Given a regular $n$-sided polygon with $n \ge 3$. Maria draws some of its diagonals in such a way that each diagonal intersects at most one of the other diagonals drawn in the interior of the polygon. Determine the maximum number of diagonals that Maria can draw in such a way. Note: Two diagonals can share a vertex of the polygon. Vertices are not part of the interior of the polygon.

An integer $n>1$ , whose positive divisors are $1=d_1<d_2< \cdots <d_k=n$, is called $\textit{southern}$ if all the numbers $d_2-d_1, d_3- d_2 , \cdots, d_k-d_{k-1}$ are divisors of $n$. a) Find a positive integer that is $\textit{not southern}$ and has exactly $2022$ positive divisors that are $\textit{southern}$. b) Show that there are infinitely many positive integers that are $\textit{not southern}$ and have exactly $2022$ positive divisors that are $\textit{southern}$.

There were $2n,$ $n\ge2,$ teams in a tournament. Each team played against every other team once without draws. A team gets 0 points for a loss and gets as many points for a win as its current number of losses. For which $n$ all the teams could end up with the same non-zero number of points?

Let $S_1$ and $S_2$ be two circumferences, with centers $O_1$ and $O_2$ respectively, and secants on $M$ and $N$. The line $t$ is the common tangent to $S_1$ and $S_2$ closer to $M$. The points $A$ and $B$ are the intersection points of $t$ with $S_1$ and $S_2$, $C$ is the point such that $BC$ is a diameter of $S_2$, and $D$ the intersection point of the line $O_1O_2$ with the perpendicular line to $AM$ through $B$. Show that $M$, $D$ and $C$ are collinear.

For every positive integer $n$ we take the greatest divisor $d$ of $n$ such that $d\leq \sqrt{n}$ and we define $a_n=\frac{n}{d}-d$. Prove that in the sequence $a_1,a_2,a_3,...$, any non negative integer $k$ its in the sequence infinitely many times.

Let $n$ be a positive integer. Two players $A$ and $B$ play a game in which they take turns choosing positive integers $k \le n$. The rules of the game are: (i) A player cannot choose a number that has been chosen by either player on any previous turn. (ii) A player cannot choose a number consecutive to any of those the player has already chosen on any previous turn. (iii) The game is a draw if all numbers have been chosen; otherwise the player who cannot choose a number anymore loses the game. The player $A$ takes the first turn. Determine the outcome of the game, assuming that both players use optimal strategies. [i]Proposed by Finland[/i]

The difference in the areas of two similar triangles is $18$ square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triange, in square feet, is an integer, and one of its sides is $3$ feet. The corresponding side of the larger triangle, in feet, is: $\textbf{(A)}\ 12\quad \textbf{(B)}\ 9\qquad \textbf{(C)}\ 6\sqrt{2}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 3\sqrt{2}$

Which of the following numbers is largest? $ \text{(A)}\ 9.12344\qquad\text{(B)}\ 9.123\overline{4}\qquad\text{(C)}\ 9.12\overline{34}\qquad\text{(D)}\ 9.1\overline{234}\qquad\text{(E)}\ 9.\overline{1234} $

Let $\triangle ABC$ be an acute triangle with $AB>AC$, let $I$ be the center of the incircle. Let $M,N$ be the midpoint of $AC$ and $AB$ respectively. $D,E$ are on $AC$ and $AB$ respectively such that $BD\parallel IM$ and $CE\parallel IN$. A line through $I$ parallel to $DE$ intersects $BC$ in $P$. Let $Q$ be the projection of $P$ on line $AI$. Prove that $Q$ is on the circumcircle of $\triangle ABC$.

In $\triangle ABC$ points $M,O$ are midpoint of $AB$ and circumcenter. It is true, that $OM=R-r$. Bisector of external $\angle A$ intersect $BC$ at $D$ and bisector of external $\angle C$ intersect $AB$ at $E$. Find possible values of $\angle CED$ [i]D. Shiryaev [/i]

Players $A$ and $B$ play a game on a blackboard that initially contains 2020 copies of the number 1 . In every round, player $A$ erases two numbers $x$ and $y$ from the blackboard, and then player $B$ writes one of the numbers $x+y$ and $|x-y|$ on the blackboard. The game terminates as soon as, at the end of some round, one of the following holds: $(1)$ one of the numbers on the blackboard is larger than the sum of all other numbers; $(2)$ there are only zeros on the blackboard. Player $B$ has to pay to player $A$ an amount in reais equivalent to the quantity of numbers left on the blackboard after the game ends. Show that player $A$ can earn at least 8 reais regardless of the moves taken by $B$ Ps.: Easier version of [url = https://artofproblemsolving.com/community/c6h2625868p22698110]ISL 2020 C8[/url]

For a positive integer $n$, let function $f(n)$ denote the number of positive integers $a\leq n$ such that $\gcd(a,n) = \gcd(a+1,n) = 1$. Find the sum of all $n$ such that $f(n)=15$. [i]Proposed by Harry Kim[/i]

Martin invented the following algorithm. Let two irreducible fractions $ \frac{s_1}{t_1}$ and $ \frac{s_2}{t_2}$ be given as inputs, with the numerators and denominators being positive integers. Divide $ s_1$ and $ s_2$ by their greatest common divisor $ c$ and obtain $ a_1$ and $ a_2$, respectively. Similarily, divide $ t_1$ and $ t_2$ by their greatest common divisor $ d$ and obtain $ b_1$ and $ b_2$, respectively. After that, form a new fraction $ \frac{a_1b_2 \plus{} a_2b_1}{t_1b_2}$, reduce it, and multiply the numerator of the result by $ c$. Martin claims that this algorithm always finds the sum of the original fractions as an irreducible fraction. Is his claim correct?

Consider an acute non-isosceles triangle. In a single step it is allowed to cut any one of the available triangles into two triangles along its median. Is it possible that after a finite number of cuttings all triangles will be isosceles? [i]Proposed by E. Bakaev[/i]

The state income tax where Kristin lives is levied at the rate of $ p\%$ of the first $ \$28000$ of annual income plus $ (p \plus{} 2)\%$ of any amount above $ \$28000$. Kristin noticed that the state income tax she paid amounted to $ (p \plus{} 0.25)\%$ of her annual income. What was her annual income? $ \textbf{(A) }\$28000\qquad\textbf{(B) }\$32000\qquad\textbf{(C) }\$35000\qquad\textbf{(D) }\$42,000\qquad\textbf{(E) }\$56000$

Two circles $O_1$ and $O_2$ intersect at $A,B$. Bisector of outer angle $\angle O_1AO_2$ intersects $O_1$ at $C$, $O_2$ at $D$. $P$ is a point on $\odot(BCD)$, $CP\cap O_1=E,DP\cap O_2=F$. Prove that $PE=PF$.

Let $(a_n)_{n=1}^{\infty}$ be a real sequence such that $a_1=1, a_3=4$ and for every $n\geq 2$, $a_{n+1}+a_{n-1}=2a_n+1$. What is $a_{2011}$? $\textbf{(A)}\ 2^{2010} \qquad\textbf{(B)}\ 2021056 \qquad\textbf{(C)}\ 1010528 \qquad\textbf{(D)}\ 3016 \qquad\textbf{(E)}\ 2011$