MathDB

3

Part of 2021 Romania National Olympiad

Problems(6)

Nice geometry prob is in fact length bash

Source: Romanian NMO 2021 grade 7 P3

4/15/2023
Let ABCABC be a scalene triangle with ∠BAC>90∘\angle BAC>90^\circ. Let DD and EE be two points on the side BCBC such that ∠BAD=∠ACB\angle BAD=\angle ACB and ∠CAE=∠ABC\angle CAE=\angle ABC. The angle-bisector of ∠ACB\angle ACB meets ADAD at NN, If MN∥BCMN\parallel BC, determine ∠(BM,CN)\angle (BM, CN).
Petru Braica
geometry
System of two equations

Source: Romania NMO 2021 grade 8

4/25/2021
Solve the system in reals: (x+x2+1)(y+y2+1)=2022(x+\sqrt{x^2+1})(y+\sqrt{y^2+1})=2022 and x+y=20212022x+y=\frac{2021}{\sqrt{2022}}
algebra
Romanian NMO 2021 inequality

Source:

4/25/2021
If a,b,c>0,a+b+c=1a,b,c>0,a+b+c=1,then:
1abc+4a2+b2+c2≥13ab+bc+ca\frac{1}{abc}+\frac{4}{a^{2}+b^{2}+c^{2}}\geq\frac{13}{ab+bc+ca}
inequalitiesalgebra
Sum of some nth roots of unity is 0

Source: Romanian NMO 2021 grade 10 P3

4/15/2023
Let n≥2n\ge 2 be a positive integer such that the set of nnth roots of unity has less than 2⌊n⌋−12^{\lfloor\sqrt n\rfloor}-1 subsets with the sum 00. Show that nn is a prime number.
Cristi Săvescu
number theoryprime numbers
Romania National Olympiad Grade 11 P3

Source:

4/28/2021
Let f:R→Rf :\mathbb R \to\mathbb R a function n≥2 n \geq 2 times differentiable so that: lim⁡x→∞f(x)=l∈R \lim_{x \to \infty} f(x) = l \in \mathbb R and lim⁡x→∞f(n)(x)=0 \lim_{x \to \infty} f^{(n)}(x) = 0. Prove that: lim⁡x→∞f(k)(x)=0 \lim_{x \to \infty} f^{(k)}(x) = 0 for all k∈{1,2,…,n−1} k \in \{1, 2, \dots, n - 1\} , where f(k)f^{(k)} is the k k - th derivative of ff.
real analysisTaylor expansionderivative
Order NT

Source: Romania NMO 2021 grade 12

4/25/2021
Given is an positive integer a>2a>2 a) Prove that there exists positive integer nn different from 11, which is not a prime, such that an=1(modn)a^n=1(mod n) b) Prove that if pp is the smallest positive integer, different from 11, such that ap=1(modp)a^p=1(mod p), then pp is a prime. c) There does not exist positive integer nn, different from 11, such that 2n=1(modn)2^n=1(mod n)
number theory