## Generalized Gossard triangles

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• According to Clark Kimberling s treatise in http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html in 1999, Paul Yiu generalized Gossard s theorem,
Message 1 of 6 , Apr 12, 2004
According to Clark Kimberling's treatise in

http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html

in 1999, Paul Yiu generalized Gossard's theorem,
triangle and perspector to arbitrary lines through
the centroid; some additional properties of the
configuration were later given by John Conway and
Steve Sigur, and finally Wilson Stothers made some
important observations in Hyacinthos message #8383.

Let G be the centroid of a triangle ABC and P an
arbitrary point, and let the line PG meet the
sidelines BC, CA, AB of triangle ABC at X, Y, Z.
Also, define a point T by the vectorial equation

--> --> -->
--> GX + GY + GZ
GT = - ---------------.
3

Obviously, this point T lies on the line PG; it is
also easily seen by vector addition that

(1) This point T lies on the line joining the
midpoints of the segments AX, BY, CZ, i. e. on
the Gauss line of the quadrilateral formed by
the lines BC, CA, AB and PG.

Okay, this was not a big deal yet. But now we can
denote by Ga, Gb, Gc the centroids of triangles AYZ,
BZX, CXY, respectively; then,

(2) The points Ga, Gb, Gc lie on the reflections of
the lines BC, CA, AB in the point T.

This is simple using vectors, too. But life is not
that easy (in general), and we can get rather
difficult theorems. If the parallel to CP through Y
meets the parallel to BP through Z at Pa, and
similarly we define the points Pb and Pc, then

(3) The lines GaPa, GbPb, GcPc are parallel to the
lines BC, CA, AB, respectively.

I have found only an awkward proof of this, so I
would appreciate a rather simple synthetic proof if
there is one.

Now, (2) and (3) together yield:

(4) The triangle formed by the lines GaPa, GbPb,
GcPc is the reflection of triangle ABC in the
point T.

These four results already contain most of what is
referred to as "Gossard stuff". I would be mainly
interested in a good proof of (3).

Darij Grinberg
• ... Hi! I m a bit confused here. T is supposed to be the symmetric of the centroid of the system of points X, Y, Z with respect to G, right? I didn t make the
Message 2 of 6 , Apr 12, 2004
--- In Hyacinthos@yahoogroups.com, "Darij Grinberg"
<darij_grinberg@w...> wrote:
> According to Clark Kimberling's treatise in
>
> http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html
>
> in 1999, Paul Yiu generalized Gossard's theorem,
> triangle and perspector to arbitrary lines through
> the centroid; some additional properties of the
> configuration were later given by John Conway and
> Steve Sigur, and finally Wilson Stothers made some
> important observations in Hyacinthos message #8383.
>
> Let G be the centroid of a triangle ABC and P an
> arbitrary point, and let the line PG meet the
> sidelines BC, CA, AB of triangle ABC at X, Y, Z.
> Also, define a point T by the vectorial equation
>
> --> --> -->
> --> GX + GY + GZ
> GT = - ---------------.
> 3
>
> Obviously, this point T lies on the line PG; it is
> also easily seen by vector addition that
>
> (1) This point T lies on the line joining the
> midpoints of the segments AX, BY, CZ, i. e. on
> the Gauss line of the quadrilateral formed by
> the lines BC, CA, AB and PG.
>
> Okay, this was not a big deal yet. But now we can
> denote by Ga, Gb, Gc the centroids of triangles AYZ,
> BZX, CXY, respectively; then,
>
> (2) The points Ga, Gb, Gc lie on the reflections of
> the lines BC, CA, AB in the point T.
>
> This is simple using vectors, too. But life is not
> that easy (in general), and we can get rather
> difficult theorems. If the parallel to CP through Y
> meets the parallel to BP through Z at Pa, and
> similarly we define the points Pb and Pc, then
>
> (3) The lines GaPa, GbPb, GcPc are parallel to the
> lines BC, CA, AB, respectively.
>
> I have found only an awkward proof of this, so I
> would appreciate a rather simple synthetic proof if
> there is one.
>
> Now, (2) and (3) together yield:
>
> (4) The triangle formed by the lines GaPa, GbPb,
> GcPc is the reflection of triangle ABC in the
> point T.
>
> These four results already contain most of what is
> referred to as "Gossard stuff". I would be mainly
> interested in a good proof of (3).
>
> Darij Grinberg

