IMO 2013 (LIV International Mathematical Olympiad)

Prove positive integer expansion product identity

Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1,m_2,\dots ,m_k$ (not necessarily different) such that $1+\frac{2^k-1}{n}=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)\cdots \left(1+\frac{1}{m_k}\right).$

Topic: Teoria dei Numeri, Algebra Metodo: Induzione, Fattorizzazione Abilita: Manipolazione algebrica, Riconoscimento di pattern, Ragionamento geometrico Area: Aritmetica e Teoria dei Numeri, Algebra e Analisi Fonte: apri PDF p.1

This is the case for the manufacturer of the product.

Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1,m_2,\dots ,m_k$ (not necessarily different) such that $1+\frac{2^k-1}{n}=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)\cdots \left(1+\frac{1}{m_k}\right).$

src_imho_2013__Q01

Colombian configuration of 4027 points; find least k lines

A configuration of $4027$ points in the plane is called Colombian if it consists of $2013$ red points and $2014$ blue points, and no three of the points are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian configuration if the following two conditions are satisfied: \begin{itemize} \item no line passes through any point of the configuration; \item no region contains points of both colours. \end{itemize} Find the least value of $k$ such that for any Colombian configuration of $4027$ points, there is a good arrangement of $k$ lines.

Topic: Combinatoria Metodo: Estremalità, Casework, Colorazione Abilita: Modellizzazione, Ragionamento geometrico, Conteggio sistematico, Lettura attenta Area: Combinatoria, Logica e Probabilita Fonte: apri PDF p.1

Colombian configuration of 4027 points; find at least k lines

A configuration of $4027$ points in the plane is called Colombian if it consists of $2013$ red points and $2014$ blue points, and no three of the points are collinear. By drawing some lines, the plane is divided into several regions. An arrangement of lines is good for a Colombian configuration if the following two conditions are satisfied: \begin{itemize} \item no line passes through any point of the configuration; \item no region contains points of both colors. Find the least value of $k$ such that for any Colombian configuration of $4027$ points, there is a good arrangement of $k$ lines.

src_imho_2013__Q02

Excircle of ABC opposite A tangent to BC; prove angle is right

Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the excircles opposite $B$ and $C$, respectively. Suppose that the circumcircle of triangle $A_1{B}_1{C}_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.

\textit{The excircle of triangle $ABC$ 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 $C$. The excircles opposite $B$ and $C$ are similarly defined.}

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

Excircle of ABC opposite A tangent to BC; prove angle is right

Let the excircle of triangle $ABC$ opposite the vertex $A$ be tangent to the side $BC$ at the point $A_1$. Define the points $B_1$ on $CA$ and $C_1$ on $AB$ analogously, using the opposite excircles $B$ and $C$, respectively. Suppose that the circumcircle of triangle $A_1{B}_1{C}_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.

\textit{The excircle of triangle $ABC$ 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 $C$. The opposite excircles $B$ and $C$ are similarly defined.}

src_imho_2013__Q03

Acute triangle with orthocentre; collinearity of X, Y, H

Let $ABC$ be an acute-angled triangle with orthocentre $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by ${\omega }_1$ the circumcircle of $BWN$, and let $X$ be the point on ${\omega }_1$ such that $WX$ is a diameter of ${\omega }_1$. Denote by ${\omega }_2$ the circumcircle of $CWM$, and let $Y$ be the point on ${\omega }_2$ such that $WY$ is a diameter of ${\omega }_2$. Prove that $X$, $Y$ and $H$ are collinear.

