on each line,
> what would the hyper-circumcenter trace out?
> 2. What determines the type of the conic?
> 3. How do we 'standardize' this property to barycentric coordinates? (Three
> lines mean three intersection points, and hence triangle geometry would be
> relevant)
> 4. Why circumcenter? How do we prove this property? (The most important
> question, actually)
>
> And for Francois, I'm sorry but I didn't really get what you meant there.
> If possible can you provide me with the 'chowchow' that you referred to?
>
>
>

[Non-text portions of this message have been removed]
• Dear bbblow As I have already said to you, if you look at a general chowchow, that is to say: 3 points (D, E, F) moving on 3 lines with constant speeds, then
Message 5 of 9 , Sep 18, 2010
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Dear bbblow
As I have already said to you, if you look at a general chowchow, that is
to say: 3 points (D, E, F) moving on 3 lines with constant speeds, then the
locus of the DEF-circumcenter is in general some cubic.
But in case of your trisector chowchow, this locus is a conic because the
line at infinity is part of the locus.
So the question is:
For what chowchows, the circumcenter locus is a conic?
Friedndly
Francois

[Non-text portions of this message have been removed]
• Dear bbblow And the answer is: The chowhow for which the locus of the DEF-circumcenter is a conic are those for which the 3 moving points D, E, F have constant
Message 6 of 9 , Sep 18, 2010
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Dear bbblow
The chowhow for which the locus of the DEF-circumcenter is a conic are those
for which the 3 moving points D, E, F have constant equal speeds.
Of course this is the case of your trisector chowchow.
In terms of areal centers, they are the chowchows of which the areal center
is the incenter or an excenter.
Friendly
Francois

>
>

[Non-text portions of this message have been removed]
• look at It is written in french Friendly Francois ... [Non-text portions of this message have been
Message 7 of 9 , Sep 18, 2010
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It is written in french
Friendly
Francois

>
>>
>>
>
>

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