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Part of 2019 District Olympiad

Problems(6)

(a+1)/3=(b+2)/4=5/(c+3) diophantine 2019 Romania District VII p1

Source:

9/1/2024
Determine the integers a,b,ca, b, c for which a+13=b+24=5c+3\frac{a+1}{3}=\frac{b+2}{4}=\frac{5}{c+3}
algebranumber theoryDiophantine equationDiophantine Equations
5(x^2+xy+y^2) = 7(x+2y), integer, rational 2019 Romania District VIII p1

Source:

9/1/2024
Determine the numbers x,yx,y, with xx integer and yy rational, for which equality holds: 5(x2+xy+y2)=7(x+2y)5(x^2+xy+y^2) = 7(x+2y)
algebranumber theory
Romanian District Olympiad 2019 - Grade 9 - Problem 1

Source: Romanian District Olympiad 2019 - Grade 9 - Problem 1

3/18/2019
Let n∈N,n≥2n \in \mathbb{N}, n \ge 2 and the positive real numbers a1,a2,…,ana_1,a_2,…,a_n and b1,b2,…,bnb_1,b_2,…,b_n such that a1+a2+…+an=b1+b2+…+bn=S.a_1+a_2+…+a_n=b_1+b_2+…+b_n=S. <spanclass=′latex−bold′>a)</span><span class='latex-bold'>a)</span> Prove that ∑k=1nak2ak+bk≥S2.\sum\limits_{k=1}^n \frac{a_k^2}{a_k+b_k} \ge \frac{S}{2}. <spanclass=′latex−bold′>b)</span><span class='latex-bold'>b)</span> Prove that ∑k=1nak2ak+bk=∑k=1nbk2ak+bk.\sum\limits_{k=1}^n \frac{a_k^2}{a_k+b_k}= \sum\limits_{k=1}^n \frac{b_k^2}{a_k+b_k}.
InequalityidentityCauchy Inequalityalgebra
Romanian District Olympiad 2019 - Grade 10 - Problem 1

Source: Romanian District Olympiad 2019 - Grade 10 - Problem 1

3/17/2019
Find the functions f:R→(0,∞)f: \mathbb{R} \to (0, \infty) which satisfy 2−x−y≤f(x)f(y)(x2+1)(y2+1)≤f(x+y)(x+y)2+1,2^{-x-y} \le \frac{f(x)f(y)}{(x^2+1)(y^2+1)} \le \frac{f(x+y)}{(x+y)^2+1}, for all x,y∈R.x,y \in \mathbb{R}.
functionFunctional inequalityalgebra
Romanian District Olympiad 2019 - Grade 11 - Problem 1

Source: Romanian District Olympiad 2019 - Grade 11 - Problem 1

3/16/2019
Let (an)n≥1(a_n)_{n \ge 1} be a sequence of positive real numbers such that the sequence (an+1−an)n≥1(a_{n+1}-a_n)_{n \ge 1} is convergent to a non-zero real number. Evaluate the limit lim⁡n→∞(an+1an)n. \lim_{n \to \infty} \left( \frac{a_{n+1}}{a_n} \right)^n.
limitSequencesConvergencecalculus
Romanian District Olympiad 2019 - Grade 12 - Problem 1

Source: Romanian District Olympiad 2019 - Grade 12 - Problem 1

3/16/2019
Let nn be a positive integer and GG be a finite group of order n.n. A function f:G→Gf:G \to G has the (P)(P) property if f(xyz)=f(x)f(y)f(z) ∀ x,y,z∈G.f(xyz)=f(x)f(y)f(z)~\forall~x,y,z \in G. <spanclass=′latex−bold′>(a)</span><span class='latex-bold'>(a)</span> If nn is odd, prove that every function having the (P)(P) property is an endomorphism. <spanclass=′latex−bold′>(b)</span><span class='latex-bold'>(b)</span> If nn is even, is the conclusion from <spanclass=′latex−bold′>(a)</span><span class='latex-bold'>(a)</span> still true?
endomorphismabtract algebraGroupssuperior algebra