MathDB

3

Part of 2006 Romania National Olympiad

Problems(6)

Acute-angled triangle with C=45

Source: Romanian NMO 2006, Grade 7, Problem 3

4/18/2006
In the acute-angle triangle ABCABC we have ∠ACB=45∘\angle ACB = 45^\circ. The points A1A_1 and B1B_1 are the feet of the altitudes from AA and BB, and HH is the orthocenter of the triangle. We consider the points DD and EE on the segments AA1AA_1 and BCBC such that A1D=A1E=A1B1A_1D = A_1E = A_1B_1. Prove that a) A1B1=A1B2+A1C22A_1B_1 = \sqrt{ \frac{A_1B^2+A_1C^2}{2} }; b) CH=DECH=DE.
geometrycircumcircle
Yet another solid geometry problem involving a cube

Source: Romanian NMO 2006, Grade 8, Problem 3

4/18/2006
Let ABCDA1B1C1D1ABCDA_1B_1C_1D_1 be a cube and PP a variable point on the side [AB][AB]. The perpendicular plane on ABAB which passes through PP intersects the line AC′AC' in QQ. Let MM and NN be the midpoints of the segments A′PA'P and BQBQ respectively. a) Prove that the lines MNMN and BC′BC' are perpendicular if and only if PP is the midpoint of ABAB. b) Find the minimal value of the angle between the lines MNMN and BC′BC'.
geometry3D geometry
The quadrilateral and beast

Source: RMO 2006

4/17/2006
We have a quadrilateral ABCDABCD inscribed in a circle of radius rr, for which there is a point PP on CDCD such that CB=BP=PA=ABCB=BP=PA=AB. (a) Prove that there are points A,B,C,D,PA,B,C,D,P which fulfill the above conditions. (b) Prove that PD=rPD=r. Virgil Nicula
geometrycircumcirclegeometry proposed
Lots of odd and even numbers

Source: RMO 2006, 10th grade

4/17/2006
Prove that among the elements of the sequence (⌊n2⌋+⌊n3⌋)n≥0\left( \left\lfloor n \sqrt 2 \right\rfloor + \left\lfloor n \sqrt 3 \right\rfloor \right)_{n \geq 0} are an infinity of even numbers and an infinity of odd numbers.
floor functionalgebra proposedalgebra
Lots of points

Source: RMO 2006

4/18/2006
We have in the plane the system of points A1,A2,…,AnA_1,A_2,\ldots,A_n and B1,B2,…,BnB_1,B_2,\ldots,B_n, which have different centers of mass. Prove that there is a point PP such that PA1+PA2+…+PAn=PB1+PB2+…+PBn. PA_1 + PA_2 + \ldots+ PA_n = PB_1 + PB_2 + \ldots + PB_n .
inequalitieslimitcomplex numbersreal analysisreal analysis unsolved
The groups are coming...

Source: RMO 2006

4/18/2006
Let G\displaystyle G be a finite group of n\displaystyle n elements (n≥2)\displaystyle ( n \geq 2 ) and p\displaystyle p be the smallest prime factor of n\displaystyle n. If G\displaystyle G has only a subgroup H\displaystyle H with p\displaystyle p elements, then prove that H\displaystyle H is in the center of G\displaystyle G. Note. The center of G\displaystyle G is the set Z(G)={a∈G∣ax=xa, ∀x∈G}\displaystyle Z(G) = \left\{ a \in G \left| ax=xa, \, \forall x \in G \right. \right\}.
group theoryabstract algebrasuperior algebrasuperior algebra unsolved