MathDB

4

Part of 2014 Romania National Olympiad

Problems(6)

perpendicular wanted, right isosceles, square (2014 Romanian NMO grade VII P4)

Source:

5/28/2020
Outside the square ABCDABCD is constructed the right isosceles triangle ABDABD with hypotenuse [AB][AB]. Let NN be the midpoint of the side [AD][AD] and M=CE∩AB{M} = CE \cap AB, P=CN∩AB{P} = CN \cap AB , F=PE∩MN{F} = PE \cap MN. On the line FPFP the point QQ is considered such that the [CE[CE is the bisector of the angle QCBQCB. Prove that MQ⊥CFMQ \perp CF.
geometrysquareperpendicularisoscelesright triangle
three discs of radius 1 cannot cover entirely a square surface of side 2

Source: 2014 Romania NMO VIII p4

8/15/2024
Prove that three discs of radius 11 cannot cover entirely a square surface of side 22, but they can cover more than 99.75%99.75\% of it.
geometrycombinatorial geometry
Vectorial geometry relating concyclic quadrilateral

Source: Romanian National Olympiad 2014, Grade IX, Problem 4

3/2/2019
Let ABCD ABCD be a quadrilateral inscribed in a circle of diameter AC. AC. Fix points E,F E,F of segments CD, CD, respectively, BC BC such that AE AE is perpendicular to DF DF and AF AF is perpendicular to BE. BE. Show that AB=AD. AB=AD.
geometry
Romania National Olympiad 2014

Source:

3/22/2018
Let n∈N,n≥2n \in \mathbb{N} , n \ge 2 and a0,a1,a2,⋯ ,an∈C;an≠0 a_0,a_1,a_2,\cdots,a_n \in \mathbb{C} ; a_n \not = 0 . Then:
P. ∣anzn+an−1zzn−1+⋯+a1z+a0∣≤∣an+a0∣|a_nz^n + a_{n-1}z^z{n-1} + \cdots + a_1z + a_0 | \le |a_n+a_0| for any z∈C,∣z∣=1z \in \mathbb{C}, |z|=1 Q. a1=a2=⋯=an−1=0a_1=a_2=\cdots=a_{n-1}=0 and a0/an∈[0,∞)a_0/a_n \in [0,\infty)
Prove that P⟺Q P \Longleftrightarrow Q
complex numbersinequalities
Characterization of some singular matrices of the form I+A²

Source: Romania National Olympiad 2014, Grade XI, Problem 4

3/3/2019
Let A∈M4(R) A\in\mathcal{M}_4\left(\mathbb{R}\right) be an invertible matrix whose trace is equal to the trace of its adjugate, which is nonzero. Show that A2+I A^2+I is singular if and only if there exists a nonzero matrix in M4(R) \mathcal{M}_4\left( \mathbb{R} \right) that anti-commutes with it.
linear algebramatrix
Subgroup of all elements of prime order and its ordinal, under some conditions

Source: Romania National Olympiad 2014, Grade XII, Problem 4

3/3/2019
Let be a finite group G G that has an element a≠1 a\neq 1 for which exists a prime number p p such that x1+p=a−1xa, x^{1+p}=a^{-1}xa, for all x∈G. x\in G.
a) Prove that the order of G G is a power of p. p. b) Show that H:={x∈G∣ord(x)=p}≤G H:=\{x\in G|\text{ord} (x)=p\}\le G and ord2(H)>ord(G). \text{ord}^2(H)>\text{ord}(G).
number theoryprime numbersgroup theoryabstract algebra