MathDB

Problems(4)

Area of marked points on an infinite grid

Source: Romanian 2018 TST Day 1 Problem 3

5/25/2020
Divide the plane into 11x11 squares formed by the lattice points. LetSS be the set-theoretic union of a finite number of such cells, and let aa be a positive real number less than or equal to 1/4.Show that S can be covered by a finite number of squares satisfying the following three conditions: 1) Each square in the cover is an array of 11x11 cells 2) The squares in the cover have pairwise disjoint interios and 3)For each square QQ in the cover the ratio of the area SQS \cap Q to the area of Q is at least aa and at most a(a1/2)2a {(\lfloor a^{-1/2} \rfloor)} ^2
combinatoricscombinatorial geometryratio
Points covered by strips

Source: Romanian TST 2018 Problem 3 Day 2

5/25/2020
Consider a 4-point configuration in the plane such that every 3 points can be covered by a strip of a unit width. Prove that: 1) the four points can be covered by a strip of length at most 2\sqrt2 and 2)if no strip of length less that 2\sqrt2 covers all the four points, then the points are vertices of a square of length 2\sqrt2
combinatorial geometrycombinatorics
Cyclic permutation of binary numbers

Source: Romania 2018 TST Problem 3 Day 3

5/25/2020
For every integer n2n \ge 2 let BnB_n denote the set of all binary nn-nuples of zeroes and ones, and split BnB_n into equivalence classes by letting two nn-nuples be equivalent if one is obtained from the another by a cyclic permutation.(for example 110, 011 and 101 are equivalent). Determine the integers n2n \ge 2 for which BnB_n splits into an odd number of equivalence classes.
Binarycombinatoricspermutation
Integral part of a sum

Source: Romanian TST 2018 Problem 3 Day 4

5/25/2020
Given an integer n2n \geq 2 determine the integral part of the number k=1n11(1+1n)(1+kn) \sum_{k=1}^{n-1} \frac {1} {({1+\frac{1} {n}}) \dots ({1+\frac {k} {n})}} - k=1n1(11n)(1kn)\sum_{k=1}^{n-1} (1-\frac {1} {n}) \dots(1-\frac{k}{n})
floor functionalgebrainequalities