Hi! I'm a bit confused here. T is supposed to be the symmetric of the
centroid of the system of points X, Y, Z with respect to G, right? I
didn't make the vector calculations, but I drew a dynamic sketch and
I don't get T to be on the Gauss line of the quadrilateral formed by
BC, CA, AB and PG. However, T' is on that line, where GT'=
(GA+GB+GC)/6 (they are all vectors). What am I doing wrong?
• ... Here s what I did: Let Y and Z be 2 points on CA and AB respectively. Let P a variable point on YZ. It s natural to ask th question: if L is the
Message 3 of 6 , Apr 12, 2004
--- In Hyacinthos@yahoogroups.com, "Darij Grinberg"
<darij_grinberg@w...> wrote:
> According to Clark Kimberling's treatise in
>
> http://faculty.evansville.edu/ck6/tcenters/recent/gosspersp.html
>
> in 1999, Paul Yiu generalized Gossard's theorem,
> triangle and perspector to arbitrary lines through
> the centroid; some additional properties of the
> configuration were later given by John Conway and
> Steve Sigur, and finally Wilson Stothers made some
> important observations in Hyacinthos message #8383.
>
> Let G be the centroid of a triangle ABC and P an
> arbitrary point, and let the line PG meet the
> sidelines BC, CA, AB of triangle ABC at X, Y, Z.
> Also, define a point T by the vectorial equation
>
> --> --> -->
> --> GX + GY + GZ
> GT = - ---------------.
> 3
>
> Obviously, this point T lies on the line PG; it is
> also easily seen by vector addition that
>
> (1) This point T lies on the line joining the
> midpoints of the segments AX, BY, CZ, i. e. on
> the Gauss line of the quadrilateral formed by
> the lines BC, CA, AB and PG.
>
> Okay, this was not a big deal yet. But now we can
> denote by Ga, Gb, Gc the centroids of triangles AYZ,
> BZX, CXY, respectively; then,
>
> (2) The points Ga, Gb, Gc lie on the reflections of
> the lines BC, CA, AB in the point T.
>
> This is simple using vectors, too. But life is not
> that easy (in general), and we can get rather
> difficult theorems. If the parallel to CP through Y
> meets the parallel to BP through Z at Pa, and
> similarly we define the points Pb and Pc, then
>
> (3) The lines GaPa, GbPb, GcPc are parallel to the
> lines BC, CA, AB, respectively.
>
> I have found only an awkward proof of this, so I
> would appreciate a rather simple synthetic proof if
> there is one.
>
> Now, (2) and (3) together yield:
>
> (4) The triangle formed by the lines GaPa, GbPb,
> GcPc is the reflection of triangle ABC in the
> point T.
>
> These four results already contain most of what is
> referred to as "Gossard stuff". I would be mainly
> interested in a good proof of (3).
>
> Darij Grinberg

Here's what I did: Let Y and Z be 2 points on CA and AB respectively.
Let P a variable point on YZ. It's natural to ask th question: if L
is the intersection between the parallel through Z to BP and the
parallel through Y to CP then what's the locus of L?

We can how that the locus is a line parallel to BC by using some
projective results. The transformation CP->BP (from the pencil of
lines with center C to the pencil with center B) is a homographic one
(in fact, it's a perspectivity). Since YL || CP and ZL || BP it means
that the transformation YL->ZL is also a homographic one, and it's
well-known that in this case the locus of L is a conic which
degenerates if the image of YZ is YZ. Since this is obviously the
case here, it means that the locus of L is a line. By putting P in
the positions Y and Z we can see that this line is also parallel to
BC.

We apply all of this to our situation, where L is Pa. The locus of Pa
is thus a line || BC, and all we need to do is show that Ga is on
this line. We do this by putting P in the positions Y and Z (we take
Y and Z to be fixed s.t. G is on YZ), and by using the following: If
ABC is a triangle, G its centroid and S and T are on AB and AC
respectively, then G is on ST iff SB/SA+TC/TA=-1 (the segments are
oriented). We can thus show that GaPa is parallel to BC for a fixed
line GP passing, and this obviously suffices in order to prove the
whole thing.
• Dear Grobber, ... I m sorry; my formula -- -- -- -- GX + GY + GZ GT = - --------------- 3 should have actually been -- -- -- -- GX +
Message 4 of 6 , Apr 12, 2004
Dear Grobber,

In Hyacinthos message #9668, you wrote:

>> Hi! I'm a bit confused here. T is supposed to
>> be the symmetric of the centroid of the system
>> of points X, Y, Z with respect to G, right?

I'm sorry; my formula

--> --> -->
--> GX + GY + GZ
GT = - ---------------
3

should have actually been

--> --> -->
--> GX + GY + GZ
GT = ---------------.
6

We also have

--> --> --> -->
3 TS = XT + YT + ZT .

>> However, T' is on that line, where
>> GT'= (GA+GB+GC)/6 (they are all vectors).

(I guess you mean GX + GY + GZ rather than
GA + GB + GC here.)

