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
Therefore, by induction, (an)2>2n for all n>2,
so (a10000)2>20000>1412, so a10000>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)2pi, 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=11-3 integers and select three with the sum 3b.
Of the remaining 5=8-3 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 6a1 and repeat
the procedure ten more times, every time at least 11 integers being
available. We obtain eleven pairwise disjoint sixtuples with the sums
6a1,..., 6a11. Of the eleven integers,
a1 through a11,
we can select six, say, a1 through a6, with the sum divisible by 6.
The six sixtuples with the sums 6a1,..., 6a6 are the required
36 integers with the sum 6(a1+...+a6) divisible by 36.