(q 1)p q. = 1 (p 1)(q 1).

Transcription

1 altic Way 001 Hamburg, November 4, 001 Problems 1. set of 8 problems was prepared for an examination. Each student was given 3 of them. No two students received more than one common problem. What is the largest possible number of students?. Let n be a positive integer. Find whether there exist n pairwise nonintersecting nonempty subsets of {1,, 3,...} such that each positive integer can be expressed in a uniue way as a sum of at most n integers, all from different subsets. 3. The numbers 1,,..., 49 are placed in a 7 7 array, and the sum of the numbers in each row and in each column is computed. Some of these 14 sums are odd while others are even. Let denote the sum of all the odd sums and the sum of all even sums. Is it possible that the numbers were placed in the array in such a way that =? 4. Let p and be two different primes. Prove that p + p + 3p ( 1)p = 1 (p 1)( 1). (Here x denotes the largest integer not greater than x.) 5. Let 001 given points on a circle be colored either red or green. In one step all points are recolored simultaneously in the following way: If both direct neighbors of a point P have the same color as P, then the color of P remains unchanged, otherwise P obtains the other color. Starting with the first coloring F 1, we obtain the colorings F, F 3,... after several recoloring steps. Prove that there is a number n such that F n0 = F n0+. Is the assertion also true if 1000 is replaced by 999? 6. The points,,,, E lie on the circle c in this order and satisfy E and E. The line tangent to the circle c at E meets the line at P. The lines and E meet at Q. Prove that = P Q. 1

2 7. Given a parallelogram. circle passing through meets the line segments, and at inner points M, K, N, respectively. Prove that M + N = K. 8. Let be a convex uadrilateral, and let N be the midpoint of. Suppose further that N = 135. Prove that Given a rhombus, find the locus of the points P lying inside the rhombus and satisfying P + P = In a triangle, the bisector of meets the side at the point. Knowing that = and = 45, determine the angles of triangle. 11. The real-valued function f is defined for all positive integers. For any integers a > 1, b > 1 with d = gcd(a, b), we have ( ( a ) ( b ) ) f(ab) = f(d) f + f, d d etermine all possible values of f(001). 1. Let a 1, a,..., a n be positive real numbers such that a 5 i = 5. Prove that a i > 3. a 3 i = 3 and 13. Let a 0, a 1, a,... be a seuence of real numbers satisfying a 0 = 1 and a n = a 7n/9 + a n/9 for n = 1,,.... Prove that there exists a positive integer k with a k < k 001!. (Here x denotes the largest integer not greater than x.) 14. There are n cards. On each card some real number x, 1 x, is written (there can be different numbers on different cards). Prove that

3 the cards can be divided into two heaps with sums s 1 and s so that n n + 1 s 1 1. s 15. Let a 0, a 1, a,... be a seuence of positive real numbers satisfying i a i (i + 1) a i 1 a i+1 for i = 1,,.... Furthermore, let x and y be positive reals, and let b i = xa i + ya i 1 for i = 1,,.... Prove that the ineuality i b i > (i+1) b i 1 b i+1 holds for all integers i. 16. Let f be a real-valued function defined on the positive integers satisfying the following condition: For all n > 1 there exists a prime divisor p of n such that ( n ) f(n) = f f(p). p Given that f(001) = 1, what is the value of f(00)? 17. Let n be a positive integer. Prove that at least n 1 + n numbers can be chosen from the set {1,, 3,..., n } such that for any two different chosen numbers x and y, x + y is not a divisor of x y. 18. Let a be an odd integer. Prove that a n + n and a m + m are relatively prime for all positive integers n and m with n m. 19. What is the smallest positive odd integer having the same number of positive divisors as 360? 0. From a seuence of integers (a, b, c, d) each of the seuences (c, d, a, b), (b, a, d, c), (a+nc, b+nd, c, d), (a+nb, b, c+nd, d), for arbitrary integer n can be obtained by one step. Is it possible to obtain (3, 4, 5, 7) from (1,, 3, 4) through a seuence of such steps? Solutions 1. nswer: 8. enote the problems by,,,, E, F, G, H, then 8 possible problem sets are, E, F G, G, F H, H, EF, EGH. Hence, there could be 8 students. 3

