Kristen Robinson

EMAT 4680

3 October 2000

Assignment 11

Proof of Conjecture 1: The 3 medians of a triangle always intersect in one point, and this point of concurreny is the centroid. AD and CE are medians of ABC that intersect at G. BG is extended through AC and intersects AC at F. c1 is a circle with center G and radius BG. L1 is a line parallel to BG through G. H is the point of intersection of c1 and L1. AG is parallel to HC and AH is parallel to GC. Therefore, AG = HC and AH = GC. Because AH = GC and GC = 2 * EG, AH = 2 * EG. Additionally, because AG = HC and AG = 2 * GD, HC = 2 * GD. By definition, AGCH is a parallelogram. Therefore, AC and GH, the diagonals of AGCH, bisect each other. Therefore AF = CF. Therefore, F is the midpoint of AC, by definition. Thus, BF also is a median of ABC. Since G lies on BF, AD, CE, and BF are concurrent at G which is the centroid of ABC.

Proof of Conjecture 3: The six small triangles of ABC have equal areas.

Because BF is a median of ABC, GF is a median of AGC. Therefore, AGF = CGF.

Because AD is a median of ABC, GD is a median of CGB. Therefore CGD = BGD.

Because CE is a median of ABC, GE is a median of BGA. Therefore, BGE = AGE.

Because BF is a median of ABC, AFB = CFB.

AFB = BGE + AGE + AGF = 2*BGE + AGF. CFB = BGD + CGD + CGF = 2*BGD + CGF.

2*BGE + AGF = 2*BGD + CGF = 2*BGD + AGF.

2*BGE = 2*BGD. BGE = BGD.

Thus BGE = AGE = BGD = CGD.

Because AD is a median of ABC, ADC = ADB.

ADC = AGF + CGF + CGD and ADB = BGD + BGE + AGE.

AGF + CGF + CGD = BGD + BGE + AGE.

2*AGF + CGD = BGD + 2*BGE.

2*AGF = 2*BGE. AGF = BGE.

Thus AGF = CGF = CGD = BGD = BGE = AGE.

The balancing points of a quadrilateral, also known as the centroid, is the intersection point of the lines through the centroids of the two triangles created by one diagonal of the quadrilateral. Given polygon ABCD, triangles ABC and ADC are created by the diagonal AC. The centroids of these two triangles are N and M, respectively. The line through both N and M is y. Triangles BAD and BCD are created by the diagonal BD. The centroids of these two triangles are H and K, respectively. The line through both H and K is z. The point of intersection of y and z is the centroid of ABCD.

If ABCD is a concave quadrilateral, the procedure still remains. Suppose D is the concave vertex of ABCD. Then the diagonal AC lies outside of ABCD. However, the triangles ABC and ADC still exist, and the centroids of these triangles can still be constructed. Therefore the centroid of the quadrilateral can still be found using the same method as before.