The keyword that Ana Viso chose for her computer has the $7$ characters of her name: A, N, A, V, I, S, O. Sorting all the different words alphabetically formed by all these $7$ characters, Ana's keyword appears in the $881$st position. What it is Ana's keyword?

Let $N$ be the sum of the first four three-digit prime numbers. Find the sum of the prime factors of $\tfrac{N}2$.

For a positive integer $n$, there are $n$ positive reals $a_1 \ge a_2 \ge a_3 \cdots \ge a_n$. For all positive reals $b_1, b_2, \cdots b_n$, prove the following inequality. $$\frac{a_1b_1+a_2b_2 + \cdots +a_nb_n}{a_1+a_2+ \cdots a_n} \le \text{max}\{ \frac{b_1}{1}, \frac{b_1+b_2}{2}, \cdots, \frac{b_1+b_2+ \cdots +b_n}{n} \}$$

In the plane, let $a$, $b$ be two closed broken lines (possibly self-intersecting), and $K$, $L$, $M$, $N$ be four points. The vertices of $a$, $b$ and the points $K$ $L$, $M$, $N$ are in general position (i.e. no three of these points are collinear, and no three segments between them concur at an interior point). Each of segments $KL$ and $MN$ meets $a$ at an even number of points, and each of segments $LM$ and $NK$ meets $a$ at an odd number of points. Conversely, each of segments $KL$ and $MN$ meets $b$ at an odd number of points, and each of segments $LM$ and $NK$ meets $b$ at an even number of points. Prove that $a$ and $b$ intersect.

The expressions $ a \plus{} b \plus{} c, ab \plus{} ac \plus{} bc,$ and $ abc$ are called the elementary symmetric expressions on the three letters $ a, b, c;$ symmetric because if we interchange any two letters, say $ a$ and $ c,$ the expressions remain algebraically the same. The common degree of its terms is called the order of the expression. Let $ S_k(n)$ denote the elementary expression on $ k$ different letters of order $ n;$ for example $ S_4(3) \equal{} abc \plus{} abd \plus{} acd \plus{} bcd.$ There are four terms in $ S_4(3).$ How many terms are there in $ S_{9891}(1989)?$ (Assume that we have $ 9891$ different letters.)

There are $11$ quadratic equations on the board, where each coefficient is replaced by a star. Initially, each of them looks like this $$\star x^2 + \star x + \star= 0.$$ Two players are playing a game making alternating moves. In one move each ofthem replaces one star with a real nonzero number. The first player tries to make as many equations as possible without roots and the second player tries to make the number of equations without roots as small as possible. What is the maximal number of equations without roots that the first player can achieve if the second player plays to her best? Describe the strategies of both players.

Let $ABCD$ be a cyclic quadrilateral with $AB$ and $CD$ not parallel. Let $M$ be the midpoint of $CD.$ Let $P$ be a point inside $ABCD$ such that $P A = P B = CM.$ Prove that $AB, CD$ and the perpendicular bisector of $MP$ are concurrent.

Let $ABC$ be a triangle. The points $K, L,$ and $M$ lie on the segments $BC, CA,$ and $AB,$ respectively, such that the lines $AK, BL,$ and $CM$ intersect in a common point. Prove that it is possible to choose two of the triangles $ALM, BMK,$ and $CKL$ whose inradii sum up to at least the inradius of the triangle $ABC$. [i]Proposed by Estonia[/i]

Given are the lines $l_1,l_2,\ldots ,l_k$ in the plane, no two of which are parallel and no three of which are concurrent. For which $k$ can one label the intersection points of these lines by $1, 2,\ldots , k-1$ so that in each of the given lines all the labels appear exactly once?

Let the inscribed circle $(I)$ of the triangle $ABC$, touches $CA, AB$ at $E, F$. $P$ moves along $EF$, $PB$ cuts $CA$ at $M, MI$ cuts the line, through $C$ perpendicular to $AC$, at $N$. Prove that the line through $N$ is perpendicular to $PC$ crosses a fixed point as $P$ moves. Tran Quang Hung, High School of Natural Sciences, Hanoi National University

A frog is placed on each cell of a $n \times n$ square inside an infinite chessboard (so initially there are a total of $n \times n$ frogs). Each move consists of a frog $A$ jumping over a frog $B$ adjacent to it with $A$ landing in the next cell and $B$ disappearing (adjacent means two cells sharing a side). Prove that at least $ \left[\frac{n^2}{3}\right]$ moves are needed to reach a configuration where no more moves are possible.

Consider a rectangle $ABCD$ with $AB = 2$ and $BC = \sqrt3$. The point $M$ lies on the side $AD$ so that $MD = 2 AM$ and the point $N$ is the midpoint of the segment $AB$. On the plane of the rectangle rises the perpendicular MP and we choose the point $Q$ on the segment $MP$ such that the measure of the angle between the planes $(MPC)$ and $(NPC)$ shall be $45^o$, and the measure of the angle between the planes $(MPC)$ and $(QNC)$ shall be $60^o$. a) Show that the lines $DN$ and $CM$ are perpendicular. b) Show that the point $Q$ is the midpoint of the segment $MP$.

