54 Vectors and geometry in three dimensions
P3
P2
6 We are going to prove a theorem in geome-
try. As in Figure 2.194, let P1, P2, Pa be vertices
of a triangle and, for each k = 1, 2, 3, let Mk be
the mid-point of the side opposite Pk. For each
k, let Ck be the point of trisection of the segment
.r M3 PkMk for which II'kCkl = siPkMd. We will prove
P h d d
Figure 2.194
tat the points Cl, C2, Ca colncl a an we may
put C = C1 = C2 = C3. The line segments PkMk
are, in geometry, called medians of the triangle
Thus, our result shows that the three medians intersect at a point C which trisects
each of them. For reasons which we shall not now discuss, the point C is the cen-
troid of the triangular region T bounded by the sides of the triangle. To prove
our result, let 0 be any point and show that
OCl = OPl + R-IM, = OP, + s(OMI - OPl)
- 2 OP2+OPa-OPl ^OP1+OP2+OP3
3\ 2 3
The way in which P1, P2, and P3 appear in the result can make us feel sure that
we must have
OCk =OP1+OP2+OP3
3
for each k. However, calculate 0C2 and OCa and show that it is so.
7 The line segment PkCk joining a vertex
Pk of a tetrahedron to the centroid Ck of the
opposite face, as in Figure 2.195, is called a
P4 median of the tetrahedron. For each k, let Qk
be the point of quadrisection of the median
PkCk for which I PkQkl = aIPkCkl. Let 0 be any
point. When k = 1, prove the formula
OP1+OP2+OP8 +OP4
OQk = 4
and then prove or guess that the formula is valid when k= 1, 2, 3, 4. The point
Q for which
OQ-OP1+OP2+OPa+OP4
4
is the centroid of the tetrahedron. Thus the four medians of a tetrahedron
intersect at the centroid, and this centroid quadrisects each median.
(^8) Prove that the line segment joining the mid-points of two opposite edges
of a tetrahedron contains and is bisected by the centroid of the tetrahedron.
(^9) Determine whether, in all cases, the two line segments joining mid-points
of opposite edges of a quadrilateral must intersect and bisect each other. Be
sure to recognize that a quadrilateral in Ea need not have all of its vertices in the
same plane.