4 Suppose that some problem (e.g., ) was given to 4 students. Then each of these 4 students should receive different supplementary problems, and there should be at least 9 problems a contradiction. Therefore each problem was given to at most 3 students, and there were at most 8 3 = 4 awardings of problems. s each student was awarded 3 problems, there were at most 8 students.. nswer: yes. Let 1 be the set of positive integers whose only non-zero digits may be the 1-st, the (n + 1)-st, the (n + 1)-st etc. from the end; be the set of positive integers whose only non-zero digits may be the -nd, the (n + )- nd, the (n + )-nd etc. from the end, and so on. The sets 1,,..., n have the reuired property. Remark. This problem is uite similar to problem 18 from altic Way nswer: no. If this were possible, then ( ) = + =. ut is even since it is the sum of even numbers, whereas = 5 49 is odd. This is a contradiction. 4. The line y = p x contains the diagonal of the rectangle with vertices (0, 0), (, 0), (, p) and (0, p) and passes through no points with integer coordinates in the interior of that rectangle. For k = 1,,..., 1 the summand kp counts the number of interior points of the rectangle lying below the diagonal y = p x and having x-coordinate eual to k. Therefore the sum in consideration counts all interior points with integer coordinates below the diagonal, which is exactly half the number of all points with integer 1 coordinates in the interior of the rectangle, i.e. (p 1)( 1). Remark. The integers p and need not be primes: in the solution we only used the fact that they are coprime. 5. nswer: no. Let the points be denoted by 1,,..., 001 such that i, j are neighbors if i j = 1 or {i, j} = {1, 001}. We say that k points form a monochromatic segment of length k if the points are consecutive on the circle and if 4

5 they all have the same color. For a coloring F let d(f ) be the maximum length of a monochromatic segment. Note that d(f n ) > 1 for all n since 001 is odd. If d(f 1 ) = 001 then all points have the same color, hence F 1 = F = F 3 =... and we can choose n 0 = 1. Thus, let 1 < d(f 1 ) < 001. elow we shall prove the following implications: If 3 < d(f n ) < 001, then d(f n+1 ) = d(f n ) ; (1) If d(f n ) = 3, then d(f n+1 ) = ; () If d(f n ) =, then d(f n+1 ) = d(f n ) and F n+ = F n ; (3) From (1) and () it follows that d(f 1000 ), hence by (3) we have F 1000 = F 100. Moreover, if F 1 is the coloring where 1 is colored red and all other points are colored green, then d(f 1 ) = 000 and thus d(f 1 ) > d(f ) >... > d(f 1000 ) = which shows that, for all n < 1000, F n F n+ and thus 1000 cannot be replaced by 999. It remains to prove (1) (3). Let (i + 1,..., i + k) be a longest monochromatic segment for F n (considering the labels of the points modulo 001). Then (i +,..., i + k 1) is a monochromatic segment for F n+1 and thus d(f n+1 ) d(f n ). Moreover, if (i + 1,..., i + k) is a longest monochromatic segment for F n+1 where k 3, then (i,..., i + k + 1) is a monochromatic segment for F n. From this and F n+1 > 1 the implications (1) and () clearly follow. For proof of (3) note that if d(f n ) then F n+1 is obtained from F n by changing the colour of all points. PSfrag replacements E Q P Figure 1 6. The arcs and E are of eual length (see Figure 1). lso, since 5

