The 24th United States of America Mathematics Olympiad
1995年第24届美国数学奥林匹克 
 The sequence a_{0}a_{1}, a_{2}, ... of nonnegative integers is defined as follows. The first p1 terms are 0, 1, 2, 3, ... , p2. Then a_{n} is the least positive integer so that there is no arithmetic progression of length p in the first n+1 terms. If p is an odd prime, show that a_{n} is the number obtained by writing n in base p1, then treating the result as a number in base p. For example, if p is 5, to get the 5th term one writes 5 as 11 in base 4, then treats this as a base 5 number to get 6 .
 A trigonometric map is any one of sin, cos, tan, arcsin, arccos and arctan. Show that given any positive rational number x, one can find a finite sequence of trigonometric maps which take 0 to x. [So we need to show that we can always find a sequence of trigonometric maps t_{i} so that: x_{1} = t_{0}(0), x_{2} = t_{1}(x_{1}), ... , x_{n} = t_{n1}(x_{n1}), x = t_{n}(x_{n}).]
 The circumcenter O of the triangle ABC does not lie on any side or median. Let the midpoints of BC, CA, AB be L, M, N respectively. Take P, Q, R on the rays OL, OM, ON respectively so that ∠OPA = ∠OAL, ∠OQB = ∠OBM and ∠ORC = ∠OCN. Show that AP, BQ and CR meet at a point.
 a_{0}, a_{1}, a_{2}, ... is an infinite sequence of integers such that a_{n}  a_{m} is divisible by n  m for all (unequal) n and m. For some polynomial p(x) we have p(n) > a_{n} for all n. Show that there is a polynomial q(x) such that q(n) = a_{n} for all n.
 A graph with n points and k edges has no triangles. Show that it has a point P such that there are at most k(1  4k/n^{2}) edges between points not joined to P (by an edge).

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