- View SourceAccording 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 - View SourceDear Grobber,

In Hyacinthos message #9671, I wrote:

>> --> --> --> -->

Sorry, another typo. It should be

>> 3 TS = XT + YT + ZT .

--> --> --> -->

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

Now I have quite a very artificial proof:

>> through B, and to YGa through C, and

>> the line YZ concur. [...] I have no

>> proof of this [...]

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