4th Austrian-Polish Mathematics Competition 1981 Problems

4th Austrian-Polish Mathematics Competition 1981 Problems

1.  Find the smallest n for which we can find 15 distinct elements a₁, a₂, ... , a₁₅ of {16, 17, ... , n} such that a_k is a multiple of k.

2.  The rational sequence a₀, a₁, a₂, ... satisfies a_n+1 = 2a_n² - 2a_n + 1. Find all a₀ for which there are four distinct integers r, s, t, u such that a_r - a_s = a_t - a_u.

3.  The diagram shows the incircle of ABC and three other circles inside the triangle, each touching the incircle and two sides of the triangle. The radius of the incircle is r and the radius of the circle nearest to A, B, C is r_A, r_B, r_C respectively. Show that r_A + r_B + r_C ≥ r, with equality iff ABC is equilateral. 

4.  n symbols are arranged in a circle. Each symbol is 0 or 1. A valid move is to change any 1 to 0 and to change the two adjacent symbols. For example one could change ... 01101 ... to ... 10001 ... . The initial configuration has just one 1. For which n can one obtain all 0s by a sequence of valid moves? 

5.  A quartic with rational coefficients has just one real root. Show that the root must be rational. 

6.  The real sequences x₁, x₂, x₃, ... , y₁, y₂, y₃, ... , z₁, z₂, z₃, ... satisfy x_n+1 = y_n + 1/z_n, y_n+1 = z_n + 1/x_n, z_n+1 = x_n + 1/y_n. Show that the sequences are all unbounded. 

7.  If N = 2ⁿ > 1 and k > 3 is odd, show that k^N - 1 has at least n+1 distinct prime factors. 

8.  A is a set of r parallel lines, B is a set of s parallel lines, and C is a set of t parallel lines. What is the smallest value of r + s + t such that the r + s + t lines divide the plane into at least 1982 regions?

9.  Let X be the closed interval [0, 1]. Let f: X → X be a function. Define f¹ = f, fⁿ⁺¹(x) = f( fⁿ(x) ). For some n we have |fⁿ(x) - fⁿ(y)| < |x - y| for all distinct x, y. Show that f has a unique fixed point.

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3rd Austrian-Polish Mathematics Competition 1980 Problems

3rd Austrian-Polish Mathematics Competition 1980 Problems

1.  A, B, C are infinite arithmetic progressions of integers. {1, 2, 3, 4, 5, 6, 7, 8} is a subset of their union. Show that 1980 also belongs to their union.

2.  1 = a₁ < a₂ < a₃ < ... is an infinite sequence of integers such that a_n < 2n-1. Show that every positive integer is the difference of two members of the sequence.

3.  P is an interior point of a tetrahedron. Show that the sum of the six angles subtended by the sides at P is greater than 540^o.

4.  [Missing]

Solutions

3rd APMC 1980 Problem 1

A, B, C are infinite arithmetic progressions of integers. {1, 2, 3, 4, 5, 6, 7, 8} is a subset of their union. Show that 1980 also belongs to their union.

Solution

If any two of 4, 6, 8 belong to the same set, then we are done. So assume (wlog) that 4 ∈ A, 6 ∈ B and 8 ∈ C. Then 5 cannot belong to A or B or we are done, so 5 ∈ C. Similarly 3 cannot belong to A or B or we are done, so 3 ∈ C. But if 3 and 5 are in C, then so is 7. Since 7 and 8 are in C, then so are all integers > 8 and hence 1980.

Thanks to Suat Namli

 

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2nd Austrian-Polish Mathematics Competition 1979 Problems

2nd Austrian-Polish Mathematics Competition 1979 Problems

1.  ABCD is a square. E is any point on AB. F is the point on BC such that BF = BE. The perpendicular from B meets EF at G. Show that ∠DGF = 90^o.

2.  Find all polynomials of degree n with real roots x₁ ≤ x₂ ≤ ... ≤ x_n such that x_k belongs to the closed interval [k, k+1] and the product of the roots is (n+1)/(n-1)! .

3.  Find all positive integers n such that for all real numbers x₁, x₂, ... , x_n we have S₂S₁ - S₃ ≥ 6P, where S_k = ∑ x_i^k, and P = x₁x₂ ... x_n.

4.  Let N₀ = {0, 1, 2, 3, ... } and R be the reals. Find all functions f: N₀ → R such that f(m + n) + f(m - n) = f(3m) for all m, n.

