Found problems: 41
2007 India IMO Training Camp, 2
Let $ S$ be a finite set of points in the plane such that no three of them are on a line. For each convex polygon $ P$ whose vertices are in $ S$, let $ a(P)$ be the number of vertices of $ P$, and let $ b(P)$ be the number of points of $ S$ which are outside $ P$. A line segment, a point, and the empty set are considered as convex polygons of $ 2$, $ 1$, and $ 0$ vertices respectively. Prove that for every real number $ x$ \[\sum_{P}{x^{a(P)}(1 \minus{} x)^{b(P)}} \equal{} 1,\] where the sum is taken over all convex polygons with vertices in $ S$.
[i]Alternative formulation[/i]:
Let $ M$ be a finite point set in the plane and no three points are collinear. A subset $ A$ of $ M$ will be called round if its elements is the set of vertices of a convex $ A \minus{}$gon $ V(A).$ For each round subset let $ r(A)$ be the number of points from $ M$ which are exterior from the convex $ A \minus{}$gon $ V(A).$ Subsets with $ 0,1$ and 2 elements are always round, its corresponding polygons are the empty set, a point or a segment, respectively (for which all other points that are not vertices of the polygon are exterior). For each round subset $ A$ of $ M$ construct the polynomial
\[ P_A(x) \equal{} x^{|A|}(1 \minus{} x)^{r(A)}.
\]
Show that the sum of polynomials for all round subsets is exactly the polynomial $ P(x) \equal{} 1.$
[i]Proposed by Federico Ardila, Colombia[/i]
2002 IMC, 8
200 students participated in a math contest. They had 6 problems to solve. Each problem was correctly solved by at least 120 participants. Prove that there must be 2 participants such that every problem was solved by at least one of these two students.
2015 Vietnam Team selection test, Problem 4
There are $100$ students who praticipate at exam.Also there are $25$ members of jury.Each student is checked by one jury.Known that every student likes $10$ jury
$a)$ Prove that we can select $7$ jury such that any student likes at least one jury.
$b)$ Prove that we can make this every student will be checked by the jury that he likes and every jury will check at most $10$ students.
2015 China Second Round Olympiad, 1
Let $a_1, a_2, \ldots, a_n$ be real numbers.Prove that you can select $\varepsilon _1, \varepsilon _2, \ldots, \varepsilon _n\in\{-1,1\}$ such that$$\left( \sum_{i=1}^{n}a_{i}\right)^2 +\left( \sum_{i=1}^{n}\varepsilon _ia_{i}\right)^2 \leq(n+1)\left( \sum_{i=1}^{n}a^2_{i}\right).$$
2006 IMO Shortlist, 3
Let $ S$ be a finite set of points in the plane such that no three of them are on a line. For each convex polygon $ P$ whose vertices are in $ S$, let $ a(P)$ be the number of vertices of $ P$, and let $ b(P)$ be the number of points of $ S$ which are outside $ P$. A line segment, a point, and the empty set are considered as convex polygons of $ 2$, $ 1$, and $ 0$ vertices respectively. Prove that for every real number $ x$ \[\sum_{P}{x^{a(P)}(1 \minus{} x)^{b(P)}} \equal{} 1,\] where the sum is taken over all convex polygons with vertices in $ S$.
[i]Alternative formulation[/i]:
Let $ M$ be a finite point set in the plane and no three points are collinear. A subset $ A$ of $ M$ will be called round if its elements is the set of vertices of a convex $ A \minus{}$gon $ V(A).$ For each round subset let $ r(A)$ be the number of points from $ M$ which are exterior from the convex $ A \minus{}$gon $ V(A).$ Subsets with $ 0,1$ and 2 elements are always round, its corresponding polygons are the empty set, a point or a segment, respectively (for which all other points that are not vertices of the polygon are exterior). For each round subset $ A$ of $ M$ construct the polynomial
\[ P_A(x) \equal{} x^{|A|}(1 \minus{} x)^{r(A)}.
\]
Show that the sum of polynomials for all round subsets is exactly the polynomial $ P(x) \equal{} 1.$
[i]Proposed by Federico Ardila, Colombia[/i]
2010 USAMO, 6
A blackboard contains 68 pairs of nonzero integers. Suppose that for each positive integer $k$ at most one of the pairs $(k, k)$ and $(-k, -k)$ is written on the blackboard. A student erases some of the 136 integers, subject to the condition that no two erased integers may add to 0. The student then scores one point for each of the 68 pairs in which at least one integer is erased. Determine, with proof, the largest number $N$ of points that the student can guarantee to score regardless of which 68 pairs have been written on the board.
