MathDB

2

Part of 2019 European Mathematical Cup

Problems(2)

Inequality on recurrence

Source: 8th European Mathematical Cup, Junior Category, Q2

12/26/2019
Let (xn)n∈N(x_n)_{n\in \mathbb{N}} be a sequence defined recursively such that x1=2x_1=\sqrt{2} and xn+1=xn+1xn for n∈N.x_{n+1}=x_n+\frac{1}{x_n}\text{ for }n\in \mathbb{N}. Prove that the following inequality holds: x122x1x2−1+x222x2x3−1+…+x201822x2018x2019−1+x201922x2019x2020−1>20192x20192+1x20192.\frac{x_1^2}{2x_1x_2-1}+\frac{x_2^2}{2x_2x_3-1}+\dotsc +\frac{x_{2018}^2}{2x_{2018}x_{2019}-1}+\frac{x_{2019}^2}{2x_{2019}x_{2020}-1}>\frac{2019^2}{x_{2019}^2+\frac{1}{x_{2019}^2}}.
Proposed by Ivan Novak
inequalitiesalgebra
Beauty of a convex board

Source: 8th European Mathematical Cup, Senior Category, Q2

12/26/2019
Let nn be a positive integer. An n×nn\times n board consisting of n2n^2 cells, each being a unit square colored either black or white, is called convex if for every black colored cell, both the cell directly to the left of it and the cell directly above it are also colored black. We define the beauty of a board as the number of pairs of its cells (u,v)(u,v) such that uu is black, vv is white, and uu and vv are in the same row or column. Determine the maximum possible beauty of a convex n×nn\times n board.
Proposed by Ivan Novak
combinatoricsExtremal combinatorics