#

#
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 *CD*^{2}+PQ^{2}=AB^{2}+AB^{2}/9 equals the sum
of the squares of the four sides *2(CP*^{2}+CQ^{2}). Therefore
*AB*^{2}+AB^{2}/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*=1999^{2000} 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.