MathDB

4

Part of 2005 District Olympiad

Problems(6)

sequence with many null terms

Source: RMO District 2005, 9th Grade, Problem 4

3/5/2005
Let {ak}k≥1\{a_k\}_{k\geq 1} be a sequence of non-negative integers, such that ak≥a2k+a2k+1a_k \geq a_{2k} + a_{2k+1}, for all k≥1k\geq 1. a) Prove that for all positive integers n≥1n\geq 1 there exist nn consecutive terms equal with 0 in the sequence {ak}k\{a_k\}_k; b) State an example of sequence with the property in the hypothesis which contains an infinite number of non-zero terms.
algebra proposedalgebra
triangle, interior angle bisector, circumcenter, incenter

Source:

3/5/2005
In the triangle ABCABC let ADAD be the interior angle bisector of ∠ACB\angle ACB, where D∈ABD\in AB. The circumcenter of the triangle ABCABC coincides with the incenter of the triangle BCDBCD. Prove that AC2=AD⋅ABAC^2 = AD\cdot AB.
geometrycircumcircleincenterratioquadraticsangle bisectorperpendicular bisector
combinatorics and paths on cubes

Source: RMO District 2005, 8th Grade, Problem 3

3/5/2005
Prove that no matter how we number the vertices of a cube with integers from 1 to 8, there exists two opposite vertices in the cube (e.g. they are the endpoints of a large diagonal of the cube), united through a broken line formed with 3 edges of the cube, such that the sum of the 4 numbers written in the vertices of this broken lines is at least 21.
geometry3D geometrypigeonhole principle
functions from {1,2, ..., n} to {1,2, ..., n}

Source: RMO District 2005, 10th Grade, Problem 4

3/5/2005
Let n≥3n\geq 3 be an integer. Find the number of functions f:{1,2,…,n}→{1,2,…,n}f:\{1,2,\ldots,n\}\to\{1,2,\ldots,n\} such that f(f(k))=f3(k)−6f2(k)+12f(k)−6,  for all k≥1. f(f(k)) = f^3(k) - 6f^2(k) + 12f(k) - 6 , \ \textrm{ for all } k \geq 1 .
functioninductionalgebrafunctional equationalgebra proposed
Romania District Olympiad 2005 - Grade XI

Source:

4/10/2011
Let f:Q→Qf:\mathbb{Q}\rightarrow \mathbb{Q} a monotonic bijective function.
a)Prove that there exist a unique continuous function F:R→RF:\mathbb{R}\rightarrow \mathbb{R} such that F(x)=f(x), (∀)x∈QF(x)=f(x),\ (\forall)x\in \mathbb{Q}.
b)Give an example of a non-injective polynomial function G:R→RG:\mathbb{R}\rightarrow \mathbb{R} such that G(Q)⊂QG(\mathbb{Q})\subset \mathbb{Q} and it's restriction defined on Q\mathbb{Q} is injective.
functionalgebrapolynomialirrational numberreal analysisreal analysis unsolved
number of perfect squares in a ring may make it a field

Source: RMO District 2005, 12th Grade, Problem 4

3/5/2005
Let (A,+,⋅)(A,+,\cdot) be a finite unit ring, with n≥3n\geq 3 elements in which there exist exactly n+12\dfrac {n+1}2 perfect squares (e.g. a number b∈Ab\in A is called a perfect square if and only if there exists an a∈Aa\in A such that b=a2b=a^2). Prove that a) 1+11+1 is invertible; b) (A,+,⋅)(A,+,\cdot) is a field. Proposed by Marian Andronache
group theoryabstract algebracalculusintegrationalgebrafunctiondomain