1999 Tuymaada Olympiad

Part of Tuymaada Olympiad

Subcontests

(4)

2

Tuymaada 1999, Q4

Prove the inequality

{x\over y^2-z}+{y\over z^2-x}+{z\over x^2-y} > 1,

2 < x, y, z < 4.

Proposed by A. Golovanov

Tuymaada 1999, Q8

A right parallelepiped (i.e. a parallelepiped one of whose edges is perpendicular to a face) is given. Its vertices have integral coordinates, and no other points with integral coordinates lie on its faces or edges. Prove that the volume of this parallelepiped is a sum of three perfect squares.
Proposed by A. Golovanov

2

Tuymaada 1999, Q3

What maximum number of elements can be selected from the set

\{1, 2, 3, \dots, 100\}

so that no sum of any three selected numbers is equal to a selected number?
Proposed by A. Golovanov

Tuymaada 1999, Q7

A sequence of integers

a_0,\ a_1,\dots a_n \dots

is defined by the following rules:

a_0=0,\ a_1=1,\ a_{n+1} > a_n

n\in \mathbb{N}

a_{n+1}

is the minimum number such that no three numbers among

a_0,\ a_1,\dots a_{n+1}

form an arithmetical progression. Prove that

a_{2^n}=3^n

n \in \mathbb{N}.

2

Tuymaada 1999, Q2

Find all polynomials

P(x)

P(x^3+1)=P(x^3)+P(x^2).

Proposed by A. Golovanov

Tuymaada 1999, Q6

Can the graphs of a polynomial of degree 20 and the function

\displaystyle y={1\over x^{40}}

have exactly 30 points of intersection?
Proposed by K. Kokhas

2

Tuymaada Olympiad 2004, Question 5

50 knights of King Arthur sat at the Round Table. A glass of white or red wine stood before each of them. It is known that at least one glass of red wine and at least one glass of white wine stood on the table. The king clapped his hands twice. After the first clap every knight with a glass of red wine before him took a glass from his left neighbour. After the second clap every knight with a glass of white wine (and possibly something more) before him gave this glass to the left neughbour of his left neighbour. Prove that some knight was left without wine.
Proposed by A. Khrabrov, incorrect translation from Hungarian

Tuymaada 1999, Q5

In the triangle

ABC

\angle ABC=100^\circ

\angle ACB=65^\circ

M\in AB

N\in AC

\angle MCB=55^\circ

\angle NBC=80^\circ

\angle NMC

.
St.Petersburg folklore