5.  A tetrahedron has circumcenter O and incenter I. If O = I, show that the faces are all congruent.

6.  k is real, n is a positive integer. Find all solutions (x₁, x₂, ... , x_n) to the n equations:

x₁ + x₂ + ... + x_n = k

x₁² + x₂² + ... + x_n² = k²

...

x₁ⁿ + x₂ⁿ + ... + x_nⁿ = kⁿ

7.  Find the number of paths from (0, 0) to (n, m) which pass through each node at most once. 

8.  ABCD is a tetrahedron. M is the midpoint of AC and N is the midpoint of BD. Show that AB² + BC² + CD² + DA² = AC² + BD² + 4 MN².

9.  Find the largest power of 2 that divides [(3 + √11)²ⁿ⁺¹].

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1st Austrian-Polish Mathematics Competition 1978 Problems

1st Austrian-Polish Mathematics Competition 1978 Problems

1.  Find all real-valued functions f on the positive reals which satisfy f(x + y) = f(x² + y²) for all positive x, y.

2.  A parallelogram has its vertices on the boundary of a regular hexagon and its center at the center of the hexagon. Show that its area is at most 2/3 the area of the hexagon.

3.  Let x = 1^o. Show that (tan x tan 2x ... tan 44x)^1/44 < √2 - 1 < (tan x + tan 2x + ... + tan 44x)/44.

4.  Given a positive rational k not equal to 1, show that we can partition the positive integers into sets A_k and B_k, so that if m and n are both in A_k or both in B_k then m/n does not equal k.

5.  The sets A₁, A₂, ... , A₁₉₇₈ each have 40 elements and the intersection of any two distinct sets has just one element. Show that the intersection of all the sets has one element.

6.  S is a set of disks in the plane. No point belongs to the interior of more than one disk. Each disk has a point in common with at least 6 other disks. Show that S is infinite.

7.  S is a finite set of lattice points in the plane such that we can find a bijection f: S → S satisfying |P - f(P)| = 1 for all P in S. Show that we can find a bijection g: S → S such that |P - g(P)| = 1 for all P in S and g(g(P) ) = P for all P in S.

8.  k is a positive integer. Define a₁ = √k, a_n+1 = √(k + a_n). Show that the sequence a_n converges. Find all k such that the limit is an integer. Show that if k is odd, then the limit is irrational.

9.  P is a convex polygon. Some of the diagonals are drawn, so that no interior point of P lies on more than one diagonal. Show that at least two vertices of P do not lie on any diagonals.

Solutions

1st APMC 1978 Problem 1

Find all real-valued functions f on the positive reals which satisfy f(x + y) = f(x² + y²) for all positive x, y.
Solution
If x + y = k and x² + y² = h then xy = (k² - h)/2, so x, y are the roots of the polynomial z² - kz + (k² - h)/2 = 0. This polynomial has positive real roots provided that (1) k > 0, (2) k² ≥ 2(k² - h) or h ≥ k²/2, and (3) k² > k² - 2(k²-h) or h < k². So if k > 0 and k²/2 ≤ h < k², then f(k) = f(h). In other words, f must be constant on the interval [k²/2, k²). Take A_n to be the interval with k = (1.2)ⁿ. The the intervals A_n and A_n+1 overlap and ... A_-2, A_-1, A₀, A₁, A₂, ... cover the positive reals, so f is constant on the positive reals.
Thanks to Suat Namli

1st APMC 1978 Problem 3

Let x = 1^o. Show that (tan x tan 2x ... tan 44x)^1/44 < √2 - 1 < (tan x + tan 2x + ... + tan 44x)/44.
Solution
Note that (3-√8) + (√8-2) = 1 and (3-√8) = (√8-2)²/4, so if the positive integers A, B, satisfy A + B = 1, A < B²/4, then A < 3-√8.
Now for 0 < y < 45^o, we have 1 = tan 45^o = (tan y + tan(45^o-y) )/(1 - tan y tan(45^o-y)), so (tan y + tan(45^o-y)) + tan y tan(45^o-y) = 1. But by AM/GM (tan y + tan(45^o-y)) < (tan y tan(45^o-y))²/4, provided y ≠ 45^o/2. So by the result above we have (tan y + tan(45^o-y)) < 3 - √8, and hence (tan x tan 2x ... tan 44x)^1/44 < (3 - √8)^22/44 = √2 - 1. Also √8 - 2 < (tan y + tan(45^o-y)) and hence √2 - 1 = (22/44)(√8 - 2) < (tan x + tan 2x + ... + tan 44x)/44.
Thanks to Suat Namli
 
 

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