The 10th All Soviet Union Mathematical Olympiad
1976年第十届全苏数学奥林匹克 
 50 watches, all keeping perfect time, lie on a table. Show that there is a moment when the sum of the distances from the center of the table to the center of each dial equals the sum of the distances from the center of the table to the tip of each minute hand.
 1000 numbers are written in line 1, then further lines are constructed as follows. If the number m occurs in line n, then we write under it in line n+1, each time it occurs, the number of times that m occurs in line n. Show that lines 11 and 12 are identical. Show that we can choose numbers in line 1, so that lines 10 and 11 are not identical.
 (1) The circles C_{1}, C_{2}, C_{3} with equal radius all pass through the point X. C_{i} and C_{j} also intersect at the point Y_{ij}. Show that angle XO_{1}Y_{12} + angle XO_{2}Y_{23} + angle XO_{3}Y_{31} = 180 deg, where O_{i} is the center of circle C_{i}.（此题可能不全，等我找到俄语原版再做修补）
 a_{1} and a_{2} are positive integers less than 1000. Define a_{n} = min{ a_{i}  a_{j} : 0 < i < j < n}. Show that a_{21} = 0.
 Can you label each vertex of a cube with a different three digit binary number so that the numbers at any two adjacent vertices differ in at least two digits ?
 a, b, c, d are vectors in the plane such that a + b + c + d = 0. Show that a + b + c + d ≥ a + d + b + d + c + d.
 S is a set of 1976 points which form a regular 1976gon. T is the set of all points which are the midpoint of at least one pair of points in S. What is the greatest number of points of T which lie on a single circle ?
 n rectangles are drawn on a rectangular sheet of paper. Each rectangle has its sides parallel to the sides of the paper. No pair of rectangles has an interior point in common. If the rectangles were removed show that the rest of the sheet would be in at most n+1 parts.
 There are three straight roads. On each road a man is walking at constant speed. At time t = 0, the three men are not collinear. Prove that they will be collinear for t > 0 at most twice.
 Initially, there is one beetle on each square in the set S. Suddenly each beetle flies to a new square, subject to the following conditions: (1) the new square may be the same as the old or different; (2) more than one beetle may choose the same new square; (3) if two beetles are initially in squares with a common vertex, then after the flight they are either in the same square or in squares with a common vertex. Suppose S is the set of all squares in the middle row and column of a 99 x 99 chess board, is it true that there must always be a beetle whose new square shares a vertex with its old square (or is identical with it)? What if S also includes all the border squares (so S is rows 1, 50 and 99 and columns 1, 50 and 99)? What if S is all squares of the board ?
 Call a triangle big if each side is longer than 1. Show that we can draw 100 big triangles inside an equilateral triangle with side length 5 so that all the triangles are disjoint. Show that you can draw 100 big triangles with every vertex inside or on an equilateral triangle with side 3, so that they cover the equilateral triangle, and any two big triangles either (1) are disjoint, or (2) have as intersection a common vertex, or (3) have as intersection a common side.
 n is a positive integer. A universal sequence of length m is a sequence of m integers each between 1 and n such that one can obtain any permutation of 1, 2, ... , n by deleting suitable members of the sequence. For example, 1, 2, 3, 1, 2, 1, 3 is a universal sequence of length 7 for n = 3. But 1, 2, 3, 2, 1, 3, 1 is not universal, because one cannot obtain the permutation 3, 1, 2. Show that one can always obtain a universal sequence for n of length n^{2}  n + 1. Show that a universal sequence for n must have length at least n(n + 1)/2. Show that the shortest sequence for n = 4 has 12 members.
[You are told, but do not have to prove, that there is a universal sequence for n of length n^{2}  2n + 4.]
 n real numbers are written around a circle. One of the numbers is 1 and the sum of the numbers is 0. Show that there are two adjacent numbers whose difference is at least n/4. Show that there is a number which differs from the arithmetic mean of its two neighbours by at least 8/n^{2}. Improve this result to some k/n^{2} with k > 8. Show that for n = 30, we can take k = 1800/113. Give an example of 30 numbers such that no number differs from the arithmetic mean of its two neighbours by more than 2/113.
 You are given a regular ngon. Each vertex is marked +1 or 1. A move consists of changing the sign of all the vertices which form a regular kgon for some 1 < k <= n. [A regular 2gon means two vertices which have the center of the ngon as their midpoint.]. For example, if we label the vertices of a regular 6gon 1, 2, 3, 4, 5, 6, then you can change the sign of {1, 4}, {2, 5}, {3, 6}, {1, 3, 5}, {2, 4, 6} or {1, 2, 3, 4, 5, 6}. Show that for (1) n = 15, (2) n = 30, (3) any n > 2, we can find some initial marking which cannot be changed to all +1 by any series of moves. Let f(n) be the largest number of markings, so that no one can be obtained from any other by any series of moves. Show that f(200) = 2^{80}.
 S is a sphere with unit radius. P is a plane through the center. For any point x on the sphere f(x) is the perpendicular distance from x to P. Show that if x, y, z are the ends of three mutually perpendicular radii, then f(x)^{2} + f(y)^{2} + f(z)^{2} = 1 (*). Now let g(x) be any function on the points of S taking nonnegative real values and satisfying (*). Regard the intersection of P and S as the equator, the poles as the points with f(x) = 1 and lines of longitude as semicircles through both poles. (1) If x and y have the same longitude and both lie on the same side of the equator with x closer to the pole, show that g(x) > g(y). (2) Show that for any x, y on the same side of the equator with x closer to the pole than y we have g(x) > g(y). (3) Show that if x and y are the same distance from the pole then g(x) = g(y). (4) Show that g(x) = f(x) for all x.

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