MathDB

2

Part of 2002 Romania Team Selection Test

Problems(5)

TST-Romania, 2002

Source: sequence of integers

10/9/2009
The sequence (an) (a_n) is defined by: a_0\equal{}a_1\equal{}1 and a_{n\plus{}1}\equal{}14a_n\minus{}a_{n\minus{}1} for all n≥1 n\ge 1. Prove that 2a_n\minus{}1 is a perfect square for any n≥0 n\ge 0.
algebrapolynomialinductionDiophantine equationnumber theory proposednumber theory
P(x)|Q(x) where both P,Q have coefficients of 1 or 2002

Source: Romanian TST 2002

2/5/2011
Let P(x)P(x) and Q(x)Q(x) be integer polynomials of degree pp and qq respectively. Assume that P(x)P(x) divides Q(x)Q(x) and all their coefficients are either 11 or 20022002. Show that p+1p+1 is a divisor of q+1q+1.
Mihai Cipu
algebrapolynomialnumber theory proposednumber theory
3 disks contain vertices of an equilateral triangle

Source: Romanian TST 2002

2/5/2011
Find the least positive real number rr with the following property:
Whatever four disks are considered, each with centre on the edges of a unit square and the sum of their radii equals rr, there exists an equilateral triangle which has its edges in three of the disks.
Radu Gologan
geometry proposedgeometry
Nice inequality with frac 45

Source: Romanian selection test 2002

8/12/2003
Let n≥4n\geq 4 be an integer, and let a1,a2,…,ana_1,a_2,\ldots,a_n be positive real numbers such that a12+a22+⋯+an2=1. a_1^2+a_2^2+\cdots +a_n^2=1 . Prove that the following inequality takes place a1a22+1+⋯+ana12+1≥45(a1a1+⋯+anan)2. \frac{a_1}{a_2^2+1}+\cdots +\frac{a_n}{a_1^2+1} \geq \frac{4}{5}\left( a_1 \sqrt{a_1}+\cdots +a_n \sqrt{a_n} \right)^2 . Bogdan Enescu, Mircea Becheanu
inequalitiesalgebraromania
AC=MC iff AM bisects angle DAB

Source: Romanian TST 2002

2/5/2011
Let ABCABC be a triangle such that AC≠BC,AB<ACAC\not= BC,AB<AC and let KK be it's circumcircle. The tangent to KK at the point AA intersects the line BCBC at the point DD. Let K1K_1 be the circle tangent to KK and to the segments (AD),(BD)(AD),(BD). We denote by MM the point where K1K_1 touches (BD)(BD). Show that AC=MCAC=MC if and only if AMAM is the bisector of the ∠DAB\angle DAB.
Neculai Roman
geometrygeometry proposed