# 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.```