2000 SOLUTIONS, PART II

If there are at most 44 colors and and most 44 cans of the same color, then
the total number of cans is 44·44=1936. This is a contradiction.

The answer is NO. Here is an example. The sides of the first triangle are
1, 1.5, 1.5·1.5=2.25. The sides of the second triangle are
1.5, 1.5·1.5=2.25, 1.5·1.5·1.5=3.375.
In both cases the triangle inequality is satisfied, so the triangles exist.
The triangles are similar.
The measures of the 3 angles of the first triangle are all different from each
other but are the same as the measures of the corresponding angles of the
second triangle.

(a_{n+1})^{2}=(a_{n})^{2}+2+(a_{n})^{2}
>(a_{n})^{2}+2.
Therefore, by induction, (a_{n})^{2}>2n for all n>2,
so (a_{10000})^{2}>20000>141^{2}, so a_{10000}>141.

Consider the 250 disks, each of radius 1/10
that are centered at each of the points. The sum of the
areas of these disks is 2.5pi, and the union of the
disks is contained inside a disk of radius 1.1.
Since 2.5pi>2(1.1)^{2}pi, there is a point
P (not necessarily in the chosen set) that is contained
in at least 3 of the small disks. Thus, the disk of
radius 1/10, centered at P contains at least three
of the original points.

a. Given any five integers, either three of them have the same remainders when divided by 3
or three of them have all different remainders. In both cases, the sum of these three is a multiple
of 3, say 3a. Take any five of the remaining 8=113 integers and select three with the sum 3b.
Of the remaining 5=83 integers select three with the sum 3c.
Two of the integers a, b, c are of the same parity, say, a and b.
The sum 3a+3b= 3(a+b) is divisible by 6.
b. Given 71 integers, select six with the sum 6a_{1} and repeat
the procedure ten more times, every time at least 11 integers being
available. We obtain eleven pairwise disjoint sixtuples with the sums
6a_{1},..., 6a_{11}. Of the eleven integers,
a_{1} through a_{11},
we can select six, say, a_{1} through a_{6}, with the sum divisible by 6.
The six sixtuples with the sums 6a_{1},..., 6a_{6} are the required
36 integers with the sum 6(a_{1}+...+a_{6}) divisible by 36.