The Art and Craft of Problem Solving

300 CHAPTER 8 GEOMETRY FOR AMERICANS

We can write AB = B -A, etc., and everything will cancel, or we can think

dynamically: The sum above means, "start at A, travel to B, then to C, then

back to A." We end up where we started; the relative motion is zero.

• Since the diagonals of a parallelogram bisect each other (8.2.9), the midpoint
  of the line segment AB has the vector position (A + B) 12.
  The next two examples show how we can use this dynamical approach to investi­
  gate more challenging problems.
  Example 8.5.2 Why is the centroid of a triangle the center of mass of its vertices?
  Solution: Let n equal point masses be placed at vector positions xi ,ii, ... ,�. We
  define their center of mass to be (Xi +i2 + ... +�)In. Thus we must show that the

centroid of a triangle with vertices A,B,C is (A + B +C)/3. In the picture below, the

centroid of triangle ABC is denoted by K, and the midpoint of AB is M.

c

We know by the centroid theorem that KM ICM = 1/3. Now let's "travel" to K. Start­

ing at M, we just go one-third of the way to C. In other words,

_ - 1�

K =M+ 3 Ml..

• 1 - -

=M+3(C-M)

2 M+C

3

A+B+C

3

where the final step used the fact that M = (A + B) 12. •

Example 8.5.3 (Hungary 1935) Prove that a finite point set S cannot have more than

one center of symmetry. (A center of symmetry of a set S means a point 0, not

necessarily in S, such that for every point A E S, there is another point BE S such that

o is the midpoint of AB. We say that B is symmetric to A with respect to 0.)

Solution: Suppose, to the contrary, that there were two centers of symmetry, P

and Q. Let S = {Al,A 2 , ... ,An}. For each k = 1,2, ... ,n, let Bk denote the point in