MathDB

1

Part of 2009 Romania Team Selection Test

Problems(5)

Covered by translates of A-A

Source: Romania TST 1 2009, Problem 1

5/4/2012
For non-empty subsets A,B⊂ZA,B \subset \mathbb{Z} define A+B={a+b:a∈A,b∈B}, A−B={a−b:a∈A,b∈B}.A+B=\{a+b:a\in A, b\in B\},\ A-B=\{a-b:a\in A, b\in B\}.
In the sequel we work with non-empty finite subsets of Z\mathbb{Z}.
Prove that we can cover BB by at most ∣A+B∣∣A∣\frac{|A+B|}{|A|} translates of A−AA-A, i.e. there exists X⊂ZX\subset Z with ∣X∣≤∣A+B∣∣A∣|X|\leq \frac{|A+B|}{|A|} such that B⊆∪x∈X(x+(A−A))=X+A−A.B\subseteq \cup_{x\in X} (x+(A-A))=X+A-A.
combinatorics proposedcombinatorics
Maximum of x_1+...+x_n subject to x_1+...+x_n=x_1...x_n

Source: Romania TST 4 2009, Problem 1

5/4/2012
Given an integer n≥2n\geq 2, determine the maximum value the sum x1+⋯+xnx_1+\cdots+x_n may achieve, as the xix_i run through the positive integers, subject to x1≤x2≤⋯≤xnx_1\leq x_2\leq \cdots \leq x_n and x1+⋯+xn=x1x2⋯xnx_1+\cdots+x_n=x_1 x_2\cdots x_n.
algebra proposedalgebra
About special rulers

Source: Romania TST 2 2009, Problem 1

5/4/2012
We call Golomb ruler a ruler of length ll, bearing k+1≥2k+1\geq 2 marks 0<a1<…<ak−1<l0<a_1<\ldots <a_{k-1}<l, such that the lengths that can be measured using marks on the ruler are consecutive integers starting with 11, and each such length be measurable between just two of the gradations of the ruler. Find all Golomb rulers.
combinatorics proposedcombinatorics
Inequality for angle

Source: Romania TST 3 2009, Problem 1

5/4/2012
Let ABCDABCD be a circumscribed quadrilateral such that AD>max⁡{AB,BC,CD}AD>\max\{AB,BC,CD\}, MM be the common point of ABAB and CDCD and NN be the common point of ACAC and BDBD. Show that 90∘<m(∠AND)<90∘+12m(∠AMD).90^{\circ}<m(\angle AND)<90^{\circ}+\frac{1}{2}m(\angle AMD).
Fixed, thank you Luis.
inequalitiesgeometry proposedgeometry
Superposing two polygonal domains

Source: Romania TST 5 2009, Problem 1

5/4/2012
Given two (identical) polygonal domains in the Euclidean plane, it is not possible in general to superpose the two using only translations and rotations. Prove that this can however be achieved by splitting one of the domains into a finite number of polygonal subdomains which then fit together, via translations and rotations in the plane, to recover the other domain.
algebrafunctiondomaingeometrygeometric transformationrotationgeometry proposed