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Difference between revisions of "2009 Zhautykov International Olympiad Problems"

(New page: ==Day 1== ===Problem 1=== Find all pairs of integers <math>(x,y)</math>, such that <center><math>x^2 - 2009y + 2y^2 = 0</math>.</center> ===Problem 2=== Find all real <math>a</math>, such...)
 
 
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Find all pairs of integers <math>(x,y)</math>, such that
 
Find all pairs of integers <math>(x,y)</math>, such that
 
<center><math>x^2 - 2009y + 2y^2 = 0</math>.</center>
 
<center><math>x^2 - 2009y + 2y^2 = 0</math>.</center>
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[[2009 Zhautykov International Olympiad Problems/Problem 1|Solution]]
  
 
===Problem 2===
 
===Problem 2===
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for all <math>x,y\in\mathbb{R}</math>.
 
for all <math>x,y\in\mathbb{R}</math>.
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[[2009 Zhautykov International Olympiad Problems/Problem 2|Solution]]
  
 
===Problem 3===
 
===Problem 3===
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<center><math>AC\cdot(BD + BF - DF) + CE\cdot(BD + DF - BF) + AE\cdot(BF + DF - BD)\geq 2\sqrt {3}</math>.</center>
 
<center><math>AC\cdot(BD + BF - DF) + CE\cdot(BD + DF - BF) + AE\cdot(BF + DF - BD)\geq 2\sqrt {3}</math>.</center>
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[[2009 Zhautykov International Olympiad Problems/Problem 3|Solution]]
  
 
==Day 2==
 
==Day 2==
Line 20: Line 26:
 
On the plane, a Cartesian coordinate system is chosen. Given points <math>A_1,A_2,A_3,A_4</math> on the parabola <math>y = x^2</math>, and points <math>B_1,B_2,B_3,B_4</math> on the parabola <math>y = 2009x^2</math>. Points <math>A_1,A_2,A_3,A_4</math> are concyclic, and points <math>A_i</math> and <math>B_i</math> have equal abscissas for each <math>i = 1,2,3,4</math>.
 
On the plane, a Cartesian coordinate system is chosen. Given points <math>A_1,A_2,A_3,A_4</math> on the parabola <math>y = x^2</math>, and points <math>B_1,B_2,B_3,B_4</math> on the parabola <math>y = 2009x^2</math>. Points <math>A_1,A_2,A_3,A_4</math> are concyclic, and points <math>A_i</math> and <math>B_i</math> have equal abscissas for each <math>i = 1,2,3,4</math>.
 
Prove that points <math>B_1,B_2,B_3,B_4</math> are also concyclic.
 
Prove that points <math>B_1,B_2,B_3,B_4</math> are also concyclic.
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[[2009 Zhautykov International Olympiad Problems/Problem 4|Solution]]
  
 
===Problem 5===
 
===Problem 5===
 
Given a quadrilateral <math>ABCD</math> with <math>\angle B = \angle D = 90^{\circ}</math>. Point <math>M</math> is chosen on segment <math>AB</math> so taht <math>AD = AM</math>. Rays <math>DM</math> and <math>CB</math> intersect at point <math>N</math>. Points <math>H</math> and <math>K</math> are feet of perpendiculars from points <math>D</math> and <math>C</math> to lines <math>AC</math> and <math>AN</math>, respectively.  
 
Given a quadrilateral <math>ABCD</math> with <math>\angle B = \angle D = 90^{\circ}</math>. Point <math>M</math> is chosen on segment <math>AB</math> so taht <math>AD = AM</math>. Rays <math>DM</math> and <math>CB</math> intersect at point <math>N</math>. Points <math>H</math> and <math>K</math> are feet of perpendiculars from points <math>D</math> and <math>C</math> to lines <math>AC</math> and <math>AN</math>, respectively.  
 
Prove that <math>\angle MHN = \angle MCK</math>.
 
Prove that <math>\angle MHN = \angle MCK</math>.
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[[2009 Zhautykov International Olympiad Problems/Problem 5|Solution]]
  
 
===Problem 6===
 
===Problem 6===
 
In a checked <math>17\times 17</math> table, <math>n</math> squares are colored in black. We call a line any of rows, columns, or any of two diagonals of the table. In one step, if at least <math>6</math> of the squares in some line are black, then one can paint all the squares of this line in black. Find the minimal value of <math>n</math> such that for some initial arrangement of <math>n</math> black squares one can paint all squares of the table in black in some steps.
 
In a checked <math>17\times 17</math> table, <math>n</math> squares are colored in black. We call a line any of rows, columns, or any of two diagonals of the table. In one step, if at least <math>6</math> of the squares in some line are black, then one can paint all the squares of this line in black. Find the minimal value of <math>n</math> such that for some initial arrangement of <math>n</math> black squares one can paint all squares of the table in black in some steps.
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[[2009 Zhautykov International Olympiad Problems/Problem 6|Solution]]

Latest revision as of 12:52, 3 February 2009

Day 1

Problem 1

Find all pairs of integers $(x,y)$, such that

$x^2 - 2009y + 2y^2 = 0$.

Solution

Problem 2

Find all real $a$, such that there exist a function $f: \mathbb{R}\rightarrow\mathbb{R}$ satisfying the following inequality:

$x + af(y)\leq y + f(f(x))$

for all $x,y\in\mathbb{R}$.

Solution

Problem 3

For a convex hexagon $ABCDEF$ with an area $S$, prove that:

$AC\cdot(BD + BF - DF) + CE\cdot(BD + DF - BF) + AE\cdot(BF + DF - BD)\geq 2\sqrt {3}$.

Solution

Day 2

Problem 4

On the plane, a Cartesian coordinate system is chosen. Given points $A_1,A_2,A_3,A_4$ on the parabola $y = x^2$, and points $B_1,B_2,B_3,B_4$ on the parabola $y = 2009x^2$. Points $A_1,A_2,A_3,A_4$ are concyclic, and points $A_i$ and $B_i$ have equal abscissas for each $i = 1,2,3,4$. Prove that points $B_1,B_2,B_3,B_4$ are also concyclic.

Solution

Problem 5

Given a quadrilateral $ABCD$ with $\angle B = \angle D = 90^{\circ}$. Point $M$ is chosen on segment $AB$ so taht $AD = AM$. Rays $DM$ and $CB$ intersect at point $N$. Points $H$ and $K$ are feet of perpendiculars from points $D$ and $C$ to lines $AC$ and $AN$, respectively. Prove that $\angle MHN = \angle MCK$.

Solution

Problem 6

In a checked $17\times 17$ table, $n$ squares are colored in black. We call a line any of rows, columns, or any of two diagonals of the table. In one step, if at least $6$ of the squares in some line are black, then one can paint all the squares of this line in black. Find the minimal value of $n$ such that for some initial arrangement of $n$ black squares one can paint all squares of the table in black in some steps.

Solution

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