2008 Polish MO Finals

Part of Polish MO Finals

Subcontests

(6)

1

Polish MO finals, problem 6

S

be a set of all positive integers which can be represented as a^2 \plus{} 5b^2 for some integers

a,b

a\bot b

p

be a prime number such that p \equal{} 4n \plus{} 3 for some integer

n

. Show that if for some positive integer

k

kp

S

2p

S

as well. Here, the notation

a\bot b

means that the integers

a

b

1

Polish MO finals, problem 1

In each cell of a matrix

n\times n

a number from a set

\{1,2,\ldots,n^2\}

is written — in the first row numbers

1,2,\ldots,n

, in the second n\plus{}1,n\plus{}2,\ldots,2n and so on. Exactly

n

of them have been chosen, no two from the same row or the same column. Let us denote by

a_i

a number chosen from row number

i

. Show that: \frac{1^2}{a_1}\plus{}\frac{2^2}{a_2}\plus{}\ldots \plus{}\frac{n^2}{a_n}\geq \frac{n\plus{}2}{2}\minus{}\frac{1}{n^2\plus{}1}

1

Polish MO finals, problem 5

R

be a parallelopiped. Let us assume that areas of all intersections of

R

with planes containing centers of three edges of

R

pairwisely not parallel and having no common points, are equal. Show that

R

1

Polish MO finals, problem 3

In a convex pentagon

ABCDE

in which BC\equal{}DE following equalities hold: \angle ABE \equal{}\angle CAB \equal{}\angle AED\minus{}90^{\circ},\qquad \angle ACB\equal{}\angle ADE Show that

BCDE

is a parallelogram.

1

Polish MO finals, problem 4

Each point of a plane with both coordinates being integers has been colored black or white. Show that there exists an infinite subset of colored points, whose points are in the same color, having a center of symmetry. [EDIT: added condition about being infinite - now it makes sense]

1

Polish MO finals, problem 2

f: R^3\rightarrow R

a,b,c,d,e

satisfies a condition: f(a,b,c)\plus{}f(b,c,d)\plus{}f(c,d,e)\plus{}f(d,e,a)\plus{}f(e,a,b)\equal{}a\plus{}b\plus{}c\plus{}d\plus{}e Show that for all reals

x_1,x_2,\ldots,x_n

n\geq 5

) equality holds: f(x_1,x_2,x_3)\plus{}f(x_2,x_3,x_4)\plus{}\ldots \plus{}f(x_{n\minus{}1},x_n,x_1)\plus{}f(x_n,x_1,x_2)\equal{}x_1\plus{}x_2\plus{}\ldots\plus{}x_n