2022 Bulgarian Spring Math Competition, Problem 11.4
Let $n \geq 2$ be a positive integer. The set $M$ consists of $2n^2-3n+2$ positive rational numbers. Prove that there exists a subset $A$ of $M$ with $n$ elements with the following property: $\forall$ $2 \leq k \leq n$ the sum of any $k$ (not necessarily distinct) numbers from $A$ is not in $A$.
2017 Czech And Slovak Olympiad III A, 4
For each sequence of $n$ zeros and $n$ units, we assign a number that is a number sections of the same digits in it. (For example, sequence $00111001$ has $4$ such sections $00, 111,00, 1$.) For a given $n$ we sum up all the numbers assigned to each such sequence. Prove that the sum total is equal to $(n+1){2n \choose n} $
1992 China Team Selection Test, 1
16 students took part in a competition. All problems were multiple choice style. Each problem had four choices. It was said that any two students had at most one answer in common, find the maximum number of problems.
2017 IMC, 4
There are $n$ people in a city, and each of them has exactly $1000$ friends (friendship is always symmetric). Prove that it is possible to select a group $S$ of people such that at least $\frac{n}{2017}$ persons in $S$ have exactly two friends in $S$.
2011 Middle European Mathematical Olympiad, 4
Let $n \geq 3$ be an integer. At a MEMO-like competition, there are $3n$ participants, there are n languages spoken, and each participant speaks exactly three different languages. Prove that at least $\left\lceil\frac{2n}{9}\right\rceil$ of the spoken languages can be chosen in such a way that no participant speaks more than two of the chosen languages.
[b]Note.[/b] $\lceil x\rceil$ is the smallest integer which is greater than or equal to $x$.
2005 Kurschak Competition, 2
A and B play tennis. The player to first win at least four points and at least two more than the other player wins. We know that A gets a point each time with probability $p\le \frac12$, independent of the game so far. Prove that the probability that A wins is at most $2p^2$.
2007 Germany Team Selection Test, 2
Let $ S$ be a finite set of points in the plane such that no three of them are on a line. For each convex polygon $ P$ whose vertices are in $ S$, let $ a(P)$ be the number of vertices of $ P$, and let $ b(P)$ be the number of points of $ S$ which are outside $ P$. A line segment, a point, and the empty set are considered as convex polygons of $ 2$, $ 1$, and $ 0$ vertices respectively. Prove that for every real number $ x$ \[\sum_{P}{x^{a(P)}(1 \minus{} x)^{b(P)}} \equal{} 1,\] where the sum is taken over all convex polygons with vertices in $ S$.
[i]Alternative formulation[/i]:
Let $ M$ be a finite point set in the plane and no three points are collinear. A subset $ A$ of $ M$ will be called round if its elements is the set of vertices of a convex $ A \minus{}$gon $ V(A).$ For each round subset let $ r(A)$ be the number of points from $ M$ which are exterior from the convex $ A \minus{}$gon $ V(A).$ Subsets with $ 0,1$ and 2 elements are always round, its corresponding polygons are the empty set, a point or a segment, respectively (for which all other points that are not vertices of the polygon are exterior). For each round subset $ A$ of $ M$ construct the polynomial
\[ P_A(x) \equal{} x^{|A|}(1 \minus{} x)^{r(A)}.
\]
Show that the sum of polynomials for all round subsets is exactly the polynomial $ P(x) \equal{} 1.$
[i]Proposed by Federico Ardila, Colombia[/i]
2016 International Zhautykov Olympiad, 3
There are $60$ towns in $Graphland$ every two countries of which are connected by only a directed way. Prove that we can color four towns to red and four towns to green such that every way between green and red towns are directed from red to green
1987 IMO Shortlist, 17
Prove that there exists a four-coloring of the set $M = \{1, 2, \cdots, 1987\}$ such that any arithmetic progression with $10$ terms in the set $M$ is not monochromatic.
[b][i]Alternative formulation[/i][/b]
Let $M = \{1, 2, \cdots, 1987\}$. Prove that there is a function $f : M \to \{1, 2, 3, 4\}$ that is not constant on every set of $10$ terms from $M$ that form an arithmetic progression.
[i]Proposed by Romania[/i]
2016 Serbia National Math Olympiad, 5
There are $2n-1$ twoelement subsets of set $1,2,...,n$. Prove that one can choose $n$ out of these such that their union contains no more than $\frac{2}{3}n+1$ elements.
