MathDB

4

Part of 2005 Romania National Olympiad

Problems(6)

2005 numbers on a circle with sum 7022

Source: Romanian Nationals RMO 2005 - grade 7, problem 4

3/31/2005
On a circle there are written 2005 non-negative integers with sum 7022. Prove that there exist two pairs formed with two consecutive numbers on the circle such that the sum of the elements in each pair is greater or equal with 8. After an idea of Marin Chirciu
yet another classical inequality

Source: Romanian Nationals RMO 2005 - grade 8, problem 4

3/31/2005
a) Prove that for all positive reals u,v,x,yu,v,x,y the following inequality takes place: ux+vy≥4(uy+vx)(x+y)2. \frac ux + \frac vy \geq \frac {4(uy+vx)}{(x+y)^2} . b) Let a,b,c,d>0a,b,c,d>0. Prove that ab+2c+d+bc+2d+a+cd+2a+b+da+2b+c≥1. \frac a{b+2c+d} + \frac b{c+2d+a} + \frac c{d+2a+b} + \frac d{a+2b+c} \geq 1. Traian Tămâian
inequalitiesalgebra
equation with rational fractional parts

Source: Romanian Nationals RMO 2005 - grade 10, problem 4

3/31/2005
For α∈(0,1)\alpha \in (0,1) we consider the equation {x{x}}=α\{x\{x\}\}= \alpha. a) Prove that the equation has rational solutions if and only if there exist m,p,q∈Zm,p,q\in\mathbb{Z}, 0<p<q0<p<q, gcd⁡(p,q)=1\gcd(p,q)=1, such that α=(pq)2+mq\alpha = \left( \frac pq\right)^2 + \frac mq. b) Find a solution for α=200420052\alpha = \frac {2004}{2005^2}.
algebra proposedalgebra
shortlist 2001 strikes back

Source: Romanian Nationals RMO 2005 - grade 9, problem 4

3/31/2005
Let x1,x2,…,xnx_1,x_2,\ldots,x_n be positive reals. Prove that 11+x1+11+x1+x2+⋯+11+x1+⋯+xn<1x1+1x2+⋯+1xn. \frac 1{1+x_1} + \frac 1{1+x_1+x_2} + \cdots + \frac 1{1+x_1+\cdots + x_n} < \sqrt { \frac 1{x_1} + \frac 1{x_2} + \cdots + \frac 1{x_n}} . Bogdan Enescu
geometryinequalitiesinequalities proposedn-variable inequality
Convex function and functional equation

Source: Romanian Nationals RMO 2005 - grade 11, problem 4

3/31/2005
Let f:R→Rf:\mathbb{R}\to\mathbb{R} be a convex function. a) Prove that ff is continous; b) Prove that there exists an unique function g:[0,∞)→Rg:[0,\infty)\to\mathbb{R} such that for all x≥0x\geq 0 we have f(x+g(x))=f(g(x))−g(x). f(x+g(x)) = f(g(x)) - g(x) .
functionlimitalgebra proposedalgebra
yet another condition for a ring to be a field

Source: Romanian Nationals RMO 2005 - grade 12, problem 4

3/31/2005
Let AA be a ring with 2n+12^n+1 elements, where nn is a positive integer and let M={k∈Z∣k≥2, xk=x, ∀ x∈A}. M = \{ k \in\mathbb{Z} \mid k \geq 2, \ x^k =x , \ \forall \ x\in A \} . Prove that the following statements are equivalent: a) AA is a field; b) MM is not empty and the smallest element in MM is 2n+12^n+1. Marian Andronache
inductiontopologyRing Theorysuperior algebrasuperior algebra unsolved