1999 SOLUTIONS, PART II

Problem 1. a) 167,166,167,166,167,166,167,166,167,166,167,168
           b) If this were possible, there would be six tables with
              even numbers of cupcakes and six tables with odd
              numbers of cupcakes. The sum of six odd numbers
              and six even numbers is even. However, 1999 is odd.

Problem 2. Complete triangle ABC to a rectangle ABCD and draw PD
           and QD. In the parallelogram CPDQ, the sum of the squares
           of the diagonals CD2+PQ2=AB2+AB2/9 equals the sum
           of the squares of the four sides 2(CP2+CQ2). Therefore
           AB2+AB2/9=2·5 and AB=3.

Problem 3. If not, then no two are equal. Without loss of generality
           assume that c is between a and b. Then |P(a)-P(b)|=|c-b|<|b-a|.
           It is easy to show that b-a is a factor of P(b)-P(a), a contradiction.

Problem 4. Since there are only 1999 colors, among every 2000 points there are
           two of the same color. There are N=19992000 ways of coloring
           a string of 2000 points. Consider the following N+1 strings of
           2000 points (k,1),(k,2),...,(k,2000), k=0,1,...,N. Since there are N+1
           of them, the coloring pattern is the same for at least two of the
           strings. Each of the two strings has at least two points of the same
           color. The four points form a rectangle.

Problem 5. Brutus wins as follows:
           (i) If Caesar leaves a single (2n+1) by 1999 rectangle, Brutus leaves
               a single (2n+1) by (2n+1) square reducing the situation to a
               smaller odd square.
          (ii) If Caesar leaves a single 2n by 1999 rectangle, Brutus leaves two
               n by 1999 rectangles and plays a mirror image strategy henceforth.
         (iii) If Caesar leaves an m by 1999 rectangle and an n by 1999
               rectangle with m>n, Brutus leaves two n by 1999 rectangles and
               plays a mirror image strategy henceforth.