IMO预选题1969

IMO Longlist 1969 IMO ShortList/LongList Project Group June 19, 2004 1. (Belgium 1) In an orthogonal system we have two parabolas with equations: P1 : x2 − 2py = 0 P2 : x2 + 2py = 0 with p > 0. The line t is a tangent line to P2. Determine the locus of M , pole of t with respect to P1. 2. (Belgium 2) Find the equation of the equilateral hyperboles which pass through the points A(α, 0), B(β, 0), C(0, γ). Show that the hyperboles pass through the orthocenter H of triangle ABC. Find the locus of the centers of those hyperboles. And verify that this locus coincides wit the nine-point circle of triangle ABC. 3. (Belgium 3) Three circles C1, C2, C3 have only one point in common. Construct a circle which is tangent to the given three circles. 4. (Belgium 4) A conic is passing through the origin O. A right angle is O intersects the conic in points A and B. Prove that the line AB passes through a fixed point which is situated on the normal on O at the conic. 5. (Belgium 5) Let G be the center of gravity of a given triangle OAB. Show that the conics which pass through the points O,A,B,G are hyperboles. Find the locus of the centers of those hyperboles. 6. (Belgium 6) Calculate ( cos (pi 4 ) + i · sin (pi 4 ))10 in two different ways and conclude that C110 − C310 + 1 3 C510 = 2 4. Global Remark: In those days analytic geometry were the highlights in the curricula of Belgium. All the questions, especially the sixth question, are questions which appeared in handbooks. 7. (Bulgaria 1) For all natural i > 2, let hi be the inradius of a regular n-gon inscribed into a circle with radius R. Prove that (n+ 1)hn+1 − nhn > R for every natural n > 2. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. 1 8. (Bulgaria 2) Prove that the equation x3 + y3 + z3 = 19692 has no solutions in integer numbers. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. 9. (Bulgaria 3) Find all functions f (x) defined for every x and satisfying the equation xf (y) + yf (x) = (x+ y) f (x) f (y) for arbitrary x and y. Show that only two such functions are continuous. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. 10. (Bulgaria 4) On a plane, consider a square with side length 38 cm, and 100 convex polygons, assuming that the area of such a polygon is always ≤ pi cm2, and the perimeter of such a polygon is always ≤ 2pi cm. Prove that there exists a circular disk of radius 1 cm having no common point with any of these 100 polygons, but completely lying in the interior of the square. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. 11. (France 1) Let a and b two positive integers. Let H(a, b) be the set of numbers n satisfying n = pa + qb with p and q being positive numbers including zero. Define H(a) = H(a, a). Show that, if a 6= b with a and b having no common divisor, it is sufficient to know a particular H(a, b) in order to know all other sets H(a, b). Show that in this case, the set H contains all the numbers greater than or equal to w = (a− 1)(b− 1) and also w 2 other numbers (exception occurs if w = 0). 12. (France 2) Let n be an integer, which has only 1 as divisor, that is a square. In the number system with base n, we consider the last digit x1 of an integer smaller than n and we consider also the last digits (denoted by xm) of the integer powers xm of that number. For which numbers x do we have xm = 0 ? Show that the sequence xm has a period t. For which numbers x do we have xt = 1 ? Show that if m,n and t have no common divisor except 1, the numbers 0m, 1m, 2m, . . . , (n− 1)m are all different one from another. Determine the minimal condition value of t as a function of n. If n does not satisfy the above conditions, show that it is possible to have xm = 0 6= x1 and that the sequence is periodic but only starting from a certain value k, independent from x. 13. (France 3) Given a polygon, not necessarily convex, with area S, of which the vertices have integer coordinates, not necessarily positive. If I is the number of points in the interior of the polygon and B the number of points on the border, determine the integer T = 2S −B − 2I + 2 c© by Orlando Do¨hring, member of the IMO ShortList/LongList Project Group, page 2/12 . 14. (France 4) We have a right triangle OAB (Bˆ = 90◦) and a circle with center located on [OB], tangent on OA. From A two tangents can be drawn at the given circle. Let T be the contact with the circle, different from that with OA. Show that the median, line of gravity, from B in the triangle OAB intersect AT in a point M such that |MB| = |MT |. 