IMO 2003 (XLIV International Mathematical Olympiad)

Disjoint translated subsets from a 101-element subset of S

$S$ is the set $\{1,2,3,\dots ,1000000\}$. Show that for any subset $A$ of $S$ with $101$ elements we can find $100$ distinct elements $x_i$ of $S$, such that the sets $\{a+x_i\mid a\in A\}$ are all pairwise disjoint.

Topic: Combinatoria Metodo: Estremalità, Principio dei cassetti Abilita: Conteggio sistematico, Modellizzazione, Ragionamento geometrico Area: Combinatoria, Logica e Probabilita Fonte: apri PDF p.1

Disjoint translated subsets from a 101-element subset of S

$S$ is the set $\{1,2,3,\dots ,1000000\}$. Show that for any subset $A$ of $S$ with $101$ elements we can find $100$ distinct elements $x_i$ of $S$, such that the sets $\{a+x_i\mid a\in A\}$ are all pairwise disjoint.

src_imho_2003__Q01

Find all pairs (m,n) making a fraction a positive integer

Find all pairs $(m,n)$ of positive integers such that $\frac{m^2}{2m{n}^2-n^3+1}$ is a positive integer.

Topic: Teoria dei Numeri, Algebra Metodo: Fattorizzazione, Congruenze, Casework Abilita: Manipolazione algebrica, Ragionamento geometrico, Lettura attenta Area: Aritmetica e Teoria dei Numeri, Algebra e Analisi Fonte: apri PDF p.1

Find all pairs (m,n) making a fraction a positive integer

Find all pairs $(m,n)$ of positive integers such that $\frac{m^2}{2m{n}^2-n^3+1}$ is a positive integer.

src_imho_2003__Q02

Convex hexagon with midpoint-distance condition has equal angles

A convex hexagon has the property that for any pair of opposite sides the distance between their midpoints is $\frac{\sqrt{3}}{2}$ times the sum of their lengths. Show that all the hexagon’s angles are equal.

Topic: Geometria piana Metodo: Trigonometria, Coordinate Abilita: Ragionamento geometrico, Manipolazione algebrica, Modellizzazione Area: Geometria Fonte: apri PDF p.1

Convex hexagon with midpoint-distance condition has equal angles

A convex hexagon has the property that for any pair of opposite sides the distance between their midpoints is $\frac{\sqrt{3}}{2}$ times the sum of their lengths. Show that all the hexagons are equal.

src_imho_2003__Q03

Cyclic quadrilateral: angle bisectors meet on AC iff RP=RQ

$ABCD$ is cyclic. The feet of the perpendicular from $D$ to the lines $AB$, $BC$, $CA$ are $P$, $Q$, $R$ respectively. Show that the angle bisectors of $\angle ABC$ and $\angle CDA$ meet on the line $AC$ iff $RP=RQ$.

Topic: Geometria piana Metodo: Trigonometria, Coordinate, Simmetria Abilita: Ragionamento geometrico, Lettura attenta, Manipolazione algebrica Area: Geometria Fonte: apri PDF p.1

Cyclic quadrilateral: angle bisectors meet on AC iff RP=RQ*

$ABCD$ is cyclic. The feet of the perpendicular from $D$ to the lines $AB$, $BC$, $CA$ are $P$, $Q$, $R$ respectively. Show that the angle bisectors of $\angle ABC$ and $\angle CDA$ meet on the line $AC$ iff $RP=RQ$.

src_imho_2003__Q04

Inequality for ordered reals with equality iff arithmetic progression

Given $n>2$ and reals $x_1\le x_2\le \cdots \le x_n$, show that ${\left({\sum }_{i,j}|x_i-x_j|\right)}^2\le \frac{2}{3}(n^2-1){\sum }_{i,j}(x_i-x_j{)}^2.$ Show that we have equality iff the sequence is an arithmetic progression.

Topic: Disuguaglianze, Algebra Metodo: Disuguaglianze, Doppio conteggio, Simmetria Abilita: Manipolazione algebrica, Stima, Conteggio sistematico Area: Algebra e Analisi Fonte: apri PDF p.1

Inequality for ordered reals with equality iff arithmetic progression

Given $n>2$ and reals $x_1\le x_2\le \cdots \le x_n$, show that ${\left({\sum }_{i,j}|x_i-x_j|\right)}^2\le \frac{2}{3}(n^2-1){\sum }_{i,j}(x_i-x_j{)}^2.$ Show that we have equality if the sequence is an arithmetic progression.

src_imho_2003__Q05

For each prime p, find prime q such that p^n - p not divisible by q

Show that for each prime $p$, there exists a prime $q$ such that $n^p-p$ is not divisible by $q$ for any positive integer $n$.

Topic: Teoria dei Numeri Metodo: Congruenze, Estremalità Abilita: Ragionamento geometrico, Manipolazione algebrica, Astrazione Area: Aritmetica e Teoria dei Numeri Fonte: apri PDF p.1

For each prime p, find prime q such that p - p not divisible by q*

Show that for each prime $p$, there exists a prime $q$ such that $n^p-p$ is not divisible by $q$ for any positive integer $n$.

src_imho_2003__Q06