MathDB

2

Part of 2003 All-Russian Olympiad

Problems(5)

Inequality with a+b+c=1 - Fractional Inequality

Source:

11/3/2010
Let a,b,ca, b, c be positive numbers with the sum 11. Prove the inequality 11−a+11−b+11−c≥21+a+21+b+21+c.\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}.
inequalitiesinequalities unsolved
KL is parallel to O1O2

Source:

11/2/2010
Two circles S1S_1 and S2S_2 with centers O1O_1 and O2O_2 respectively intersect at AA and BB. The tangents at AA to S1S_1 and S2S_2 meet segments BO2BO_2 and BO1BO_1 at KK and LL respectively. Show that KL∥O1O2.KL \parallel O_1O_2.
trigonometrygeometryangle bisectortrig identitiesLaw of Sinesgeometry proposed
Russia 2003,hard

Source: help me

11/10/2009
Let a0 a_0 be a natural number. The sequence (an) (a_n) is defined by a_{n\plus{}1}\equal{}\frac{a_n}{5} if an a_n is divisible by 5 5 and a_{n\plus{}1}\equal{}[a_n \sqrt{5}] otherwise . Show that the sequence an a_n is increasing starting from some term.
functionnumber theory unsolvednumber theory
Integers in Infinite Chessboard

Source: All-Russian MO 2003 Grade 11 #6

1/2/2012
Is it possible to write a positive integer in every cell of an infinite chessboard, in such a manner that, for all positive integers m,nm, n, the sum of numbers in every m×nm\times n rectangle is divisible by m+nm + n ?
geometryrectangleanalytic geometrycombinatorics unsolvedcombinatorics
Geometry Problem (15)

Source:

8/7/2010
The diagonals of a cyclic quadrilateral ABCDABCD meet at OO. Let S1,S2S_1, S_2 be the circumcircles of triangles ABOABO and CDOCDO respectively, and O,KO,K their intersection points. The lines through OO parallel to ABAB and CDCD meet S1S_1 and S2S_2 again at LL and MM, respectively. Points PP and QQ on segments OLOL and OMOM respectively are taken such that OP:PL=MQ:QOOP : PL = MQ : QO. Prove that O,K,P,QO,K, P,Q lie on a circle.
geometrycircumcirclecyclic quadrilateralgeometric transformationgeometry proposedSpiral SimilarityRussian