MathDB

2

Part of 2011 District Olympiad

Problems(6)

QR =\perp AD in orthodiagonal isosceles trapezoid (2011 Romania District VII P2)

Source:

5/19/2020
The isosceles trapezoid ABCDABCD has perpendicular diagonals. The parallel to the bases through the intersection point of the diagonals intersects the non-parallel sides [BC][BC] and [AD][AD] in the points PP, respectively RR. The point QQ is symmetric of the point PP with respect to the midpoint of the segment [BC][BC]. Prove that:
a) QR=ADQR = AD, b) QR⊥ADQR \perp AD.
geometrytrapezoidperpendiculardiagonals
p^2 + p +1=(r^2 + r + 1)(q^2 + q + 1), perfect square

Source: 2011 Romania District VIII p2

9/1/2024
a) Show that m2−m+1m^2- m +1 is an element of the set {n2+n+1∣n∈N}\{n^2 + n +1 | n \in N\}, for any positive integer m m.
b) Let pp be a perfect square, p>1p> 1. Prove that there exists positive integers rr and qq such that p2+p+1=(r2+r+1)(q2+q+1).p^2 + p +1=(r^2 + r + 1)(q^2 + q + 1).
number theoryPerfect Square
How many numbers of the form ±1±2...±n are there?

Source: Romanian District Olympiad 2011, Grade IX, Problem 2

10/8/2018
Let n n be a natural number. How many numbers of the form ±1±2±3±⋯±n \pm 1\pm 2\pm 3\pm\cdots\pm n are there?
countingalgebra
Romania District Olympiad 2011 - Grade XI

Source:

3/12/2011
Consider the matrices A∈Mm,n(C)A\in \mathcal{M}_{m,n}(\mathbb{C}) and B∈Mn,m(C)B\in \mathcal{M}_{n,m}(\mathbb{C}) with n≤mn\le m. It is given that rank(AB)=n\text{rank}(AB)=n and (AB)2=AB(AB)^2=AB.
a)Prove that (BA)3=(BA)2(BA)^3=(BA)^2. b)Find BABA.
linear algebramatrixalgebrapolynomiallinear algebra unsolved
Sufficient condition for 4 complex numbers to represent a rectangle

Source: Romanian District Olympiad 2011, Grade X, Problem 2

10/8/2018
a) Show that if four distinct complex numbers have the same absolute value and their sum vanishes, then they represent a rectangle.
b) Let x,y,z,t x,y,z,t be four real numbers, and k k be an integer. Prove the following implication: ∑j∈{x,y,z,t}sin⁡j=0=∑j∈{x,y,z,t}cos⁡j  ⟹  ∑j∈{x,y,z,t}sin⁡(1+2n)j. \sum_{j\in\{ x,y,z,t\}} \sin j = 0 = \sum_{j\in\{ x,y,z,t\}} \cos j\implies \sum_{j\in\{ x,y,z,t\}} \sin (1+2n)j.
complex numbersabsolute valuegeometryrectangletrigonometry
There are no proper morphisms from a given group to C_7

Source: Romanian District Olympiad 2011, Grade XII, Problem 2

10/8/2018
Let G G be the set of matrices of the form (ab01), \begin{pmatrix} a&b\\0&1 \end{pmatrix} , with a,b∈Z7,a≠0. a,b\in\mathbb{Z}_7,a\neq 0.
a) Verify that G G is a group. b) Show that Hom((G,⋅);(Z7,+))={0} \text{Hom}\left( (G,\cdot) ; \left( \mathbb{Z}_7,+ \right) \right) =\{ 0\}
superior algebramorphismsgroup theory