Professor Moriarty has designed a “prime-testing trail.” The trail has $2002$ stations, labeled $1,... , 2002$. Each station is colored either red or green, and contains a table which indicates, for each of the digits $0, ..., 9$, another station number. A student is given a positive integer $n$, and then walks along the trail, starting at station $1$. The student reads the first (leftmost) digit of $n,$ and looks this digit up in station $1$’s table to get a new station location. The student then walks to this new station, reads the second digit of $n$ and looks it up in this station’s table to get yet another station location, and so on, until the last (rightmost) digit of $n$ has been read and looked up, sending the student to his or her final station. Here is an example that shows possible values for some of the tables. Suppose that $n = 19$: [img]https://cdn.artofproblemsolving.com/attachments/f/3/db47f6761ca1f350e39d53407a1250c92c4b05.png[/img] Using these tables, station $1$, digit $1$ leads to station $29$m station $29$, digit $9$ leads to station $1429$, and station $1429$ is green. Professor Moriarty claims that for any positive integer $n$, the final station (in the example, $1429$) will be green if and only if $n$ is prime. Is this possible?

We have three functions. The first one is $y=\phi(x)$. The second one is the inverse function of the first one. The figure of the third funcion is symmetrical to the second one about line $x+y=0$. Then, the third function is $\text{(A)}y=-\phi(x)\qquad\text{(B)}y=-\phi(-x)\qquad\text{(C)}y=-\phi^{-1}(x)\qquad\text{(D)}y=-\phi^{-1}(x)$

The vigilantes are a group of five superheroes, such that each one has one and only one of the following powers: hypnosis, super speed, shadow manipulation, immortality and super strength (each has a different power). On an adventure to the island of Philippines, they meet the sorcerer Vicencio, an old wise man who offers them the following ritual to help them: The ritual consists of a superhero $A$ acquiring the gift (s) of $B$ without $B$ acquiring the gift (s) of $A$. Determine the fewest number of rituals to be performed by the sorcerer Vicencio so that each superhero controls each of the five gifts. Clarification: At the end of the ritual, a superhero $A$ will have his gifts and those of a superhero $B$, but $B$ does not acquire those of $A$, but it does keep its own.

Let the sequence $a_i$ be defined as $a_{i+1} = 2^{a_i}$. Find the number of integers $1 \le n \le 1000$ such that if $a_0 = n$, then $100$ divides $a_{1000} - a_1$.

Kilua and Ndoti play the following game in a square $ABCD$: Kilua chooses one of the sides of the square and draws a point $X$ at this side. Ndoti chooses one of the other three sides and draws a point Y. Kilua chooses another side that hasn't been chosen and draws a point Z. Finally, Ndoti chooses the last side that hasn't been chosen yet and draws a point W. Each one of the players can draw his point at a vertex of $ABCD$, but they have to choose the side of the square that is going to be used to do that. For example, if Kilua chooses $AB$, he can draws $X$ at the point $B$ and it doesn't impede Ndoti of choosing $BC$. A vertex cannot de chosen twice. Kilua wins if the area of the convex quadrilateral formed by $X$, $Y$, $Z$, and $W$ is greater or equal than a half of the area of $ABCD$. Otherwise, Ndoti wins. Which player has a winning strategy? How can he play?

Find the maximal positive number $r$ with the following property: If all altitudes of a tetrahedron are $\geq 1$, then a sphere of radius $r$ fits into the tetrahedron.

A line in the plane of a triangle $ABC$ intersects the sides $AB$ and $AC$ respectively at points $X$ and $Y$ such that $BX = CY$ . Find the locus of the center of the circumcircle of triangle $XAY .$

For real numbers $a$ and $b$, define $$f(a,b) = \sqrt{a^2+b^2+26a+86b+2018}.$$ Find the smallest possible value of the expression $$f(a, b) + f (a,-b) + f(-a, b) + f (-a, -b).$$

Let $x_1, x_2, \cdots , x_n$ be positive numbers. Prove that \[\frac{x_1^2}{x_1^2+x_2x_3} + \frac{x_2^2}{x_2^2+x_3x_4} + \cdots +\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1} +\frac{x_n^2}{x_n^2+x_1x_2} \leq n-1\]

Prove that in any set of $2000$ distinct real numbers there exist two pairs $a>b$ and $c>d$ with $a \neq c$ or $b \neq d $, such that \[ \left| \frac{a-b}{c-d} - 1 \right|< \frac{1}{100000}. \]

Some number of coins is firstly separated into 200 groups and then to 300 groups. One coin is [i]special[/i], if on the second grouping it is in a group that has less coins than the previous one, in the first grouping, that it was in. Find the least amount of [i]special[/i] coins we can have.

Consider a checkered $3m\times 3m$ square, where $m$ is an integer greater than $1.$ A frog sits on the lower left corner cell $S$ and wants to get to the upper right corner cell $F.$ The frog can hop from any cell to either the next cell to the right or the next cell upwards. Some cells can be [i]sticky[/i], and the frog gets trapped once it hops on such a cell. A set $X$ of cells is called [i]blocking[/i] if the frog cannot reach $F$ from $S$ when all the cells of $X$ are sticky. A blocking set is [i] minimal[/i] if it does not contain a smaller blocking set.[list=a][*]Prove that there exists a minimal blocking set containing at least $3m^2-3m$ cells. [*]Prove that every minimal blocking set containing at most $3m^2$ cells.

Let $n \geq 5$ be a given integer. Determine the greatest integer $k$ for which there exists a polygon with $n$ vertices (convex or not, with non-selfintersecting boundary) having $k$ internal right angles. [i]Proposed by Juozas Juvencijus Macys, Lithuania[/i]