Found problems: 85335
2003 ITAMO, 4
There are two sorts of people on an island: [i]knights[/i], who always talk truth, and [i]scoundrels[/i], who always lie. One day, the people establish a council consisting of $2003$ members. They sit around a round table, and during the council each member said: "Both my neighbors are scoundrels". In a later day, the council meets again, but one member could not come due to illness, so only $2002$ members were present. They sit around the round table, and everybody said: "Both my neighbors belong to the sort different from mine". Is the absent member a knight or a scoundrel?
2011 Germany Team Selection Test, 1
Two circles $\omega , \Omega$ intersect in distinct points $A,B$ a line through $B$ intersects $\omega , \Omega$ in $C,D$ respectively such that $B$ lies between $C,D$ another line through $B$ intersects $\omega , \Omega$ in $E,F$ respectively such that $E$ lies between $B,F$ and $FE=CD$. Furthermore $CF$ intersects $\omega , \Omega$ in $P,Q$ respectively and $M,N$ are midpoints of the arcs $PB,QB$. Prove that $CNMF$ is a cyclic quadrilateral.
2003 National High School Mathematics League, 7
The solution set for inequality $|x|^3-2x^2-4|x|+3<0$ is________.
2007 Pan African, 2
For which positive integers $n$ is $231^n-222^n-8 ^n -1$ divisible by $2007$?
2020-21 KVS IOQM India, 5
Find the number of positive integers $n$ such that the highest power of $7$ dividing $n!$ is $8$.
2025 Benelux, 4
Let $a_0, a_1, \ldots, a_{10}$ be integers such that, for each $i \in \{0,1,\ldots,2047\}$, there exists a subset $S \subseteq \{0,1,\ldots,10\}$ with
\[
\sum_{j \in S} a_j \equiv i \pmod{2048}.
\]
Show that for each $i \in \{0,1,\ldots,10\}$, there is exactly one $j \in \{0,1,\ldots,10\}$ such that $a_j$ is divisible by $2^i$ but not by $2^{i+1}$.
Note: $\sum_{j \in S} a_j$ is the summation notation, for instance, $\sum_{j \in \{2,5\}} a_j = a_2 + a_5$, while for the empty set $\varnothing$, one defines $\sum_{j \in \varnothing} a_j = 0$.
1998 IMO Shortlist, 6
Let $ABCDEF$ be a convex hexagon such that $\angle B+\angle D+\angle F=360^{\circ }$ and \[ \frac{AB}{BC} \cdot \frac{CD}{DE} \cdot \frac{EF}{FA} = 1. \] Prove that \[ \frac{BC}{CA} \cdot \frac{AE}{EF} \cdot \frac{FD}{DB} = 1. \]
2010 Sharygin Geometry Olympiad, 5
Let $AH$, $BL$ and $CM$ be an altitude, a bisectrix and a median in triangle $ABC$. It is known that lines $AH$ and $BL$ are an altitude and a bisectrix of triangle $HLM$. Prove that line $CM$ is a median of this triangle.
2023 Myanmar IMO Training, 6
Let points $M$ and $N$ lie on sides $AB$ and $BC$ of triangle $ABC$ in such a way that $MN||AC$. Points $M'$ and $N'$ are the reflections of $M$ and $N$ about $BC$ and $AB$ respectively. Let $M'A$ meet $BC$ at $X$, and let $N'C$ meet $AB$ at $Y$. Prove that $A,C,X,Y$ are concyclic.
V Soros Olympiad 1998 - 99 (Russia), 11.3
For what a from the interval $[0,\pi]$ do there exist $a$ and $b$ that are not simultaneously equal to zero, for which the inequality
$$a \cos x + b \cos 2x \le 0$$ is satisfied for all $x$ belonging to the segment $[a, \pi]$?
1994 India Regional Mathematical Olympiad, 4
Solve the system of equations for real $x$ and $y$: \begin{eqnarray*} 5x \left( 1 + \frac{1}{x^2 + y^2}\right) &=& 12 \\ 5y \left( 1 - \frac{1}{x^2+y^2} \right) &=& 4 . \end{eqnarray*}
2018 Germany Team Selection Test, 1
A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even.
[i]Proposed by Jeck Lim, Singapore[/i]
2025 District Olympiad, P3
Determine all positive real numbers $a,b,c,d$ such that $a+b+c+d=80$ and $$a+\frac{b}{1+a}+\frac{c}{1+a+b}+\frac{d}{1+a+b+c}=8.$$
2011 ELMO Shortlist, 1
Let $S$ be a finite set, and let $F$ be a family of subsets of $S$ such that
a) If $A\subseteq S$, then $A\in F$ if and only if $S\setminus A\notin F$;
b) If $A\subseteq B\subseteq S$ and $B\in F$, then $A\in F$.
Determine if there must exist a function $f:S\to\mathbb{R}$ such that for every $A\subseteq S$, $A\in F$ if and only if
\[\sum_{s\in A}f(s)<\sum_{s\in S\setminus A}f(s).\]
[i]Evan O'Dorney.[/i]
2016 All-Russian Olympiad, 4
There is three-dimensional space. For every integer $n$ we build planes $ x \pm y\pm z = n$. All space is divided on octahedrons and tetrahedrons.
