2005 Tuymaada Olympiad

Part of Tuymaada Olympiad

Subcontests

(8)

1

sum of squares = 1 kind of ineq

a,b,c

be positive reals s.t.

a^2+b^2+c^2=1

. Prove the following inequality

\sum \frac{a}{a^3+bc} >3 .

Proposed by A. Khrabrov

1

Nagel cevians are sides of triangle

ABC

A_{1}

B_{1}

C_{1}

be the points where the excircles touch the sides

BC

CA

AB

respectively. Prove that

A A_{1}

B B_{1}

C C_{1}

are the sidelenghts of a triangle.
Proposed by L. Emelyanov

1

fractions, integers

Given are a positive integer

n

and an infinite sequence of proper fractions

x_0 = \frac{a_0}{n}

\ldots

x_i=\frac{a_i}{n+i}

a_i < n+i

. Prove that there exist a positive integer

k

c_1

\ldots

c_k

c_1 x_1 + \ldots + c_k x_k = 1.

Proposed by M. Dubashinsky

1

A rather unfair team leader

Six members of the team of Fatalia for the International Mathematical Olympiad are selected from

13

candidates. At the TST the candidates got

a_1,a_2, \ldots, a_{13}

a_i \neq a_j

i \neq j

.
The team leader has already

6

candidates and now wants to see them and nobody other in the team. With that end in view he constructs a polynomial

P(x)

and finds the creative potential of each candidate by the formula

c_i = P(a_i)

.
For what minimum

n

can he always find a polynomial

P(x)

of degree not exceeding

n

such that the creative potential of all

6

candidates is strictly more than that of the

7

others?
Proposed by F. Petrov, K. Sukhov

1

Rooks

2

11

squares in the middle, in the right and in the left you have columns of

9

squares (centered on the ones of

11

squares), then columns of

7,5,3,1

squares. (This is the way it was explained in the original thread, http://www.artofproblemsolving.com/Forum/viewtopic.php?t=44430 ; anyway, i think you can understand how it looks)
Several rooks stand on the table and beat all the squares ( a rook beats the square it stands in, too). Prove that one can remove several rooks such that not more than

11

rooks are left and still beat all the table.
Proposed by D. Rostovsky, based on folklore

1

121 numbers

The positive integers

1,2,...,121

are arranged in the squares of a

11 \times 11

table. Dima found the product of numbers in each row and Sasha found the product of the numbers in each column. Could they get the same set of

11

numbers?
Proposed by S. Berlov

1

Russian geometry

I

be the incentre of triangle

ABC

. A circle containing the points

B

C

meets the segments

BI

CI

P

Q

respectively. It is known that

BP\cdot CQ=PI\cdot QI

. Prove that the circumcircle of the triangle

PQI

is tangent to the circumcircle of

ABC

.
Proposed by S. Berlov

1

Round tables on graph theory

The organizers of a mathematical congress found that if they accomodate any participant in a room the rest can be accomodated in double rooms so that 2 persons living in each room know each other. Prove that every participant can organize a round table on graph theory for himself and an even number of other people so that each participant of the round table knows both his neigbours.
Proposed by S. Berlov, S. Ivanov