MathDB

1

Part of 2005 Romania National Olympiad

Problems(6)

ABCD parallelogram, intersections

Source: Romanian Nationals RMO 2005 - grade 7, problem 1

3/31/2005
Let ABCDABCD be a parallelogram. The interior angle bisector of ∠ADC\angle ADC intersects the line BCBC in EE, and the perpendicular bisector of the side ADAD intersects the line DEDE in MM. Let F=AM∩BCF= AM \cap BC. Prove that: a) DE=AFDE=AF; b) AD⋅AB=DE⋅DMAD\cdot AB = DE\cdot DM. Daniela and Marius Lobaza, Timisoara
geometryparallelogramtrigonometryAMCUSA(J)MOUSAMOangle bisector
tetrahedron inscribed in a cube of side-length 1

Source: Romanian Nationals RMO 2005 - grade 8, problem 1

3/31/2005
We consider a cube with sides of length 1. Prove that a tetrahedron with vertices in the set of the vertices of the cube has the volume 16\dfrac 16 if and only if 3 of the vertices of the tetrahedron are vertices on the same face of the cube. Dinu Serbanescu
geometry3D geometrytetrahedronanalytic geometrytrigonometry
convex quadrilateral and centroids

Source: Romanian Nationals RMO 2005 - grade 9, problem 1

3/31/2005
Let ABCDABCD be a convex quadrilateral with AD∦BCAD\not\parallel BC. Define the points E=AD∩BCE=AD \cap BC and I=AC∩BDI = AC\cap BD. Prove that the triangles EDCEDC and IABIAB have the same centroid if and only if AB∥CDAB \parallel CD and IC2=IA⋅ACIC^{2}= IA \cdot AC. Virgil Nicula
ratioquadraticsgeometryanalytic geometryvectorGauss
a little ugly sequences and trig sums

Source: Romanian Nationals RMO 2005 - grade 10, problem 1

3/31/2005
Let nn be a positive integer, n≥2n\geq 2. For each t∈Rt\in \mathbb{R}, t≠kπt\neq k\pi, k∈Zk\in\mathbb{Z}, we consider the numbers xn(t)=∑k=1nk(n−k)cos⁡(tk) and yn(t)=∑k=1nk(n−k)sin⁡(tk). x_n(t) = \sum_{k=1}^n k(n-k)\cos{(tk)} \textrm{ and } y_n(t) = \sum_{k=1}^n k(n-k)\sin{(tk)}. Prove that if xn(t)=yn(t)=0x_n(t) = y_n(t) =0 if and only if tan⁡nt2=ntan⁡t2\tan {\frac {nt}2} = n \tan {\frac t2}. Constantin Buse
trigonometrycalculusderivativealgebra proposedalgebra
harazi's diminished radical matrices

Source: Romanian Nationals RMO 2005 - grade 11, problem 1

3/31/2005
Let n≥2n\geq 2 a fixed integer. We shall call a n×nn\times n matrix AA with rational elements a radical matrix if there exist an infinity of positive integers kk, such that the equation Xk=AX^k=A has solutions in the set of n×nn\times n matrices with rational elements. a) Prove that if AA is a radical matrix then det⁡A∈{−1,0,1}\det A \in \{-1,0,1\} and there exists an infinity of radical matrices with determinant 1; b) Prove that there exist an infinity of matrices that are not radical and have determinant 0, and also an infinity of matrices that are not radical and have determinant 1. After an idea of Harazi
linear algebramatrixlinear algebra unsolved
continous morphism from (C,+) to (C,+)

Source: Romanian Nationals RMO 2005 - grade 12, problem 1

3/31/2005
Prove that the group morphisms f:(C,+)→(C,+)f: (\mathbb{C},+)\to(\mathbb{C},+) for which there exists a positive λ\lambda such that ∣f(z)∣≤λ∣z∣|f(z)| \leq \lambda |z| for all z∈Cz\in\mathbb{C}, have the form f(z)=αz+βz‾ f(z) = \alpha z + \beta \overline{z} for some complex α\alpha, β\beta. Cristinel Mortici
superior algebrasuperior algebra solved