6 E and E, we have = E and the arcs and are of eual length. Since P E is tangent to c and E =, then P E = = Q. s is inscribed in c, we have Q = 180 E = P E. lso, is an isosceles trapezium, whence E =. So the triangles P E and Q are congruent, and Q = P. Now P Q is a uadrilateral with a pair of opposite sides eual and parallel. So P Q is a parallelogram, and P Q =. 7. Let X be the point on segment such that X = KN, then X = NK = 180 MK (see Figure ). Triangles NK and X are similar, having two pairs of N PSfrag replacements eual angles, hence X =. Since triangles MK and X K M M are also similar, we have X = = and K K M + N =( X + X ) K = K. E Q P N K X M Figure 8. Let X be the point symmetric to with respect to N, and let Y be the point symmetric to with respect to N (see Figure 3). Then XNY = 180 ( ) = 90 and NX = NY =. Therefore, XY =. Moreover, we have 6

7 X = and Y =. onseuently, X + XY + Y = + +. replacements Y X N E Figure 3 9. Qnswer: the locus of the points P is the union of the diagonals and P. Let Q be a point such that P Q is a parallelogram (see Figure 4). Then QP is also a parallelogram. From the euality P + P = 180 it follows that Q + P = 180, so the points, Q,, P lie on a common circle. Therefore, P = P Q = P, and since M =, we obtain that P = P or P + P = 180. NHence, the point P lies on the segment or on the segment. K X P Q P Q Figure 4 onversely, any point P lying on the diagonal satisfies the euation P = P. Therefore, P + P = 180. nalogously, we show that the last euation holds if the point P lies on the diagonal. 10. nswer: = 60, = 105 and = 15. Suppose the line meets the circumcircle of triangle at and E (see Figure 5). Let M be the midpoint of and O the circumcentre of triangle. Since the arcs E and E are eual, then the points O, M, E are collinear and OE is perpendicular to. From the euality 7

8 P E M = = 45 it follows that EO = 45. Since O = EO, we have N OE = 90 and O M. K From the X euality = we obtain = E, which implies that OM = ME. Therefore O = E and also O = EO. Hence the triangle OE is euilateral. This gives E = 30, so = 60. Summing up the angles of the triangle we obtain = 105 and from this = 15. P Q O M E Figure nswer: 0 and 1. Obviously the constant functions f(n) = 0 and f(n) = 1 provide solutions. We show that there are no other solutions. ssume f(001) 0. Since 001 = and gcd(3, 667) = 1, then f(001) = f(1) (f(3) + f(667)), and f(1) 0. Since gcd(001, 001) = 001 then f(001 ) = f(001)( f(1)) 0. lso gcd(001, ) = 001, so f(001 4 )=f(001) (f(1) + f(001 ))=f(1)f(001)(1 + f(001)). On the other hand, gcd(001, 001 ) = 001 and f(001 4 )=f(001 ) (f(1)+f(1))=f(1)f(001 )=4f(1) f(001). So 4f(1) = 1 + f(001) and f(001) = f(1) 1. Exactly the same 8

9 argument starting from f(001 ) 0 instead of f(001) shows that f(001 ) = f(1) 1. So f(1) 1 ( = f(1) f(1) 1 ). Since f(1) 1 = f(001) 0, we have f(1) = 1, which implies f(001) = f(1) 1 = y Hölder s ineuality, ( n ) 3/5 ( n ) /5 a 3 = (a i a i ) a 5/3 i (a i ) 5/. We will show that ( n ) 5/3 a i. (4) a 5/3 i Let S = ( ai S a i, then (4) is euivalent to ) 5/3 1 = n a i S, which holds since 0 < a i S 1 and 5 ( 3 > 1 yield ai ) 5/3 a i S S. So, ( n ) ( n ) /5 a 3 i a i a 5 i, which gives 13. onsider the euation ( 7 ) x ( 1 ) x + = a i 3 5 /5 > 3, since 5 > 5 and hence > 5 /5. 9