Point $(x_0,y_0,z_0)$ has rational coordinates but not lies on any plane. Prove, that there is such natural $k$ , that point $(kx_0,ky_0,kz_0)$ lies strictly inside the octahedron of partition.
2021 AIME Problems, 10
Consider the sequence $(a_k)_{k\ge 1}$ of positive rational numbers defined by $a_1 = \frac{2020}{2021}$ and for $k\ge 1$, if $a_k = \frac{m}{n}$ for relatively prime positive integers $m$ and $n$, then
\[a_{k+1} = \frac{m + 18}{n+19}.\]
Determine the sum of all positive integers $j$ such that the rational number $a_j$ can be written in the form $\frac{t}{t+1}$ for some positive integer $t$.
2017 NIMO Problems, 1
In the diagram below, how many rectangles can be drawn using the grid lines which contain none of the letters $N$, $I$, $M$, $O$?
[asy]
size(4cm);
for(int i=0;i<6;++i)draw((i,0)--(i,5)^^(0,i)--(5,i));
label("$N$", (1.5, 2.5));
label("$I$", (2.5, 3.5));
label("$M$", (3.5, 2.5));
label("$O$", (2.5, 1.5));
[/asy]
[i]Proposed by Michael Tang[/i]
1987 Traian Lălescu, 1.1
Let be a natural number $ n. $ Show that:
[b]a.[/b] $ \left( 1+1/n \right)^k<1+k/n+k^2/n^2, $ for all naturals $ k $ with $ 1\le k\le n. $
[b]b.[/b] $ 3^n\cdot n!>(1+n)^n. $
1991 AMC 12/AHSME, 12
The measures (in degrees) of the interior angles of a convex hexagon form an arithmetic sequence of positive integers. Let $m^{\circ}$ be the measure of the largest interior angle of the hexagon. The largest possible value of $m^{\circ}$ is
$ \textbf{(A)}\ 165^{\circ}\qquad\textbf{(B)}\ 167^{\circ}\qquad\textbf{(C)}\ 170^{\circ}\qquad\textbf{(D)}\ 175^{\circ}\qquad\textbf{(E)}\ 179^{\circ} $
1997 Mexico National Olympiad, 5
Let $P,Q,R$ be points on the sides $BC,CA,AB$ respectively of a triangle $ABC$. Suppose that $BQ$ and $CR$ meet at $A', AP$ and $CR$ meet at $B'$, and $AP$ and $BQ$ meet at $C'$, such that $AB' = B'C', BC' =C'A'$, and $CA'= A'B'$. Compute the ratio of the area of $\triangle PQR$ to the area of $\triangle ABC$.
1993 AMC 12/AHSME, 28
How many triangles with positive area are there whose vertices are points in the $xy$-plane whose coordinates are integers $(x,y)$ satisfying $1 \le x \le 4$ and $1 \le y \le 4$?
$ \textbf{(A)}\ 496 \qquad\textbf{(B)}\ 500 \qquad\textbf{(C)}\ 512 \qquad\textbf{(D)}\ 516 \qquad\textbf{(E)}\ 560 $
2016 Taiwan TST Round 3, 2
Determine all functions $f:\mathbb{R}^+\rightarrow \mathbb{R}^+$ satisfying
$f(x+y+f(y))=4030x-f(x)+f(2016y), \forall x,y \in \mathbb{R}^+$.
2011 Indonesia MO, 1
For a number $n$ in base $10$, let $f(n)$ be the sum of all numbers possible by removing some digits of $n$ (including none and all). For example, if $n = 1234$, $f(n) = 1234 + 123 + 124 + 134 + 234 + 12 + 13 + 14 + 23 + 24 + 34 + 1 + 2 + 3 + 4 = 1979$; this is formed by taking the sums of all numbers obtained when removing no digit from $n$ (1234), removing one digit from $n$ (123, 124, 134, 234), removing two digits from $n$ (12, 13, 14, 23, 24, 34), removing three digits from $n$ (1, 2, 3, 4), and removing all digits from $n$ (0). If $p$ is a 2011-digit integer, prove that $f(p)-p$ is divisible by $9$.
Remark: If a number appears twice or more, it is counted as many times as it appears. For example, with the number $101$, $1$ appears three times (by removing the first digit, giving $01$ which is equal to $1$, removing the first two digits, or removing the last two digits), so it is counted three times.
2025 Thailand Mathematical Olympiad, 6
Find all function $f: \mathbb{R}^+ \rightarrow \mathbb{R}$,such that the inequality $$f(x) + f\left(\frac{y}{x}\right) \leqslant \frac{x^3}{y^2} + \frac{y}{x^3}$$ holds for all positive reals $x,y$ and for every positive real $x$, there exist positive reals $y$, such that the equality holds.
2007 Mathematics for Its Sake, 3
Solve in the real numbers the equation $ \lfloor ax \rfloor -\lfloor (1+a)x \rfloor = (1+a)(1-x) . $
[i]Dumitru Acu[/i]