In Hyacinthos message, #9669, you wrote:

>> Here's what I did: [...]

Thanks, I'll have to learn projective
geometry at last... Here is the main idea
of my proof for (3):

Let's prove that the line GaPa is parallel
to BC. Obviously, this is equivalent to
saying that the parallels to CP through Y,
to BP through Z, and to BC through Ga
concur. Now, after the well-known theorem
on reciprocal triangles ("If the lines
through the vertices of one triangle T1
parallel to the respective sidelines of
another triangle T2 concur, then the lines
through the vertices of T2 parallel to the
sidelines of T1 also concur"), the
assertion we have to prove is equivalent
to the following one: The parallels to ZGa
through B, to YGa through C, and to YZ
through P concur. In other words, the
parallels to ZGa through B, and to YGa
through C, and the line YZ concur. This
should be easier than the initial
statement, since the point P is not
involved anymore; however, I have no proof
of this other than an ugly proof using
Menelaos. Do you have something
elementary?

Sincerely,
Darij Grinberg
• ... Yeah, I mean GX+GY+GZ, of course. I guess I couldn t help myself and I wrote A, B, and C being more familiar with these letters :). ... Not really. I tried
Message 5 of 6 , Apr 12, 2004
--- In Hyacinthos@yahoogroups.com, "Darij Grinberg"
<darij_grinberg@w...> wrote:
> (I guess you mean GX + GY + GZ rather than
> GA + GB + GC here.)

Yeah, I mean GX+GY+GZ, of course. I guess I couldn't help myself and
I wrote A, B, and C being more familiar with these letters :).

> Do you have something
> elementary?

Not really. I tried something with Menelaus as well, but I abandoned
it quickly..
• Dear Grobber, ... Sorry, another typo. It should be -- -- -- -- 3 TG = XT + YT + ZT . But the main property is still that T lies on the Gauss line
Message 6 of 6 , Apr 15, 2004
Dear Grobber,

In Hyacinthos message #9671, I wrote:

>> --> --> --> -->
>> 3 TS = XT + YT + ZT .

Sorry, another typo. It should be

--> --> --> -->
3 TG = XT + YT + ZT .

But the main property is still that T lies on
the Gauss line of the quadrilateral formed by
the lines BC, CA, AB and PG; even more can be
said: T is the centroid of the points X', Y',
Z', where X', Y', Z' are the midpoints of the
segments AX, BY, CZ. (The best way to prove this
is to define T as the centroid of the points X',
Y', Z'; then, all the rest follows by simple
vector calculation.)

>> In other words, the parallels to ZGa
>> through B, and to YGa through C, and
>> the line YZ concur. [...] I have no
>> proof of this [...]

Now I have quite a very artificial proof:

I rename Ga as D, so that D is the centroid of
triangle AYZ, and I have to show that the
parallels to ZD through B, and to YD through C,
and the line YZ concur.

Well, let the parallel to YZ through A meet the
parallel to ZD through B at B', the parallel to
YD through C at C', the parallel to ZD through Y
at M, the line ZD at K, and the line YD at L.
Then, YZKM is a parallelogram (trivial), and
AZYK and AYZL are parallelograms (since the
lines ZD and YD pass through the midpoints of
the segments AY and AZ, respectively). Hence, we
have MK = YZ, KA = YZ and AL = YZ with directed
segments. Now,

LC' LC' YC
--- = --- = -- (Thales)
YZ AL AY

and

B'K B'K BZ
--- = --- = -- (Thales).
YZ KA ZA

However, with d(R; l) denoting the directed
distance from a point R to a line l, we have

YC BZ d(C; PG) d(B; PG)
-- + -- = - -------- - -------- = 1,
AY ZA d(A; PG) d(A; PG)

since it is well-known that
d(A; PG) + d(B; PG) + d(C; PG) = 0 (this is a
property of any line passing through the
centroid G of triangle ABC).

Hence,

LC' B'K
--- + --- = 1,
YZ YZ

and LC' + B'K = YZ, or LC' + B'K = MK. Therefore,

MB' = MK - B'K = LC',

-->
so that the translation by the vector MB' maps M
to B' and L to C'. This translation maps the
point Y to a point W such that B'W || MY,
-->
C'W || LY, and, of course, YW || MB' (since MB'
is the translation vector). From B'W || MY and
C'W || LY, it follows that W lies on the
parallel to ZD through B and on the parallel to
YD through C. From YW || MB', it follows that W
lies on the line YZ. Hence, the parallel to ZD
through B, the parallel to YD through C, and the
line YZ concur. Qed.!

How do you like this proof? :-)

Sincerely,
Darij Grinberg
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