MathDB

3

Part of 2014 Belarus Team Selection Test

Problems(6)

min no of cells covered by a 3-1 figure in a nxn table

Source: 2014 Belarus TST 1.3

12/30/2020
NN cells are marked on an n×nn\times n table so that at least one marked cel is among any four cells of the table which form the figure https://cdn.artofproblemsolving.com/attachments/2/2/090c32eb52df31eb81b9a86c63610e4d6531eb.png (tbe figure may be rotated). Find the smallest possible value of NN.
(E. Barabanov)
combinatorics
incircle trisects a segment

Source: 2014 Belarus TST 2.3

12/29/2020
Point LL is marked on the side ABAB of a triangle ABCABC. The incircle of the triangle ABCABC meets the segment CLCL at points PP and QQ .Is it possible that the equalities CP=PQ=QLCP = PQ = QL hold if CLCL is a) the median? b) the bisector? c) the altitude? d) the segment joining vertex CC with the point LL of tangency of the excircle of the triangie ABCABC with ABAB ?
(I. Gorodnin)
geometrytrisectequal segmentsincircle
f(x + f(y)) = y^2 + g(x)

Source: 2014 Belarus TST 4.3

12/29/2020
Do there exist functions ff and gg, f:R→Rf : R \to R, g:R→Rg : R \to R such that f(x+f(y))=y2+g(x)f(x + f(y)) = y^2 + g(x) for all real xx and yy ?
(I. Gorodnin)
algebrafunctional equationTST
paint degment by n points using 4 colours

Source: 2014 Belarus TST 3.3

12/30/2020
nn points are marked on a plane. Each pair of these points is connected with a segment. Each segment is painted one of four different colors. Find the largest possible value of nn such that one can paint the segments so that for any four points there are four segments (connecting these four points) of four different colors.
(E. Barabanov)
combinatorial geometrycombinatoricsColoring
sum a^2/(a^2-a+1) <=3, if a + b + c = ab+bc+ca and 0<a,b,c<2

Source: 2014 Belarus TST 6.3

12/29/2020
Given a,b,ca,b,c ,(a,b,c∈(0,2)(a, b,c \in (0,2)), with a+b+c=ab+bc+caa + b + c = ab+bc+ca, prove that a2a2−a+1+b2b2−b+1+c2c2−c+1≤3\frac{a^2}{a^2-a+1}+\frac{b^2}{b^2-b+1}+\frac{c^2}{c^2-c+1} \le 3 (D. Pirshtuk)
algebrainequalities
China Mathematical Olympiad 1992 problem5

Source: China Mathematical Olympiad 1992 problem5

9/30/2013
Find the maximum possible number of edges of a simple graph with 88 vertices and without any quadrilateral. (a simple graph is an undirected graph that has no loops (edges connected at both ends to the same vertex) and no more than one edge between any two different vertices.)
floor functioncombinatorics unsolvedcombinatorics