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Difference between revisions of "2013 IMO Problems"

(Created page with "==Problem 1== Prove that for any pair of positive integers <math>k</math> and <math>n</math>, there exist <math>k</math> positive integers <math>m_1,m_2,...,m_k</math> (not ne...")
 
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Prove that for any pair of positive integers <math>k</math> and <math>n</math>, there exist <math>k</math> positive integers <math>m_1,m_2,...,m_k</math> (not necessarily different) such that
 
Prove that for any pair of positive integers <math>k</math> and <math>n</math>, there exist <math>k</math> positive integers <math>m_1,m_2,...,m_k</math> (not necessarily different) such that
  
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<math>1+\frac{2^k-1}{n}=(1+\frac{1}{m_1})(1+\frac{1}{m_2})...(1+\frac{1}{m_k})</math>.
+
<math>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).</math>
  
 
[[2013 IMO Problems/Problem 1|Solution]]
 
[[2013 IMO Problems/Problem 1|Solution]]

Latest revision as of 04:43, 17 February 2021

Problem 1

Prove that for any pair of positive integers $k$ and $n$, there exist $k$ positive integers $m_1,m_2,...,m_k$ (not necessarily different) such that

$1+\frac{2^k-1}{n}=\left(1+\frac{1}{m_1}\right)\left(1+\frac{1}{m_2}\right)\cdots\left(1+\frac{1}{m_k}\right).$

Solution

Problem 2

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 of the configuration 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:

  • no line passes through any point of the configuration;
  • no region contains points of both colours.

Find the least value of $k$ such that for any Colombian configuration of $4027$ points, there is a good arrangement of $k$ lines.

Solution

Problem 3

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 circumcentre of triangle $A_1B_1C_1$ lies on the circumcircle of triangle $ABC$. Prove that triangle $ABC$ is right-angled.

Solution

Problem 4

Let $ABC$ be an acute triangle with orthocenter $H$, 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 $\omega_1$ is [sic] the circumcircle of $BWN$, and let $X$ be the point on $\omega_1$ such that $WX$ is a diameter of $\omega_1$. Analogously, denote by $\omega_2$ the circumcircle of triangle $CWM$, and let $Y$ be the point such that $WY$ is a diameter of $\omega_2$. Prove that $X, Y$ and $H$ are collinear.

Solution

Problem 5

Let $\mathbb Q_{>0}$ be the set of all positive rational numbers. Let $f:\mathbb Q_{>0}\to\mathbb R$ be a function satisfying the following three conditions:

(i) for all $x,y\in\mathbb Q_{>0}$, we have $f(x)f(y)\geq f(xy)$; (ii) for all $x,y\in\mathbb Q_{>0}$, we have $f(x+y)\geq f(x)+f(y)$; (iii) there exists a rational number $a>1$ such that $f(a)=a$.

Prove that $f(x)=x$ for all $x\in\mathbb Q_{>0}$.

Solution

Problem 6

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 $0, 1, ... , n$ 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 labelings, 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 \[M = N + 1.\]

Solution

2013 IMO (Problems) • Resources
Preceded by
2012 IMO Problems 		1 • 2 • 3 • 4 • 5 • 6 Followed by
2014 IMO Problems
All IMO Problems and Solutions
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