MathDB

3

Part of 2012 Kazakhstan National Olympiad

Problems(6)

Inequality

Source:

5/9/2012
Let a,b,c,d>0 a,b,c,d>0 for which the following conditions:: a)a) (a−c)(b−d)=−4(a-c)(b-d)=-4 b)b) a+c2≥a2+b2+c2+d2a+b+c+d\frac{a+c}{2}\geq\frac{a^{2}+b^{2}+c^{2}+d^{2}}{a+b+c+d} Find the minimum of expression a+ca+c
inequalitiesinequalities unsolvedKazakhstanKanat Satylkhanov
Squares of a majority color in both row and column

Source: Kazakhstan 2012 ( Grade 9 problem 6 )

5/20/2012
The cell of a (2m+1)×(2n+1)(2m +1) \times (2n +1) board are painted in two colors - white and black. The unit cell of a row (column) is called dominant on the row (the column) if more than half of the cells that row (column) have the same color as this cell. Prove that at least m+n−1m + n-1 cells on the board are dominant in both their row and column.
geometryrectanglecombinatorics unsolvedcombinatorics
Balls into boxes with restricted choice for each ball

Source: Kazakhstan 2012 ( Grade 10 problem 3)

5/20/2012
There are nn balls numbered from 11 to nn, and 2n−12n-1 boxes numbered from 11 to 2n−12n-1. For each ii, ball number ii can only be put in the boxes with numbers from 11 to 2i−12i-1. Let kk be an integer from 11 to nn. In how many ways we can choose kk balls, kk boxes and put these balls in the selected boxes so that each box has exactly one ball?
functioninductioncombinatorics unsolvedcombinatorics
Kazakhstan 2012 problems 6(grade 10)

Source:

5/9/2012
The sequence ana_{n} defined as follows: a1=4,a2=17a_{1}=4, a_{2}=17 and for any k≥1k\geq1 true equalities a2k+1=a2+a4+...+a2k+(k+1)(22k+3−1)a_{2k+1}=a_{2}+a_{4}+...+a_{2k}+(k+1)(2^{2k+3}-1) a2k+2=(22k+2+1)a1+(22k+3+1)a3+...+(23k+1+1)a2k−1+ka_{2k+2}=(2^{2k+2}+1)a_{1}+(2^{2k+3}+1)a_{3}+...+(2^{3k+1}+1)a_{2k-1}+k Find the smallest mm such that (a1+...am)20122012−1(a_{1}+...a_{m})^{2012^{2012}}-1 divided 2201220122^{2012^{2012}}
inductionnumber theory unsolvednumber theory
Kazakhstan 2012 ( Grade 11 problem 3)

Source:

5/20/2012
Line PQPQ is tangent to the incircle of triangle ABCABC in such a way that the points PP and QQ lie on the sides ABAB and ACAC, respectively. On the sides ABAB and ACAC are selected points MM and NN, respectively, so that AM=BPAM = BP and AN=CQAN = CQ. Prove that all lines constructed in this manner MNMN pass through one point
geometryincentergeometric transformationreflectiongeometry unsolved
Kazakhstan 2012 ( Grade 11 problem 6)

Source:

5/20/2012
Consider the equation ax2+by2=1ax^{2}+by^{2}=1, where a,ba,b are fixed rational numbers. Prove that either such an equation has no solutions in rational numbers, or it has infinitely many solutions.
number theory unsolvednumber theory