1992 China Team Selection Test, 1
16 students took part in a competition. All problems were multiple choice style. Each problem had four choices. It was said that any two students had at most one answer in common, find the maximum number of problems.
2023 China Team Selection Test, P11
Let $n\in\mathbb N_+.$ For $1\leq i,j,k\leq n,a_{ijk}\in\{ -1,1\} .$ Prove that: $\exists x_1,x_2,\cdots ,x_n,y_1,y_2,\cdots ,y_n,z_1,z_2,\cdots ,z_n\in \{-1,1\} ,$ satisfy
$$\left| \sum\limits_{i=1}^n\sum\limits_{j=1}^n\sum\limits_{k=1}^na_{ijk}x_iy_jz_k\right| >\frac {n^2}3.$$
[i]Created by Yu Deng[/i]
2016 Azerbaijan BMO TST, 2
There are $100$ students who praticipate at exam.Also there are $25$ members of jury.Each student is checked by one jury.Known that every student likes $10$ jury
$a)$ Prove that we can select $7$ jury such that any student likes at least one jury.
$b)$ Prove that we can make this every student will be checked by the jury that he likes and every jury will check at most $10$ students.
2001 USA Team Selection Test, 3
For a set $S$, let $|S|$ denote the number of elements in $S$. Let $A$ be a set of positive integers with $|A| = 2001$. Prove that there exists a set $B$ such that
(i) $B \subseteq A$;
(ii) $|B| \ge 668$;
(iii) for any $u, v \in B$ (not necessarily distinct), $u+v \not\in B$.
2015 China Team Selection Test, 2
Let $X$ be a non-empty and finite set, $A_1,...,A_k$ $k$ subsets of $X$, satisying:
(1) $|A_i|\leq 3,i=1,2,...,k$
(2) Any element of $X$ is an element of at least $4$ sets among $A_1,....,A_k$.
Show that one can select $[\frac{3k}{7}] $ sets from $A_1,...,A_k$ such that their union is $X$.
2009 Germany Team Selection Test, 3
Let $ S \equal{} \{x_1, x_2, \ldots, x_{k \plus{} l}\}$ be a $ (k \plus{} l)$-element set of real numbers contained in the interval $ [0, 1]$; $ k$ and $ l$ are positive integers. A $ k$-element subset $ A\subset S$ is called [i]nice[/i] if
\[ \left |\frac {1}{k}\sum_{x_i\in A} x_i \minus{} \frac {1}{l}\sum_{x_j\in S\setminus A} x_j\right |\le \frac {k \plus{} l}{2kl}\]
Prove that the number of nice subsets is at least $ \dfrac{2}{k \plus{} l}\dbinom{k \plus{} l}{k}$.
[i]Proposed by Andrey Badzyan, Russia[/i]
2015 Romania Team Selection Tests, 3
Let $n$ be a positive integer . If $\sigma$ is a permutation of the first $n$ positive integers , let $S(\sigma)$ be the set of all distinct sums of the form $\sum_{i=k}^{l} \sigma(i)$ where $1 \leq k \leq l \leq n$ .
[b](a)[/b] Exhibit a permutation $\sigma$ of the first $n$ positive integers such that $|S(\sigma)|\geq \left \lfloor{\frac{(n+1)^2}{4}}\right \rfloor $.
[b](b)[/b] Show that $|S(\sigma)|>\frac{n\sqrt{n}}{4\sqrt{2}}$ for all permutations $\sigma$ of the first $n$ positive integers .
1996 All-Russian Olympiad, 4
In the Duma there are 1600 delegates, who have formed 16000 committees of 80 persons each. Prove that one can find two committees having no fewer than four common members.
[i]A. Skopenkov[/i]
2009 Belarus Team Selection Test, 3
Let $ S \equal{} \{x_1, x_2, \ldots, x_{k \plus{} l}\}$ be a $ (k \plus{} l)$-element set of real numbers contained in the interval $ [0, 1]$; $ k$ and $ l$ are positive integers. A $ k$-element subset $ A\subset S$ is called [i]nice[/i] if
\[ \left |\frac {1}{k}\sum_{x_i\in A} x_i \minus{} \frac {1}{l}\sum_{x_j\in S\setminus A} x_j\right |\le \frac {k \plus{} l}{2kl}\]
Prove that the number of nice subsets is at least $ \dfrac{2}{k \plus{} l}\dbinom{k \plus{} l}{k}$.
[i]Proposed by Andrey Badzyan, Russia[/i]