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

Acute triangle with orthocentre; collinearity of X, Y, H

Let $ABC$ be an acute-angled triangle with orthocentre $H$, and let $W$ be a point on the side $BC$, lying strictly between $B$ and $C$. The points $M$ and $N$ are the feet of the altitudes from $B$ and $C$, respectively. Denote by ${\omega }_1$ the circumcircle of $BWN$, and let $X$ be the point on ${\omega }_1$ such that $WX$ is a diameter of ${\omega }_1$. Denote by ${\omega }_2$ the circumcircle of $CWM$, and let $Y$ be the point on ${\omega }_2$ such that $WY$ is a diameter of ${\omega }_2$. Prove that $X$, $Y$ and $H$ are collinear.

src_imho_2013__Q04

Functional equation on positive rationals with f(x)=x

Let ${\mathbb{Q}}_{>0}$ be the set of positive rational numbers. Let $f:{\mathbb{Q}}_{>0}\to \mathbb{R}$ be a function satisfying the following three conditions: \begin{itemize} \item[(i)] for all $x,y\in {\mathbb{Q}}_{>0}$, we have $f(x)\cdot f(y)\ge f(xy)$; \item[(ii)] for all $x,y\in {\mathbb{Q}}_{>0}$, we have $f(x+y)\ge f(x)+f(y)$; \item[(iii)] there exists a rational number $a>1$ such that $f(a)=a$. \end{itemize} Prove that $f(x)=x$ for all $x\in {\mathbb{Q}}_{>0}$.

Topic: Equazioni funzionali, Algebra Metodo: Induzione, Disuguaglianze Abilita: Manipolazione algebrica, Ragionamento geometrico, Lettura attenta, Stima Area: Algebra e Analisi Fonte: apri PDF p.1

Functional equation on positive rationals with f(x)=x*

Let${\mathbb{Q}}_{>0}$be the set of positive rational numbers. Let $f:{\mathbb{Q}}_{>0}\to \mathbb{R}$ be a function satisfying the following three conditions: \begin{itemize} \item[(i) ] for all $x,y\in {\mathbb{Q}}_{>0}$, we have $f(x)\cdot f(y)\ge f(xy)$; \item[(ii)] for all $x,y\in {\mathbb{Q}}_{>0}$, we have $f(x+y)\ge f(x)+f(y)$; \item[(iii)] there exists a rational number $a>1$ such that $f(a)=a$. \end{itemize} Prove that for all $x\in {\mathbb{Q}}_{>0}$.

src_imho_2013__Q05

Beautiful labellings of circle points; prove M = N + 1

Let $n\ge 3$ be an integer, and consider a circle with $n+1$ equally spaced points marked on it. Consider all labellings of these points with the numbers $0,1,\dots ,n$ such that each label is used exactly once; two such labellings are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels $a<b<c<d$ with $a+d=b+c$, the chord joining the points labelled $a$ and $d$ does not intersect the chord joining the points labelled $b$ and $c$.

Let $M$ be the number of beautiful labellings, and let $N$ be the number of ordered pairs $(x,y)$ of positive integers such that $x+y\le n$ and $\gcd (x,y)=1$. Prove that $M=N+1.$

Topic: Combinatoria, Teoria dei Numeri Metodo: Biiezione, Doppio conteggio, Conteggio, Induzione Abilita: Conteggio sistematico, Riconoscimento di pattern, Modellizzazione, Astrazione, Manipolazione algebrica Area: Combinatoria, Logica e Probabilita, Aritmetica e Teoria dei Numeri Fonte: apri PDF p.1

Beautiful labelling of circle points; proofs M = N + 1

Let $n\ge 3$ be an integer, and consider a circle with $n+1$ equally spaced points marked on it. Consider all labelling of these points with the numbers $0,1,\dots ,n$ such that each label is used exactly once; two such labelling are considered to be the same if one can be obtained from the other by a rotation of the circle. A labelling is called beautiful if, for any four labels $a<b<c<d$ with $a+d=b+c$, the chord joining the points labelled $a$ and $d$ does not intersect the chord joining the points labelled $b$ and $c$.

Let $M$ be the number of beautiful labelling, and let $N$ be the number of ordered pairs $(x,y)$ of positive integers such that $x+y\le n$ and $\gcd (x,y)=1$. Prove that $M=N+1.$

src_imho_2013__Q06