If $T_{n} = 1 + 2 + 3 + \ldots + n$ and \[P_{n} = \frac{T_{2}}{T_{2} - 1} \cdot \frac{T_{3}}{T_{3} - 1} \cdot \frac{T_{4}}{T_{4} - 1} \cdot\,\, \cdots \,\,\cdot \frac{T_{n}}{T_{n} - 1}\quad\text{for }n = 2,3,4,\ldots,\] then $P_{1991}$ is closest to which of the following numbers? $ \textbf{(A)}\ 2.0\qquad\textbf{(B)}\ 2.3\qquad\textbf{(C)}\ 2.6\qquad\textbf{(D)}\ 2.9\qquad\textbf{(E)}\ 3.2 $

There are $ n$ markers, each with one side white and the other side black. In the beginning, these $ n$ markers are aligned in a row so that their white sides are all up. In each step, if possible, we choose a marker whose white side is up (but not one of the outermost markers), remove it, and reverse the closest marker to the left of it and also reverse the closest marker to the right of it. Prove that, by a finite sequence of such steps, one can achieve a state with only two markers remaining if and only if $ n \minus{} 1$ is not divisible by $ 3$. [i]Proposed by Dusan Dukic, Serbia[/i]

Find all primes $p,q,r$ and natural numbers $n$ such that $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{n}$.

Let $U$ be an open subset of the complex plane $\mathbb{C}$ including $\mathbb{D}=\{z \in \mathbb{C} : |z| \le 1\}$ and $f$ be analytic over $U$. Prove that if for every $z$ with a complex norm equal to $1$($|z|=1$) we have $0<Re(\bar{z}f(z))$, then $f$ has only one root in $\mathbb{D}$ and that's simple.

In a conference $2n$ personalities take apart. Each person has at least $n$ acquaintaces among the others. Prove that it is possible to quarter the participants into two-person rooms, so that each participant would share the room with his/her acquaintace.

A piece of paper in the shape of a square $FBHD$ with side $a$ is given. Points $G,A$ on $FB$ and $E,C$ on $BH$ are marked so that $FG=GA=AB$ and $BE=EC=CH$. The paper is folded along $DG,DA,DC$ and $AC$ so that $G$ overlaps with $B$, and $F$ and $H$ overlap with $E$. Compute the volume of the obtained tetrahedron $ABCD$.

Let $x, y \in \mathbb{R}^+$. Prove that: \[ \left( 1 + \frac{1}{x} \right) \left( 1 + \frac{1}{y} \right) \geq \left( 1 + \frac{2}{x + y} \right)^2. \]

Points $L$ and $H$ are marked on the sides $AB$ of an acute-angled triangle ABC so that $CL$ is a bisector and $CH$ is an altitude. Let $P,Q$ be the feet of the perpendiculars from $L$ to $AC$ and $BC$ respectively. Prove that $AP \cdot BH = BQ \cdot AH$. I. Gorodnin

Prove that $$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \ge \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+2(a+b+c)$$ for the all $a,b,c$ positive real numbers satisfying $a^2+b^2+c^2+2abc \le 1$.

Let $n$ be a positive integer. Given a sequence $\varepsilon_1$, $\dots$, $\varepsilon_{n - 1}$ with $\varepsilon_i = 0$ or $\varepsilon_i = 1$ for each $i = 1$, $\dots$, $n - 1$, the sequences $a_0$, $\dots$, $a_n$ and $b_0$, $\dots$, $b_n$ are constructed by the following rules: \[a_0 = b_0 = 1, \quad a_1 = b_1 = 7,\] \[\begin{array}{lll} a_{i+1} = \begin{cases} 2a_{i-1} + 3a_i, \\ 3a_{i-1} + a_i, \end{cases} & \begin{array}{l} \text{if } \varepsilon_i = 0, \\ \text{if } \varepsilon_i = 1, \end{array} & \text{for each } i = 1, \dots, n - 1, \\[15pt] b_{i+1}= \begin{cases} 2b_{i-1} + 3b_i, \\ 3b_{i-1} + b_i, \end{cases} & \begin{array}{l} \text{if } \varepsilon_{n-i} = 0, \\ \text{if } \varepsilon_{n-i} = 1, \end{array} & \text{for each } i = 1, \dots, n - 1. \end{array}\] Prove that $a_n = b_n$. [i]Proposed by Ilya Bogdanov, Russia[/i]

Show that \[ (2a-1) \sin x + (1-a) \sin(1-a)x \geq 0 \] for $0 \leq a \leq 1$ and $0 \leq x \leq \pi$.

Consider a latin square of size $n$. We are allowed to choose a $1 \times 1$ square in the table, and add $1$ to any number on the same row and column as the chosen square (the original square will be counted aswell) , or we can add $-1$ to all of them instead. Can we with doing finitly many operation , reach any latin square of size $n?$

Manzi has $n$ stamps and an album with $10$ pages. He distributes the $n$ stamps in the album such that each page has a distinct number of stamps. He finds that, no matter how he does this, there is always a set of $4$ pages such that the total number of stamps in these $4$ pages is at least $\frac{n}{2}$. Determine the maximum possible value of $n$.

