Proposed solution

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 pointG:

The lineg intersects the line BC at a point X:

The parallel to the lineBG throughA intersects the line g at a pointX_b: The parallel to the lineCG throughA intersects the line g at a point X_c: Prove that 1

GX + 1

GX_b + 1

GX_c = 0; where the segments are directed. (See Fig. 1.)

B

A

C G

X

X

X

g

Fig. 1

Solution. (See Fig. 2.) Let C⁰ be the midpoint of the segment AB: Then, the line CC⁰ is a median of triangleABC; and thus passes through the centroid G of this triangle.

Let the parallel to the line CG throughB intersect the lineg at a point Y_c: 1

Then, AX_c k CG and BY_c k CG; so that AX_c k BY_c k CG: Thus, the points G;

X_c; Y_c are the images of the points C⁰; A; B under a parallel projection from the line ABonto the lineg:Since parallel projection preserves ratios of signed lengths, we thus have GX_c

GYc

= C⁰A

C⁰B: But C⁰ is the midpoint of AB; so that C⁰A = C⁰B; and thus C⁰A

C⁰B = 1: Hence, GX_c

GY_c = 1;so that GX_c = GY_c:

B

A

C G

C'

X

X

g

Y

Fig. 2

Now, BY_c k CG; so that the Thales theorem yields GX

GY_c = CX

CB: Hence, GX GX_c = GX

GY_c = GX

GY_c = CX

CB = CX

CB = CX

BC:Similarly, GX

GX_b = BX

CB: Thus, GX

GXb

+ GX GXc

= BX

CB + CX

BC = XB

BC +CX

BC = CX +XB

BC = CB

BC = BC

BC = 1:

Dividing this equation by GX; we obtain 1

GX_b + 1

GX_c = 1

GX;so that 1

GX + 1 GX_b + 1

GXc

= 0;and the problem is solved.

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. LetG be the centroid of a triangleABC; and letg be a line through the pointG:

Let the line g intersect the lines BC; CA; AB at three points X; Y; Z:

Then, 1

GX + 1

GY + 1

GZ = 0; where the segments are directed.

B

A

C G

X

g Y

Z

Fig. 3

Proof of Theorem 1. (See Fig. 4.) Let the parallel to the line CG throughA meet the lineg at a point X_c:

Let the parallel to the line BG through A meet the lineg at a point X_b: Let the parallel to the line CG throughB meet the line g at a point Y_c: Let the parallel to the line AG throughC meet the lineg at a point Z_a: According to the problem, we have 1

GX + 1

GX_b + 1

GX_c = 0; so that 1 GX = 1

GX_b + 1

GX_c : But during the solution of the problem, we have also shown that

3

GX_c = GY_c:Thus, 1

GX = 1

GX_b + 1

GX_c = 1

GX_b + 1

GY_c = 1 GX_b

1

GY_c = 1 GY_c

1 GX_b: Similarly,

1

GY = 1 GZa

1 GYc

and 1

GZ = 1 GXb

1 GZa

: Thus,

1

GX + 1

GY + 1

GZ = 1 GY_c

1

GX_b + 1 GZ_a

1

GY_c + 1 GX_b

1

GZ_a = 0;

and Theorem 1 is proven.

B

A

C G

X

X

g

Y

Z

Y

Z

Fig. 4

References

[1] H. S. M. Coxeter, S. L. Greitzer,Geometry Revisited, Mathematical Association of America: New Mathematical Library, volume 19.

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