10 It has a root 1 < α < 1, because = > 1 and < 1. We will prove that a n M n α for some M > 0 since nα will be n arbitrarily small for large enough n, the claim follows from this immediately. We choose M so that the ineuality a n M n α holds for 1 n 8; since for n 9 we have 1 < [7n/9] < n and 1 [n/9] < n, it follows by induction that [ 7n ] α [ n ] α a n = a [7n/9] + a [n/9] M + M 9 9 ( 7n ) α ( n ) ( α (7 α ( 1 ) ) α M + M = M nα + = M n 9 9 9) α Let the numbers be x 1 x... x n 1 x n. We will show that the choice s 1 = x 1 + x 3 + x x n 1 and s = x + x x n solves the problem. Indeed, the ineuality s 1 s 1 is obvious and we have s 1 s = x 1 + x 3 + x x n 1 x + x 4 + x x n = (x 3 + x x n 1 ) + x 1 (x + x x n ) + x n (x 3 + x x n 1 ) + 1 (x + x x n ) + (x + x x n ) + 1 (x + x x n ) + = 1 = 1 (x + x x n ) (n 1) + = n n Let i. We are given the ineualities (i 1) a i 1 i a i a i (5) and i a i (i + 1) a i+1 a i 1. (6) Multiplying both sides of (6) by x, we obtain i x a i (i + 1) x a i+1 a i 1. (7) 10

11 y (5), a i 1 i a i a i i 1 = i 1 > i = i + 1, i which implies i y a i 1 > (i + 1) y a i a i. (8) Multiplying (5) and (6), and dividing both sides of the resulting ineuality by ia i a i 1, we get (i 1) a i a i 1 (i + 1) a i+1 a i. dding (i + 1)a i a i 1 to both sides of the last ineuality and multiplying both sides of the resulting ineuality by xy gives i xy a i a i 1 (i + 1) xy (a i+1 a i + a i a i 1 ). (9) Finally, adding up (7), (8) and (9) results in i (xa i + ya i 1 ) > (i + 1) (xa i+1 + ya i )(xa i 1 + ya i ), which is euivalent to the claim. 16. nswer:. For any prime p we have f(p) = f(1) f(p) and thus f(p) = f(1). If n is a product of two primes p and, then f(n) = f(p) f() or f(n) = f() f(p), so f(n) = 0. y the same reasoning we find that if n is a product of three primes, then there is a prime p such that ( n ) f(n) = f f(p) = f(p) = f(1). p y simple induction we can show that if n is the product of k primes, then f(n) = ( k) f(1). In particular, f(001) = f(3 3 9) = 1 so f(1) =. Therefore, f(00) = f( ) = f(1) =. 17. We choose the numbers 1, 3, 5,..., n 1 and, 4, 8, 16,..., n, i.e. all odd numbers and all powers of. onsider the three possible cases. 11

12 (1) If x = a 1 and y = b 1, then x+y = (a 1)+(b 1) = (a+b 1) is even and does not divide xy = (a 1)(b 1) which is odd. () If x = k and y = m where k < m, then x + y = k ( m k + 1) has an odd divisor greater than 1 and hence does not divide xy = a+b. (3) If x = k and y = b 1, then x + y = k + (b 1) > (b 1) is odd and hence does not divide xy = k (b 1) which has b 1 as its largest odd divisor. 18. Rewriting a n + n = a n n + n and making repeated use of the identity a n n = (a n 1 n 1 ) (a n 1 + n 1 ) we get a n + n = (a n 1 + n 1 ) (a n + n )... (a m + m ) (a + ) (a + ) (a ) + n. For n > m, assume that a n + n and a m + m have a common divisor d > 1. Then an odd integer d divides n, a contradiction. 19. nswer: n integer with the prime factorization p r1 1 pr... pr k k (where p 1, p,..., p k are distinct primes) has precisely (r 1 + 1) (r + 1)... (r k + 1) distinct positive divisors. Since 360 = 3 3 5, it follows that 360 has 4 3 = 4 positive divisors. Since 4 = 3, it is easy to check that the smallest odd number with 4 positive divisors is = nswer: no. Under all transformations (a, b, c, d) (a, b, c, d ) allowed in the problem we have ad bc = a d b c, but = 1 = Remark. The transformations allowed in the problem are in fact the elementary transformations of the determinant a b c d and the invariant ad bc is the absolute value of the determinant which is preserved under these transformations. 1