15. (France 5) Let α(n) be the number of integer pairs (x, y) such that x + y = n, 0 ≤ y ≤ x. Let β(n) be the number integer triplets (x, y, z) such that x + y + z = n, 0 ≤ z ≤ y ≤ x. Find a simple relation between α(n) and ⌊ n+2 2 ⌋ , and another relation among β(n), β(n − 3) and α(n). Finally represent β(n) as a result of the values of the rest of the division of n by 6. What can be said about β(n) and the number 1 + n(n+ 6) 12 , and β(n) and the number (n+ 3)2 12 ? Determine the number of triplets (x, y, z) in a function divisible by 6. What can be said about this number regarding (n+ 6)(2n2 + 9n+ 12) 72 ? Global Remark: All these questions are made by Mr. Warusfel, who was a fanatic supporter of the new math. Secondly he follows the French tradition of the questions for the exam ”Compo- sition des Matheme´matiques” and this means that every question is a complex investigation of a problem and not a problem on its own. 16. (Great Britain 1) Consider the polynomial P (x) = a0xk + a1xk−1 + . . .+ ak with a0, a1, . . . , ak being integers. Such a polynomial is said to be divisible by m if P (x) is a multiple for every integer x. Show that a0k! is a multiple of m if P (k) is divisible by m. Prove also the following property: If a0, k and m are positive integers with a0k! a multiple of m, then it is possible to find a polynomial divisible by m of which a0xk has the highest degree. (Darij Grinberg’s translation from Morozova/Petrakov:) A polynomial P (x) = a0xk + a1xk−1 + ...+ ak, where a0, a1, ..., ak are integers, is said to be divisible by m, if m divides the number P (x) for every integer x. Show that if the polynomial P (x) is divisible by m, then the number k!a0 is divisible by m. Also show that if the numbers a0, k and m are positive integers such that k!a0 is divisible by m, then you can always find a polynomial P (x) = a0xk + a1xk−1 + ...+ ak divisible by m. 17. (Great Britain 2) Given are the negative integers a, b, x, y with the assumption that a, b have no common factor. Prove that ab− a− b represents the greatest number which cannot be written in the form ax+ by. c© by Orlando Do¨hring, member of the IMO ShortList/LongList Project Group, page 3/12 18. (Great Britain 3) Consider a solid composed of a right circular cylinder, with height h and radius r, and the hemisphere place on the top of it, radius r and center O. That solid is placed on a horizontal table. A wire is attached on one side at a point of the base of the cylinder, stretched vertically and attached at the top at point P of the hemisphere. Let α be the angle which OP makes with the horizontal table. Show that, if α is sufficiently small and the wire is a little bit displaced, not any more in the vertical plane, the wire is coming loose. Remark: Seems to be rather a science olympiad problem than a mathematical olympiad problem. 19. (Great Britain 4) Let AB be perpendicular to CD in a pointX in such a way that |CX| = |XD|. Show that with some restrictions there exists a truncated cone on which A,B,C,D are ranged. What are those restrictions ? 20. (Great Britain 5) We have u0 = 1, u1 = 1 and un+2 = aun+1 + bun, n ≥ 0, with a and b being natural numbers. Write un as a polynomial in a and b and prove your result. If b is a prime, show that b divides a(u9 − 1). 21. (Great Britain 6) Given a point O and three lengths x, y, z. Show that an equilateral triangle ABC with vertices satisfying OA = x, OB = y, OC = z exists if and only if the inequalities y + z ≥ x, z + x ≥ y, x+ y ≥ z are fulfilled. (Hereby, the points O, A, B, C are supposed to lie on one plane.) Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. Global Remark: The question proposed by Great Britain are- in the whole - rather peculiar. The questions are totally influenced by the traditions of the A − level exams in England. It must honestly be said that the proposals of a lot of western countries were analogous to the curriculum situation in their country. 22. (Hungary 1) 126. In an isosceles triangle ABC, we have AB = AC and ]BAC = 20◦. Let D be the point on the edge AB such that AD = CD, and let E be the point on the edge AC such that BC = CE. Find ]CDE. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. 23. (Hungary 2) Prove the inequality for any n ∈ N n∑ k=1 1 k3 < 5 4 . Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. c© by Orlando Do¨hring, member of the IMO ShortList/LongList Project Group, page 4/12 24. (Hungary 3) Given 4000 points in the plane such that no three of these points are collinear. Show the existence of 1000 pairwise disjunct quadrilaterals whose vertices are the given points. (Two quadrilaterals are called disjunct if there is no point lying in the interior of both quadrilat- erals.) Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. 25. (Mongolia 1) Find the circumradius of the isosceles triangle whose side lengths are the roots of a quadratic equation x2 − ax+ b = 0. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. 26. (Netherlands 1) The vertices of a polygon with (n + 1) sides are chosen on the perimeter of a given regular polygon with n sides in such a way that they divide the perimeter in two equal parts. How can this be done so that the area of the polygon with (n+1) sides is (i) maximal and (ii) minimal. (Darij Grinberg’s translation from Morozova/Petrakov:) Consider an (n+ 1)-gon with vertices lying on the sides of a fixed regular n-gon and dividing the perimeter of this n-gon into n + 1 equal parts. [Of course, two certain vertices of the (n+ 1)-gon will have to lie on one side of the n-gon.] How should the vertices of the (n+ 1)-gon be placed on the sides of the n-gon in order to have the area of the (n+ 1)-gon a) maximal ? b) minimal ? 27. (Netherlands 2) A semicircular arc γ is drawn on AB as diameter. C is a point on γ other than A and B, and D is the foot of the perpendicular from C to AB. We consider the three circles γ1, γ2, γ3, all tangent to the line AB. Of these, γ1 is inscribed in triangle ABC, while γ2 and γ3 are both tangent to CD and to γ, one on each side of CD. Prove that γ1, γ2, γ3 have a second tangent in common. Remark: This question was chosen as the fourth IMO question. c© by Orlando Do¨hring, member of the IMO ShortList/LongList Project Group, page 5/12 28. (Netherlands 3) The positive integers x1, x2, x3, x4, x5 which satisfy the following relations: x1 + x2 + x3 + x4 + x5 = 1000 x1 − x2 + x3 − x4 + x5 > 0 x1 + x2 − x3 + x4 − x5 > 0 −x1 + x2 + x3 − x4 + x5 > 0 x1 − x2 + x3 + x4 − x5 > 0 −x1 + x2 − x3 + x4 + x5 > 0 Determine the maximum value of (x1 + x3)x2+x4 . In how many different ways can you choose x1, x2, x3, x4, x5 so that the maximum value is attained. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. 29. (Netherlands 4) A boy plays with his trains and has constructed a special type of closed circuit, which has no intersections and no linear segments. At his disposal he has a great number of standard (equal) rails, all having the shape of a quarter of a circle. Some of them oblige the train to ride in the clockwise sense, others oblige the trains to move anti-clockwise. Prove that an even number of standard rails of both types is used and that the total number of used rails is a multiple of four. (Darij Grinberg’s translation from Morozova/Petrakov:) A boy has constructed a railway line for his model train. The railway line is closed, it has no self-intersections, and it consists of a number of congruent rail segments; each of these segments has the form of a quarter circle (i. e. a circular arc with central angle 90◦. Some of the rail segments are placed in such a way that the train, passing through them, drives in counterclockwise direction; these rail segments will be called positive segments. Other rail segments are placed in such a way that the train, passing through them, drives in clockwise direction; these rail segments will be referred to as negative segments. Show that the number of all positive segments in the railway line is even, the number of all negative segments in the railway line is also even, and the number of all segments in the railway line is divisible by 4. 30. (Netherlands 5) The bisectors of the exterior angels of a given pentagon B1, B2, B3, B4, B5 pro- vide a second pentagon A1, A2, A3, A4, A5. Resolve the inverse question: Construct the pentagon B1, B2, B3, B4, B5 when the pentagon A1, A2, A3, A4, A5 is given. 31. (Netherlands 6) The curve defined by the equation y = √ x2 − 10x+ 52, 0 ≤ x ≤ 100 and drawn on ”millimeter-paper”. In the drawn network of that paper the side of a square is chosen as unit of measure and the two axes x′x and y′y coincide with two lines from the network. De- termine the number of squares where the curve is passing through. 