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Difference between revisions of "2012 IMO Problems"
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== Day 1 ==
 
== Day 1 ==
 
=== Problem 1. ===
 
=== Problem 1. ===
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Given triangle ABC the point <math>J</math> is the centre of the excircle opposite the vertex <math>A</math>.
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Given <math>\triangle ABC</math> the point <math>J</math> is the centre of the excircle opposite the vertex <math>A</math>. This excircle is tangent to the side <math>BC</math> at <math>M</math>, and to the lines <math>AB</math> and <math>AC</math> at <math>K</math> and <math>L</math>, respectively. The lines <math>LM</math> and <math>BJ</math> meet at <math>F</math>, and the lines <math>KM</math> and <math>CJ</math> meet at <math>G</math>. Let <math>S</math> be the point of intersection of the lines <math>AF</math> and <math>BC</math>, and let <math>T</math> be the point of intersection of the lines <math>AG</math> and <math>BC</math>. Prove that <math>M</math> is the midpoint of <math>ST</math>.
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This excircle is tangent to the side <math>BC</math> at <math>M</math>, and to the lines <math>AB</math> and <math>AC</math> at <math>K</math> and <math>L</math>, respectively.
 
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The lines <math>LM</math> and <math>BJ</math> meet at <math>F</math>, and the lines <math>KM</math> and <math>CJ</math> meet at <math>G</math>. Let <math>S</math> be the point of
 
−
intersection of the lines <math>AF</math> and <math>BC</math>, and let <math>T</math> be the point of intersection of the lines <math>AG</math> and <math>BC</math>.
 
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Prove that <math>M</math> is the midpoint of <math>ST</math>.
 
 
(The excircle of <math>ABC</math> opposite the vertex <math>A</math> is the circle that is tangent to the line segment <math>BC</math>,
 
(The excircle of <math>ABC</math> opposite the vertex <math>A</math> is the circle that is tangent to the line segment <math>BC</math>,
 
to the ray <math>AB</math> beyond <math>B</math>, and to the ray <math>AC</math> beyond <math>C</math>.)
 
to the ray <math>AB</math> beyond <math>B</math>, and to the ray <math>AC</math> beyond <math>C</math>.)
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[[2012 IMO Problems/Problem 6 | Solution]]
 
[[2012 IMO Problems/Problem 6 | Solution]]
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== Resources ==
 
== Resources ==
 
* [[2012 IMO]]
 
* [[2012 IMO]]
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=2012&sid=0b159fdf0134b865758167cc5cf255dd 2012 IMO Problems on the Resources page]
 
* [http://www.artofproblemsolving.com/Forum/resources.php?c=1&cid=16&year=2012&sid=0b159fdf0134b865758167cc5cf255dd 2012 IMO Problems on the Resources page]
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{{IMO box|year=2012|before=[[2011 IMO Problems]]|after=[[2013 IMO Problems]]}}

Latest revision as of 10:00, 24 February 2021

Problems of the 53st IMO 2012 in Mar del Plata, Argentina.

Day 1

Problem 1.

Given $\triangle ABC$ the point $J$ is the centre of the excircle opposite the vertex $A$. This excircle is tangent to the side $BC$ at $M$, and to the lines $AB$ and $AC$ at $K$ and $L$, respectively. The lines $LM$ and $BJ$ meet at $F$, and the lines $KM$ and $CJ$ meet at $G$. Let $S$ be the point of intersection of the lines $AF$ and $BC$, and let $T$ be the point of intersection of the lines $AG$ and $BC$. Prove that $M$ is the midpoint of $ST$. (The excircle of $ABC$ opposite the vertex $A$ is the circle that is tangent to the line segment $BC$, to the ray $AB$ beyond $B$, and to the ray $AC$ beyond $C$.)

Author: Evangelos Psychas, Greece

Solution

Problem 2.

Let ${{a}_{2}}, {{a}_{3}}, \cdots, {{a}_{n}}$ be positive real numbers that satisfy ${{a}_{2}}\cdot {{a}_{3}}\cdots {{a}_{n}}=1$ . Prove that \[\left(a_2+1\right)^2\cdot \left(a_3+1\right)^3\cdots \left(a_n+1\right)^n\gneq n^n\]

Author: Angelo di Pasquale, Australia

Solution

Problem 3.

The liar’s guessing game is a game played between two players $A$ and $B$. The rules of the game depend on two positive integers $k$ and $n$ which are known to both players. At the start of the game A chooses integers $x$ and $N$ with $1\le x\le N$. Player $A$ keeps $x$ secret, and truthfully tells $N$ to player $B$. Player $B$ now tries to obtain information about $x$ by asking player $A$ questions as follows: each question consists of $B$ specifying an arbitrary set $S$ of positive integers (possibly one specified in some previous question), and asking $A$ whether $x$ belongs to $S$. Player $B$ may ask as many such questions as he wishes. After each question, player $A$ must immediately answer it with yes or no, but is allowed to lie as many times as she wants; the only restriction is that, among any $k + 1$ consecutive answers, at least one answer must be truthful. After $B$ has asked as many questions as he wants, he must specify a set $X$ of at most $n$ positive integers. If $x$ belongs to $X$, then $B$ wins; otherwise, he loses. Prove that:

  1. If $n\ge {{2}^{k}}$, then $B$ can guarantee a win.
  2. For all sufficiently large $k$, there exists an integer $n\ge {1.99^k}$ such that $B$ cannot guarantee a win.

Author: David Arthur, Canada

Solution


Day 2

Problem 4.

Find all functions $f:\mathbb{Z}\to \mathbb{Z}$ such that, for all integers $a$, $b$, $c$ that satisfy $a+b+c = 0$, the following equality holds: \[f(a)^2 + f(b)^2 + f(c)^2 = 2f(a)f(b) + 2f(b)f(c) + 2f(c)f(a).\] (Here $\mathbb{Z}$ denotes the set of integers.)

Author: Liam Baker, South Africa

Solution

Problem 5.

Let $ABC$ be a triangle with $\angle BCA=90{}^\circ$, and let $D$ be the foot of the altitude from $C$. Let $X$ be a point in the interior of the segment $CD$. Let K be the point on the segment $AX$ such that $BK = BC$. Similarly, let $L$ be the point on the segment $BX$ such that $AL = AC$. Let $M$ be the point of intersection of $AL$ and $BK$. Show that $MK = ML$.

Author: Josef Tkadlec, Czech Republic

Solution

Problem 6.

Find all positive integers n for which there exist non-negative integers $a_1$, $a_2$, $\ldots$ , $a_n$ such that $\frac{1}{{{2}^{{{a}_{1}}}}}+\frac{1}{{{2}^{{{a}_{2}}}}}+\cdots +\frac{1}{{{2}^{{{a}_{n}}}}}=\frac{1}{{{3}^{{{a}_{1}}}}}+\frac{2}{{{3}^{{{a}_{2}}}}}+\cdots +\frac{n}{{{3}^{{{a}_{n}}}}}=1$

Author: Dušan Djukić, Serbia

Solution

Resources

2012 IMO (Problems) • Resources
Preceded by
2011 IMO Problems 		1 • 2 • 3 • 4 • 5 • 6 Followed by
2013 IMO Problems
All IMO Problems and Solutions
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