MathDB

3

Part of 2009 Ukraine National Mathematical Olympiad

Problems(6)

Girls, boys, and candies - [UKRMO 2009 Grade 8]

Source:

1/20/2011
On the party every boy gave 11 candy to every girl and every girl gave 11 candy to every boy. Then every boy ate 22 candies and every girl ate 33 candies. It is known that 14\frac 14 of all candies was eaten. Find the greatest possible number of children on the party.
2009 × 4018 rectangular board - [UKRMO 2009 Grade 8]

Source:

1/21/2011
Given 2009×40182009 \times 4018 rectangular board. Frame is a rectangle n×nn \times n or n×(n+2)n \times(n + 2) for (n≥3) ( n \geq 3 ) without all cells which don’t have any common points with boundary of rectangle. Rectangles 1×1,1×2,1×31\times1,1\times 2,1\times 3 and 2×4 2\times 4 are also frames. Two players by turn paint all cells of some frame that has no painted cells yet. Player that can't make such move loses. Who has a winning strategy?
geometryrectangle
Prove that ∠PNA = ∠AMB - [UKRMO 2009 Grade 9]

Source:

1/23/2011
In triangle ABCABC points M,NM, N are midpoints of BC,CABC, CA respectively. Point PP is inside ABCABC such that ∠BAP=∠PCA=∠MAC.\angle BAP = \angle PCA = \angle MAC . Prove that ∠PNA=∠AMB.\angle PNA = \angle AMB .
geometrycircumcircleparallelogramcyclic quadrilateralgeometric transformationgeometry proposed
Who has a winning strategy? - [UKRMO 2009 Grade 10]

Source:

1/23/2011
Given a n×nn \times n square board. Two players by turn remove some side of unit square if this side is not a bound of n×nn \times n square board. The player lose if after his move n×nn \times n square board became broken into two parts. Who has a winning strategy?
combinatorics unsolvedcombinatorics
Geometric Inequality - [UKRMO 2009 Grade 11]

Source:

1/23/2011
Point OO is inside triangle ABCABC such that ∠AOB=∠BOC=∠COA=120∘.\angle AOB = \angle BOC = \angle COA = 120^\circ . Prove that AO2BC+BO2CA+CO2AB≥AO+BO+CO3.\frac{AO^2}{BC}+\frac{BO^2}{CA}+\frac{CO^2}{AB} \geq \frac{AO+BO+CO}{\sqrt 3}.
inequalitiestrigonometryfunctionalgebradomaingeometry unsolvedgeometry
ABC is isosceles triangle - [UKRMO 2009 Grade 11]

Source:

1/23/2011
In triangle ABCABC let MM and NN be midpoints of BCBC and AC,AC, respectively. Point PP is inside ABCABC such that ∠BAP=∠PBC=∠PCA.\angle BAP = \angle PBC = \angle PCA . Prove that if ∠PNA=∠AMB,\angle PNA = \angle AMB, then ABCABC is isosceles triangle.
geometry proposedgeometry