MathDB

Problems(6)

Triangle sidelengths

Source: RMO 2004, Grade 7, Problem 2

2/26/2006
The sidelengths of a triangle are a,b,ca,b,c. (a) Prove that there is a triangle which has the sidelengths a,b,c\sqrt a,\sqrt b,\sqrt c. (b) Prove that ab+bc+caa+b+c<2ab+2bc+2ca\displaystyle \sqrt{ab}+\sqrt{bc}+\sqrt{ca} \leq a+b+c < 2 \sqrt{ab} + 2 \sqrt{bc} + 2 \sqrt{ca}.
inequalities
Diophantic

Source: RMO 2004, Grade 8, Problem 2

2/26/2006
Prove that the equation x2+y2+z2+t2=22004x^2+y^2+z^2+t^2=2^{2004}, where 0xyzt0 \leq x \leq y \leq z \leq t, has exactly 22 solutions in Z\mathbb Z. Mihai Baluna
modular arithmeticinequalities
Second degree f

Source: RMO 2004, 9th grade, problem 2

3/6/2005
Let P(n)P(n) be the number of functions f:RRf: \mathbb{R} \to \mathbb{R}, f(x)=ax2+bx+cf(x)=a x^2 + b x + c, with a,b,c{1,2,,n}a,b,c \in \{1,2,\ldots,n\} and that have the property that f(x)=0f(x)=0 has only integer solutions. Prove that n<P(n)<n2n<P(n)<n^2, for all n4n \geq 4. Laurentiu Panaitopol
functionalgebrapolynomiallogarithmscalculusintegrationinequalities
The inequality of the ranks

Source: Romanian MO 2004, Final Round, 11th Grade, Problem 2

2/28/2006
Let nNn \in \mathbb N, n2n \geq 2. (a) Give an example of two matrices A,BMn(C)A,B \in \mathcal M_n \left( \mathbb C \right) such that rank(AB)rank(BA)=n2. \textrm{rank} \left( AB \right) - \textrm{rank} \left( BA \right) = \left\lfloor \frac{n}{2} \right\rfloor . (b) Prove that for all matrices X,YMn(C)X,Y \in \mathcal M_n \left( \mathbb C \right) we have rank(XY)rank(YX)n2. \textrm{rank} \left( XY \right) - \textrm{rank} \left( YX \right) \leq \left\lfloor \frac{n}{2} \right\rfloor . Ion Savu
inequalitiesfloor functionlinear algebralinear algebra unsolved
Regular tetrahedron

Source: Romanian MO 2004, 10th grade, Problem 2

3/6/2005
Let ABCDABCD be a tetrahedron in which the opposite sides are equal and form equal angles. Prove that it is regular.
geometry3D geometrytetrahedrontrigonometrygeometry solved
Good old zn

Source: RMO 2004, Grade 12, Problem 2

2/26/2006
Let fZ[X]f \in \mathbb Z[X]. For an nNn \in \mathbb N, n2n \geq 2, we define fn:Z/nZZ/nZf_n : \mathbb Z / n \mathbb Z \to \mathbb Z / n \mathbb Z through fn(x^)=f(x)^f_n \left( \widehat x \right) = \widehat{f \left( x \right)}, for all xZx \in \mathbb Z. (a) Prove that fnf_n is well defined. (b) Find all polynomials fZ[X]f \in \mathbb Z[X] such that for all nNn \in \mathbb N, n2n \geq 2, the function fnf_n is surjective. Bogdan Enescu
algebrapolynomialfunctionsuperior algebrasuperior algebra solved