Sergei chooses two different natural numbers $a$ and $b$. He writes four numbers in a notebook: $a$, $a+2$, $b$ and $b+2$. He then writes all six pairwise products of the numbers of notebook on the blackboard. Let $S$ be the number of perfect squares on the blackboard. Find the maximum value of $S$. [i]S. Berlov[/i]

Each cell of a $100$ x $100$ board is painted in one of $100$ different colors so that there are exactly $100$ cells of each color. Prove that there is a row or column in which there are at least $10$ cells of different colors.

Integers $x,y,z,t$ satisfy $x^2+y^2=z^2+t^2$and$xy=2zt$ prove that $xyzt=0$ Proposed by $M. Abduvaliev$

Consider constructing a tower of tables of numbers as follows. The first table is a one by one array containing the single number $1$. The second table is a two by two array formed underneath the first table and built as followed. For each entry, we look at the terms in the previous table that are directly up and to the left, up and to the right, and down and to the right of the entry, and we fill that entry with the sum of the numbers occurring there. If there happens to be no term at a particular location, it contributes a value of zero to the sum. [img]https://cdn.artofproblemsolving.com/attachments/d/8/ab56dddfc23e84348e205f031001d157cb8386.png[/img] The diagram above shows how we compute the second table from the first. The diagram below shows how to then compute the third table from the second. [img]https://cdn.artofproblemsolving.com/attachments/9/3/e1d8cf0fd0b71b970625a4fa97bc2912492a78.png[/img] For example, the entry in the middle row and middle column of the third table is equal the sum of the top left entry $1$, the top right entry $0$, and the bottom right entry $1$ from the second table, which is just $2$. Similarly, to compute the bottom rightmost entry in the third table, we look above it to the left and see that the entry in the second table’s bottom rightmost entry is $1$. There are no entries from the second table above it and to the right or below it and to the right, so we just take this entry in the third table to be $1$. We continue constructing the tower by making more tables from the previous tables. Find the entry in the third (from the bottom) row of the third (from the left) column of the tenth table in this resulting tower.

$AB,AC$ are tangent to $\Omega$ at $B$ and $C$, respectively. $D,E,F$ lie on segments $BC,CA,AB$ such that $AF<AE$ and $\angle FDB= \angle EDC$. The circumcircle of $\triangle FEC$ intersects $\Omega$ at $G$ and $C$. Show that $ \angle AEF= \angle BGD$

Let $ABC$ be a triangle in which $\angle BAC = 60^{\circ}$ . Let $P$ (similarly $Q$) be the point of intersection of the bisector of $\angle ABC$(similarly of $\angle ACB$) and the side $AC$(similarly $AB$). Let $r_1$ and $r_2$ be the in-radii of the triangles $ABC$ and $AP Q$, respectively. Determine the circum-radius of $APQ$ in terms of $r_1$ and $r_2$.

In a $44\times 44$ board, some of the cells are blue, and the rest are red. No blue cells borders another blue cell on the side. The red cells, on the other hand, form a connected component (one may get from any red cell to any other red cell only by traversing edge-adjacent red cells). Prove that less than one third of the cells on the board are blue. [i]Proposed by B. Frenkin[/i]

Point $X$ is chosen inside a convex $ABCD$ so that $\angle XBC = \angle XAD, \angle XCB = \angle XDA$. Rays $AB, DC$ intersect at point $O$, circumcircles of triangles $BCO, ADO$ intersect at point $T$. Prove that line $TX$ and the line through $O$ perpendicular to $BC$ intersect on the circumcircle of $\triangle AOD$. [i]Proposed by Anton Trygub[/i]

The diagonals of a convex quadrilateral $ABCD$ intersect at $M$. The bisector of $\angle ACD$ intersects the ray $BA$ at $K$. Prove that if $MA\cdot MC + MA\cdot CD = MB \cdot MD $, then $\angle BKC = \angle BDC$

The set of numbers $M=\{4k-3 | k\in N\}$ is considered. A number of of this set is called “simple” if it is impossible to put in the form of a product of numbers from $M$ other than $1$. Show that in this set, the decomposition of numbers in the product of "simple" factors is ambiguous.

On a party with $99$ guests, hosts Ann and Bob play a game (the hosts are not regarded as guests). There are $99$ chairs arranged in a circle; initially, all guests hang around those chairs. The hosts take turns alternately. By a turn, a host orders any standing guest to sit on an unoccupied chair $c$. If some chair adjacent to $c$ is already occupied, the same host orders one guest on such chair to stand up (if both chairs adjacent to $c$ are occupied, the host chooses exactly one of them). All orders are carried out immediately. Ann makes the first move; her goal is to fulfill, after some move of hers, that at least $k$ chairs are occupied. Determine the largest $k$ for which Ann can reach the goal, regardless of Bob's play.

Given are two integers $a>b>1$ such that $a+b \mid ab+1$ and $a-b \mid ab-1$. Prove that $a<\sqrt{3}b$.