olymmath

IrMO 1989

Irish Mathematical Olympiad 1989.

Paper 1

Problem 1

  • Paper 1

A quadrilateral ABCDABCD is inscribed, as shown, in a square of area one unit. Prove that 2≤∣AB∣2+∣BC∣2+∣CD∣2+∣DA∣2≤42\le |AB|^2+|BC|^2+|CD|^2+|DA|^2\le 4

DD AA BB CC

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Problem 5

  • Paper 1

Let x=a1a2…anx = a_1a_2 \dots a_n be an n-digit number, where a1,a2,…,an(a1≠0)a_1, a_2, \dots , an (a_1 \neq 0) are the digits. The nn numbers x1=x=a1a2...an,x_1 = x = a_1 a_2 ... a_n, x2=ana1...an−1,x_2 = a_n a_1 ... a_{n-1}, x3=an−1ana1...an−2x_3 = a_{n-1} a_n a _1 ... a_{n-2} , x4=an−2an−1ana1,...an−3,x_4 = a_{n-2} a_{n-1} a_n a_1 , ... a_{n-3} , ...,xn=a2a3...ana1... , x_n = a_2 a_3 ... a_n a_1 are said to be obtained from xx by the cyclic permutation of digits. [For example, if n=5n = 5 and x=37001x = 37001 , then the numbers are x1=37001,x2=13700,x_1 = 37001, x_2 = 13700, x3=01370(=1370),x4=00137(=137),x_3 = 01370(= 1370), x_4 = 00137(= 137), x5=70013.]x_5 = 70013.] Find, with proof, (i) the smallest natural number n for which there exists an n-digit number x such that the n numbers obtained from x by the cyclic permutation of digits are all divisible by 1989; and (ii) the smallest natural number x with this property.

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Paper 2

Problem 6

  • Paper 2

Suppose LL is a fixed line, and AA is a fixed point not on LL . Let kk be a fixed nonzero real number. For PP a point on LL , let QQ be a point on the line APAP with ∣AP∣⋅∣AQ∣=k2|AP|\cdot |AQ|=k^2 . Determine the locus of QQ as PP varies along the line LL .

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Problem 10

  • Paper 2

(i): Prove that if nn is a positive integer, then (2nn)=(2n)!(n!)2\binom{2n}{n}=\frac{(2n)!}{(n!)^2} is a positive integer that is divisible by all prime numbers pp with n<p≤2nn<p\le 2n , and that (2nn)<22n.\binom{2n}{n}<2^{2n}. (ii): For xx a positive real number, let π(x)\pi(x) denote the number of prime numbers p≤xp \le x . [Thus, π(10)=4\pi(10) = 4 since there are 44 primes, viz., 22 , 33 , 55 , and 77 , not exceeding 1010 .]Prove that if n≥3n \ge 3 is an integer, then (a) π(2n)<π(n)+2nlog⁡2(n);\pi(2n) < \pi(n) + {{2n}\over{\log_2(n)}}; (b) π(2n)<2n+1log⁡2(n−1)n;\pi(2^n) < {{2^{n+1}\log_2(n-1)}\over{n}}; (c) Deduce that, for all real numbers x≥8x \ge 8 , π(x)<4xlog⁡2(log⁡2(x))log⁡2(x).\pi(x) < {{4x \log_2(\log_2(x))}\over{\log_2(x)}}.

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