32. (Poland 1) Prove that it is impossible to divide a regular polygon with an odd number of sides, in four parts with equal area by two lines passing through the center of the polygon. 33. (Poland 2) Consider two line segments [AB] and [CD] not lying in the same plane. Find the locus of the point M satisfying |MA|2 + |MB|2 = |MC|2 + |MD|2 c© by Orlando Do¨hring, member of the IMO ShortList/LongList Project Group, page 6/12 34. (Poland 3) A polynomial in x with integer coefficients has the property that f(x) is a multiple of three for three values k, k+1, k+2 of x. Show that f(m) is a multiple of three for every integer m. 35. (Poland 4) For each value of k = 1, 2, 3, 4, 5 find a necessary and sufficient condition for the number a > 0 so that there exist a tetrahedron with k edges of length a, and the remaining 6− k edges of length 1. Remark: This question was chosen as the third question in the IMO itself. 36. (Poland 5) Let a, b be two natural numbers with the same number n of digits and such that the first m digits (from left to right) of a and b are the same. Let m > n 2 , then prove that a 1 n − b 1n < 1 n . 37. (Poland 6) In a triangle ABC, the points K and L lie on [AB] and [CD], respectively, and |KB| |AK| + LC AL = 1. Show that KL passes through the center of gravity of triangle ABC. 38. (Poland 7) Given a polynomial f (x) with integer coefficients, such that there exists an integer k with 3 | f (k) , 3 | f (k + 1) and 3 | f (k + 2) . Show that 3 | f (m) for every integer m. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. 39. (Poland 8) Let a and b be positive real numbers and m an integer. Show( 1 + a b )m + ( 1 + b a )m ≥ 2m+1. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. But I wonder why there seem to be 8 proposed problems by Poland (usually at most 6 problems can be proposed for IMO by a country.) 40. (Socialists Republic Of Czechoslovakia 1) Given a cube with side length 1. Find the locus of the centers of gravity of all the tetrahedrons whose summits are situated in the interior of four different faces of the given cube. 41. (Socialists Republic Of Czechoslovakia 2) Given an odd prime number p. Can you find p−1 natural numbers n+1, n+2m. . . , n+ p− 1 such that the sum of their squares is divisible by the sum of those numbers ? Remark: This problem can also be found in one of the volumes by Morozova/Petrakov. c© by Orlando Do¨hring, member of the IMO ShortList/LongList Project Group, page 7/12 42. (Socialists Republic Of Czechoslovakia 3) Let a, b be two real numbers. If x is a real so- lution of an equation x2 + px + q = 0 with real p, q such that (1) |p| ≤ a, |q| ≤ b then (2) x = 1 2 ( a+ √ a2 + 4b ) . Reversely, when x is a real number satisfying (2), one can find an equa- tion (1) with |p| ≤ a, |q| ≤ b so that x is one of the solutions. 43. (Socialists Republic Of Czechoslovakia 4) Let k1, k2, . . . , kn, n ≥ 2, be n non-negative numbers. Prove the following inequality: n∏ k=1 ki! ≥   n∑ k=1 ki n !  n . Remark: This problem can also be found in one of the volumes by Morozova/Petrakov. 44. (Socialists Republic Of Czechoslovakia 5) Let ABCD be a convex quadrilateral with sides |AB| = a, |BC| = b, |CD| = c, |DA| = d and interior angles α = ∠DAB, β = ∠ABC, γ = ∠BCD, δ = ∠CDA. Let s = 1 2 (a+ b+ c+ d) and P the area of the quadrilateral. Prove that P 2 = (s− a)(s− b)(s− c)(s− d)− abcd · cos2 ( α+ β 2 ) . Remark: This is a generalization of the formula of Heron for triangles. 45. (Socialists Republic Of Czechoslovakia 6) Let d and p be two real numbers. Find the first element of an arithmetical sequence a1, a2, a3, . . . with difference d such that a1a2a3a4 = p. What is the numbers of solutions as a function in terms of d, p ? 46. (Socialists Republic Of Czechoslovakia 7) Given n points in space such that any three of them form a triangle with one angle > 120◦. Show that it is possible to label these points by A1, A2, ..., An such that the angle AiAjAk is > 120◦ for any three i, j, k with i < j < k. Remark By Orlando: This problem was added to the IMO LongList from one of the Moro- zova/Petrakov volumes. The logical order how the problems are listed in the book makes me assume that the problem was proposed for the IMO 1969. But I wonder why there seem to be 7 proposed problems by Socialists Republic Of Czechoslovakia (usually at most 6 problems can be proposed for IMO by a country.) 47. (Soviet Union 1) Prove that