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Problems(6)

<AQC =< PQB, <DRQ = 45^o, right isosceles (2004 Romania District VII P4)

Source:

5/24/2020
Consider the isosceles right triangle ABCABC (AB=ACAB = AC) and the points M,P[AB]M, P \in [AB] so that AM=BPAM = BP. Let DD be the midpoint of the side BCBC and R,QR, Q the intersections of the perpendicular from AA onCM CM with CMCM and BCBC respectively. Prove that
a) AQC=PQB\angle AQC = \angle PQB b) DRQ=45o\angle DRQ = 45^o
geometryanglesequal anglesisoscelesright triangle
cosine of angles between planes , distance of point to plane, right trapezoid

Source: 2004 Romania District VIII P4

5/24/2020
In the right trapezoid ABCDABCD with ABCD,B=90oAB \parallel CD, \angle B = 90^o and AB=2DCAB = 2DC. At points AA and DD there is therefore a part of the plane (ABC)(ABC) perpendicular to the plane of the trapezoid, on which the points NN and PP are taken, (APAP and PDPD are perpendicular to the plane) such that DN=aDN = a and AP=a2AP = \frac{a}{2} . Knowing that MM is the midpoint of the side BCBC and the triangle MNPMNP is equilateral, determine:
a) the cosine of the angle between the planes MNPMNP and ABCABC. b) the distance from DD to the plane MNPMNP
trigonometry3D geometrygeometrypalnesanglestrapezoidright angle
14 specific points on a ’chocolate’ tablet

Source: Romanian District Olympiad, Grade IX, Problem 4

10/7/2018
Divide a 2×4 2\times 4 rectangle into 8 8 unit squares to obtain a set of 15 15 vertices denoted by M. \mathcal{M} . Find the points AM A\in\mathcal{M} that have the property that the set M{A} \mathcal{M}\setminus \{ A\} can form 7 7 pairs (A1,B1),(A2,B2),,(A7,B7)M×M \left( A_1,B_1\right) ,\left( A_2,B_2\right) ,\ldots ,\left( A_7,B_7\right)\in\mathcal{M}\times\mathcal{M} such that A1B1+A2B2++A7B7=O. \overrightarrow{A_1B_1} +\overrightarrow{A_2B_2} +\cdots +\overrightarrow{A_7B_7} =\overrightarrow{O} .
geometryrectanglevectorial geometryanalytic geometry
complex numbers, algebra

Source:

1/10/2020
If x,y(0,π2)x,y \in (0, \frac{\pi}{2}) such as (cosx+isiny)n=cos(nx)+isin(ny) (cosx+isiny)^n=cos(nx)+isin(ny) for two consecutive positive integers, then the relation is true for all positive integers.
complex numbersalgebra
Romania District Olympiad 2004 - Grade XI

Source:

4/10/2011
Let A=(aij)Mp(C)A=(a_{ij})\in \mathcal{M}_p(\mathbb{C}) such that a12=a23==ap1,p=1a_{12}=a_{23}=\ldots=a_{p-1,p}=1 and aij=0a_{ij}=0 for any other entry.
a)Prove that Ap1OpA^{p-1}\neq O_p and Ap=OpA^p=O_p.
b)If XMp(C)X\in \mathcal{M}_{p}(\mathbb{C}) and AX=XAAX=XA, prove that there exist a1,a2,,apCa_1,a_2,\ldots,a_p\in \mathbb{C} such that:
X=(a1a2a3ap0a1a2ap100a1ap2000a1)X=\left( \begin{array}{ccccc} a_1 & a_2 & a_3 & \ldots & a_p \\ 0 & a_1 & a_2 & \ldots & a_{p-1} \\ 0 & 0 & a_1 & \ldots & a_{p-2} \\ \ldots & \ldots & \ldots & \ldots & \ldots \\ 0 & 0 & 0 & \ldots & a_1 \end{array} \right)
c)If there exist B,CMp(C)B,C\in \mathcal{M}_p(\mathbb{C}) such that (Ip+A)n=Bn+Cn, ()nN(I_p+A)^n=B^n+C^n,\ (\forall)n\in \mathbb{N}^*, prove that B=OpB=O_p or C=OpC=O_p.
linear algebramatrixcalculusderivativealgebrapolynomiallinear algebra unsolved
integral of f is equal to two integrals of f

Source: Romanian District Olympiad 2004, Grade XII, Problem 4

10/7/2018
Let a,b(0,1) a,b\in (0,1) and a continuous function f:[0,1]R f:[0,1]\longrightarrow\mathbb{R} with the property that \int_0^x f(t)dt=\int_0^{ax} f(t)dt +\int_0^{bx} f(t)dt, \forall x\in [0,1] .
a) Show that if a+b<1, a+b<1, then f=0. f=0. b) Show that if a+b=1, a+b=1, then f f is constant.
Integralreal analysiscalculusintegration