MathDB

2

Part of 2013 District Olympiad

Problems(6)

|ax+by|+ |bx + ay| = 2|x| + 2|y| 2013 Romania District VII p2

Source:

9/1/2024
Find all pairs of real numbers (a,b)(a, b) such that the equality ∣ax+by∣+∣bx+ay∣=2∣x∣+2∣y∣|ax+by|+ |bx + ay| = 2|x| + 2|y| holds for all reals xx and yy.
algebra
(2x + 1)/(x^2 + 2x + 3) is integer

Source: 2013 Romania District VIII p2

9/1/2024
Find all real numbers xx for which the number a=2x+1x2+2x+3a =\frac{2x + 1}{x^2 + 2x + 3} is an integer.
number theoryInteger
Concurrent

Source: Romania District Olympiad 2013,grade IX,(problem 2)

3/14/2013
Given triangle ABCABC and the pointsD,E∈(BC)D,E\in \left( BC \right), F,G∈(CA)F,G\in \left( CA \right), H,I∈(AB)H,I\in \left( AB \right) so that BD=CEBD=CE, CF=AGCF=AG and AH=BIAH=BI. Note with M,N,PM,N,P the midpoints of [GH]\left[ GH \right], [DI]\left[ DI \right] and [EF]\left[ EF \right] and with M′{M}' the intersection of the segments AMAMand BCBC. a) Prove that BM′CM′=AGAH⋅ABAC\frac{B{M}'}{C{M}'}=\frac{AG}{AH}\cdot \frac{AB}{AC}. b) Prove that the segmentsAMAM, BNBN and CPCP are concurrent.
trigonometryratiogeometry proposedgeometry
Inequality

Source: Romania District Olympiad 2013,grade X(problem 2)

3/14/2013
Let a,b∈Ca,b\in \mathbb{C}. Prove that ∣az+bzˉ∣≤1\left| az+b\bar{z} \right|\le 1, for every z∈Cz\in \mathbb{C}, with ∣z∣=1\left| z \right|=1, if and only if ∣a∣+∣b∣≤1\left| a \right|+\left| b \right|\le 1.
inequalitiestrigonometryfunctioninequalities proposed
matrices

Source: Romania District Olympiad 2013,grade XI(problem 2)

3/14/2013
Let the matrices of order 2 with the real elements AA and BB so that AB=A2B2−(AB)2AB={{A}^{2}}{{B}^{2}}-{{\left( AB \right)}^{2}} and det⁡(B)=2\det \left( B \right)=2. a) Prove that the matrix AA is not invertible. b) Calculate det⁡(A+2B)−det⁡(B+2A)\det \left( A+2B \right)-\det \left( B+2A \right).
linear algebramatrixalgebra proposedalgebra
group

Source: Romania District Olympiad 2013,grade XII(problem 2)

3/14/2013
Problem 2. A group (G,⋅)\left( G,\cdot \right) has the propriety(P)\left( P \right), if, for any automorphism f for G,there are two automorphisms g and h in G, so that f(x)=g(x)⋅h(x)f\left( x \right)=g\left( x \right)\cdot h\left( x \right), whatever x∈Gx\in Gwould be. Prove that: (a) Every group which the property (P)\left( P \right) is comutative. (b) Every commutative finite group of odd order doesn’t have the (P)\left( P \right) property. (c) No finite group of order 4n+2,n∈N4n+2,n\in \mathbb{N}, doesn’t have the (P)\left( P \right)property. (The order of a finite group is the number of elements of that group).
superior algebrasuperior algebra unsolved