olymmath

NL-NWO 1990

Dutch Mathematical Olympiad Final 1990.

Other problems

Problem 1

Prove that for every integer n>1,1⋅3⋅5⋅...⋅(2n−1)<nn.n>1, 1 \cdot 3 \cdot 5 \cdot ... \cdot (2n-1)<n^n.

Problem 2

Consider the sequence a1=32,an+1=3an2+4an−34an2.a_1=\frac{3}{2}, a_{n+1}=\frac{3a_n^2+4a_n-3}{4a_n^2}. (a)(a) Prove that 1<an1<a_n and an+1<ana_{n+1}<a_n for all nn . (b)(b) From (a)(a) it follows that lim⁡n→∞an\lim_{n\to\infty}a_n exists. Find this limit. (c)(c) Determine lim⁡n→∞a1a2a3...an\lim_{n\to\infty}a_1a_2a_3...a_n .

Problem 3

A polynomial f(x)=ax4+bx3+cx2+dxf(x)=ax^4+bx^3+cx^2+dx with a,b,c,d>0a,b,c,d>0 is such that f(x)f(x) is an integer for x∈{−2,−1,0,1,2}x \in \{ -2,-1,0,1,2 \} and f(1)=1f(1)=1 and f(5)=70f(5)=70 . (a)(a) Show that a=124,b=14,c=1124,d=14a=\frac{1}{24}, b=\frac{1}{4},c=\frac{11}{24},d=\frac{1}{4} . (b)(b) Prove that f(x)f(x) is an integer for all x∈Zx \in \mathbb{Z} .

Problem 4

If ABCDEFGABCDEFG is a regular 77 -gon with side 11 , show that: 1AC+1AD=1\frac{1}{AC}+\frac{1}{AD}=1 .