Difference between revisions of "2004 IMO Problems"
(Created page with "Problems of the 46th IMO 2004 Mérida, Mexico. == Day 1 == === Problem 1 === Six points are chosen on the sides of an equilateral triangle <math>ABC</math>: <math>A_1, A...") |
(→Problem 4) |
||
(2 intermediate revisions by the same user not shown) | |||
Line 1: | Line 1: | ||
− | Problems of the | + | Problems of the 45th [[IMO]] 2004 Athens, Greece. |
== Day 1 == | == Day 1 == | ||
=== Problem 1 === | === Problem 1 === | ||
− | + | Let <math>ABC</math> be an acute-angled triangle with <math>AB\neq AC</math>. The circle with diameter <math>BC</math> intersects the sides <math>AB</math> and <math>AC</math> at <math>M</math> and <math>N</math> respectively. Denote by <math>O</math> the midpoint of the side <math>BC</math>. The bisectors of the angles <math>\angle BAC</math> and <math>\angle MON</math> intersect at <math>R</math>. Prove that the circumcircles of the triangles <math>BMR</math> and <math>CNR</math> have a common point lying on the side <math>BC</math>. | |
[[2004 IMO Problems/Problem 1 | Solution]] | [[2004 IMO Problems/Problem 1 | Solution]] | ||
=== Problem 2 === | === Problem 2 === | ||
− | + | Find all polynomials <math>f</math> with real coefficients such that for all reals <math>a,b,c</math> such that <math>ab + bc + ca = 0</math> we have the following relations | |
+ | |||
+ | <cmath>f(a - b) + f(b - c) + f(c - a) = 2f(a + b + c).</cmath> | ||
[[2004 IMO Problems/Problem 2 | Solution]] | [[2004 IMO Problems/Problem 2 | Solution]] | ||
=== Problem 3 === | === Problem 3 === | ||
− | + | Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure. | |
+ | |||
+ | <asy> | ||
+ | unitsize(0.5 cm); | ||
+ | |||
+ | draw((0,0)--(1,0)); | ||
+ | draw((0,1)--(1,1)); | ||
+ | draw((2,1)--(3,1)); | ||
+ | draw((0,2)--(3,2)); | ||
+ | draw((0,3)--(3,3)); | ||
+ | draw((0,0)--(0,3)); | ||
+ | draw((1,0)--(1,3)); | ||
+ | draw((2,1)--(2,3)); | ||
+ | draw((3,1)--(3,3)); | ||
+ | </asy> | ||
+ | |||
+ | Determine all <math>m \times n</math> rectangles that can be covered without gaps and without overlaps with hooks such that; | ||
+ | (a) the rectangle is covered without gaps and without overlaps, | ||
+ | (b) no part of a hook covers area outside the rectangle. | ||
[[2004 IMO Problems/Problem 3 | Solution]] | [[2004 IMO Problems/Problem 3 | Solution]] | ||
Line 21: | Line 41: | ||
=== Problem 4 === | === Problem 4 === | ||
− | + | Let <math>n \geq 3</math> be an integer. Let <math>t_1, t_2, \dots ,t_n</math> be positive real numbers such that | |
+ | |||
+ | <cmath>n^2 + 1 > \left( t_1 + t_2 + ... + t_n \right) \left( \frac {1}{t_1} + \frac {1}{t_2} + ... + \frac {1}{t_n} \right).</cmath> | ||
+ | Show that <math>t_i</math>, <math>t_j</math>, <math>t_k</math> are side lengths of a triangle for all <math>i</math>, <math>j</math>, <math>k</math> with <math>1 \leq i < j < k \leq n</math>. | ||
[[2004 IMO Problems/Problem 4 | Solution]] | [[2004 IMO Problems/Problem 4 | Solution]] | ||
=== Problem 5 === | === Problem 5 === | ||
− | + | In a convex quadrilateral <math>ABCD</math>, the diagonal <math>BD</math> bisects neither the angle <math>ABC</math> nor the angle <math>CDA</math>. The point <math>P</math> lies inside <math>ABCD</math> and satisfies<cmath>\angle PBC = \angle DBA \text{ and } \angle PDC = \angle BDA.</cmath> | |
+ | Prove that <math>ABCD</math> is a cyclic quadrilateral if and only if <math>AP = CP.</math> | ||
[[2004 IMO Problems/Problem 5 | Solution]] | [[2004 IMO Problems/Problem 5 | Solution]] | ||
=== Problem 6 === | === Problem 6 === | ||
− | + | We call a positive integer alternating if every two consecutive digits in its decimal representation have a different parity. | |
+ | |||
+ | Find all positive integers <math>n</math> such that <math>n</math> has a multiple which is alternating. | ||
[[2004 IMO Problems/Problem 6 | Solution]] | [[2004 IMO Problems/Problem 6 | Solution]] |
Latest revision as of 04:36, 21 February 2021
Problems of the 45th IMO 2004 Athens, Greece.
Contents
Day 1
Problem 1
Let be an acute-angled triangle with
. The circle with diameter
intersects the sides
and
at
and
respectively. Denote by
the midpoint of the side
. The bisectors of the angles
and
intersect at
. Prove that the circumcircles of the triangles
and
have a common point lying on the side
.
Problem 2
Find all polynomials with real coefficients such that for all reals
such that
we have the following relations
Problem 3
Define a "hook" to be a figure made up of six unit squares as shown below in the picture, or any of the figures obtained by applying rotations and reflections to this figure.
Determine all rectangles that can be covered without gaps and without overlaps with hooks such that;
(a) the rectangle is covered without gaps and without overlaps,
(b) no part of a hook covers area outside the rectangle.
Day 2
Problem 4
Let be an integer. Let
be positive real numbers such that
Show that
,
,
are side lengths of a triangle for all
,
,
with
.
Problem 5
In a convex quadrilateral , the diagonal
bisects neither the angle
nor the angle
. The point
lies inside
and satisfies
Prove that
is a cyclic quadrilateral if and only if
Problem 6
We call a positive integer alternating if every two consecutive digits in its decimal representation have a different parity.
Find all positive integers such that
has a multiple which is alternating.
Resources
2004 IMO (Problems) • Resources | ||
Preceded by 2003 IMO Problems |
1 • 2 • 3 • 4 • 5 • 6 | Followed by 2005 IMO Problems |
All IMO Problems and Solutions |