MathDB

2

Part of 2016 Romania National Olympiad

Problems(6)

angle bisector, 135-30-15 triangle (2016 Romanian NMO grade VII P2)

Source:

6/1/2020
Consider the triangle ABCABC, where ∠B=30o,∠C=15o\angle B= 30^o, \angle C = 15^o, and MM is the midpoint of the side [BC][BC]. Let point N∈(BC)N \in (BC) be such that [NC]=[AB][NC] = [AB]. Show that [AN[AN is the angle bisector of MACMAC
geometryangle bisectorequal segmentsangles
perp. common in cube, D'M/D'C=DN/DA'=1/3 (2016 Romanian NMO grade VII P2)

Source:

6/1/2020
In a cube ABCDA′B′C′D′ABCDA'B'C'D' two points are considered, M∈(CD′)M \in (CD') and N∈(DA′)N \in (DA'). Show that the MNMN is common perpendicular to the lines CD′CD' and DA′DA' if and only if D′MD′C=DNDA′=13.\frac{D'M}{D'C}=\frac{DN}{DA'} =\frac{1}{3}.
3D geometrygeometrycubeperpendicularratio
Inequality involving the terms of a fast increasing finite series

Source: Romania National Olympiad 2016, grade ix, p.2

8/25/2019
Let be a natural number n≥2 n\ge 2 and n n positive real numbers a1,an,…,an a_1,a_n,\ldots ,a_n that satisfy the inequalities \sum_{j=1}^i a_j\le a_{i+1} ,  \forall i\in\{ 1,2,\ldots ,n-1 \} . Prove that ∑k=1n−1akak+1≤n/2. \sum_{k=1}^{n-1} \frac{a_k}{a_{k+1}}\le n/2 .
inequalitiesn-variable inequality
Equalities and inequalities of ranks

Source: Romanian National Olympiad 2016, grade xi, p.2

8/25/2019
Consider a natural number, n≥2, n\ge 2, and three n×n n\times n complex matrices A,B,C A,B,C such that A A is invertible, B B is formed by replacing the first line of A A with zeroes, and C C is formed by putting the last n−1 n-1 lines of A A above a line of zeroes. Prove that:
a) rank(A−1B)=rank((A−1B)2)=⋯=rank((A−1B)n) \text{rank} \left( A^{-1} B \right) = \text{rank} \left( \left( A^{-1} B\right)^2 \right) =\cdots =\text{rank} \left( \left( A^{-1} B\right)^n \right) b) rank(A−1C)>rank((A−1C)2)>⋯>rank((A−1C)n) \text{rank} \left( A^{-1} C \right) > \text{rank} \left( \left( A^{-1} C\right)^2 \right) >\cdots >\text{rank} \left( \left( A^{-1} C\right)^n \right)
linear algebrarank
Inequality results about some function

Source: Romania National Olympiad 2016, grade x, p.2

8/25/2019
Let be a function f:R⟶R f:\mathbb{R}\longrightarrow\mathbb{R} satisfying the conditions: {f(x+y)≤f(x)+f(y)f(tx+(1−t)y)≤t(f(x))+(1−t)f(y), \left\{\begin{matrix} f(x+y) &\le & f(x)+f(y) \\ f(tx+(1-t)y) &\le & t(f(x)) +(1-t)f(y) \end{matrix}\right. , for all real numbers x,y,t x,y,t with t∈[0,1]. t\in [0,1] .
Prove that: a) f(b)+f(c)≤f(a)+f(d), f(b)+f(c)\le f(a)+f(d) , for any real numbers a,b,c,d a,b,c,d such that a≤b≤c≤d a\le b\le c\le d and d−c=b−a. d-c=b-a. b) for any natural number n≥3 n\ge 3 and any n n real numbers x1,x2,…,xn, x_1,x_2,\ldots ,x_n, the following inequality holds. f(∑1≤i≤nxi)+(n−2)∑1≤i≤nf(xi)≥∑1≤i<j≤nf(xi+xj) f\left( \sum_{1\le i\le n} x_i \right) +(n-2)\sum_{1\le i\le n} f\left( x_i \right)\ge \sum_{1\le i<j\le n} f\left( x_i+x_j \right)
functioninequalitiesinduction
Romanian National Olympiad 2016 (correction)

Source: Romanian National Olympiad 2016, grade 12, problem 2

12/29/2023
Let AA be a ring and let DD be the set of its non-invertible elements. If a2=0a^2=0 for any a∈D,a \in D, prove that: a) axa=0axa=0 for all a∈Da \in D and x∈Ax \in A; b) if DD is a finite set with at least two elements, then there is a∈D,a \in D, a≠0,a \neq 0, such that ab=ba=0,ab=ba=0, for every b∈D.b \in D.
Ioan Băetu
abstract algebraRing Theory