Problem proposal for "Mathematical Re‡ections": A line through the
centroid
Darij Grinberg
Darij Grinberg, Problem S64, Mathematical Re‡ections 5/2007.
Problem. Let G be the centroid of a triangle ABC; and let g be a line
through the point G:
The line g intersects the line BC at a point X:
The parallel to the line BG through A intersects the line g at a point Xb :
The parallel to the line CG through A intersects the line g at a point Xc :
1
1
1
Prove that
+
+
= 0; where the segments are directed. (See
GX GXb GXc
Fig. 1.)
Xb
g
A
Xc
G
B
C
X
Fig. 1
Solution. (See Fig. 2.) Let C 0 be the midpoint of the segment AB: Then, the
line CC 0 is a median of triangle ABC; and thus passes through the centroid G of this
triangle.
Let the parallel to the line CG through B intersect the line g at a point Yc :
1
Then, AXc k CG and BYc k CG; so that AXc k BYc k CG: Thus, the points G;
Xc ; Yc are the images of the points C 0 ; A; B under a parallel projection from the line
AB onto the line g: Since parallel projection preserves ratios of signed lengths, we thus
C 0A
GXc
have
= 0 : But C 0 is the midpoint of AB; so that C 0 A = C 0 B; and thus
GYc
CB
C 0A
GXc
=
1:
Hence,
= 1; so that GXc = GYc :
C 0B
GYc
A
g
Xc
C'
G
X
B
C
Yc
Fig. 2
GX
CX
GX
=
: Hence,
=
GYc
CB
GXc
CX
CX
CX
GX
BX
=
=
: Similarly,
=
: Thus,
CB
CB
BC
GXb
CB
Now, BYc k CG; so that the Thales theorem yields
GX
=
GYc
GX
=
GYc
GX
GX
BX CX
XB CX
CX + XB
CB
BC
+
=
+
=
+
=
=
=
=
GXb GXc
CB
BC
BC
BC
BC
BC
BC
Dividing this equation by GX; we obtain
1
= 0; and the problem is solved.
GXc
1:
1
1
1
1
1
+
=
; so that
+
+
GXb GXc
GX
GX GXb
2
Remark. Using the above problem and its solution, we can give a new proof to
the following fact ([1], §2.1, problem 8):
Theorem 1. Let G be the centroid of a triangle ABC; and let g be a line
through the point G:
Let the line g intersect the lines BC; CA; AB at three points X; Y; Z:
1
1
1
Then,
+
+
= 0; where the segments are directed.
GX GY
GZ
A
g
Y
G
B
X
C
Z
Fig. 3
Proof of Theorem 1. (See Fig. 4.) Let the parallel to the line CG through A meet
the line g at a point Xc :
Let the parallel to the line BG through A meet the line g at a point Xb :
Let the parallel to the line CG through B meet the line g at a point Yc :
Let the parallel to the line AG through C meet the line g at a point Za :
1
1
1
1
According to the problem, we have
+
+
= 0; so that
=
GX
GXb
GXc
GX
1
1
+
: But during the solution of the problem, we have also shown that
GXb GXc
3
GXc =
GYc : Thus,
1
=
GX
1
1
+
GXb GXc
1
+
GXb
=
1
GYc
=
1
GXb
1
GYc
=
1
GYc
1
:
GXb
Similarly,
1
1
=
GY
GZa
1
GYc
1
1
=
GZ
GXb
and
1
:
GZa
Thus,
1
1
1
+
+
=
GX GY
GZ
1
GYc
1
GXb
1
GZa
+
1
GYc
1
GXb
+
1
GZa
= 0;
and Theorem 1 is proven.
Xb
g
A
Za
Y
G
B
X
Yc
Z
Fig. 4
References
[1] H. S. M. Coxeter, S. L. Greitzer, Geometry Revisited, Mathematical Association
of America: New Mathematical Library, volume 19.
4
C