MathDB

3

Part of 2006 Romania Team Selection Test

Problems(4)

Describe the numbers

Source: Romanian IMO TST 2006, day 2, problem 3

4/22/2006
For which pairs of positive integers (m,n)(m,n) there exists a set AA such that for all positive integers x,yx,y, if ∣x−y∣=m|x-y|=m, then at least one of the numbers x,yx,y belongs to the set AA, and if ∣x−y∣=n|x-y|=n, then at least one of the numbers x,yx,y does not belong to the set AA?
Adapted by Dan Schwarz from A.M.M.
modular arithmeticinductionnumber theory proposednumber theory
Circles and tangents

Source: Romanian IMO TST 2006, day 3, problem 3

5/16/2006
Let γ\gamma be the incircle in the triangle A0A1A2A_0A_1A_2. For all i∈{0,1,2}i\in\{0,1,2\} we make the following constructions (all indices are considered modulo 3): γi\gamma_i is the circle tangent to γ\gamma which passes through the points Ai+1A_{i+1} and Ai+2A_{i+2}; TiT_i is the point of tangency between γi\gamma_i and γ\gamma; finally, the common tangent in TiT_i of γi\gamma_i and γ\gamma intersects the line Ai+1Ai+2A_{i+1}A_{i+2} in the point PiP_i. Prove that a) the points P0P_0, P1P_1 and P2P_2 are collinear; b) the lines A0T0A_0T_0, A1T1A_1T_1 and A2T2A_2T_2 are concurrent.
geometryincentercircumcirclepower of a pointradical axisgeometry proposed
Rare sets

Source: Romanian IMO TST 2006, day 4, problem 3

5/19/2006
Let n>1n>1 be an integer. A set S⊂{0,1,2,…,4n−1}S \subset \{ 0,1,2, \ldots, 4n-1\} is called rare if, for any k∈{0,1,…,n−1}k\in\{0,1,\ldots,n-1\}, the following two conditions take place at the same time (1) the set S∩{4k−2,4k−1,4k,4k+1,4k+2}S\cap \{4k-2,4k-1,4k, 4k+1, 4k+2 \} has at most two elements; (2) the set S∩{4k+1,4k+2,4k+3}S\cap \{4k+1,4k+2,4k+3\} has at most one element. Prove that the set {0,1,2,…,4n−1}\{0,1,2,\ldots,4n-1\} has exactly 8⋅7n−18 \cdot 7^{n-1} rare subsets.
algebrapolynomiallinear algebracombinatoricsSet systems
Approximation of a sequence of real numbers

Source: Romanian IMO TST 2006, day 5, problem 3

5/23/2006
Let x1=1x_1=1, x2x_2, x3x_3, …\ldots be a sequence of real numbers such that for all n≥1n\geq 1 we have xn+1=xn+12xn. x_{n+1} = x_n + \frac 1{2x_n} . Prove that ⌊25x625⌋=625. \lfloor 25 x_{625} \rfloor = 625 .
floor functionlogarithmsfunctionalgebra proposedalgebra