Given is a triangle $ABC$.On the extensions of the side $AB$ we consider points $A_1,B_1$ such that $AB_1=BA_1$ (with $A_1$ lying closer to $B$).On the extensions of the side $BC$ we consider points $B_4,C_4$ such that $CB_4=BC_4$ (with $B_4$ lying closer to $C$).On the extensions of the side $AC$ we consider points $C_1,A_4$ such that $AC_1=CA_4$ (with $C_1$ lying closer to $A$).On the segment $A_1A_4$ we consider points $A_2,A_3$ such that $A_1A_2=A_3A_4=mA_1A_4$ where $0<m<\frac{1}{2}$.Points $B_2,B_3$ and $C_2,C_3$ are defined similarly,on the segments $B_1B_4,C_1C_4$ respectively.If $D\equiv BB_2\cap CC_2 \ , \ E\equiv AA_3\cap CC_2 \ , \ F\equiv AA_3\cap BB_3$, $\ G\equiv BB_3\cap CC_3 \ , \ H\equiv AA_2\cap CC_3$ and $I\equiv AA_2\cap BB_2$,prove that the diagonals $DG,EH,FI$ of the hexagon $DEFGHI$ are concurrent. [hide=Diagram][asy]import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -7.984603447540051, xmax = 21.28710511372557, ymin = -6.555010307713199, ymax = 10.006614273002825; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); draw((1.1583842866003107,4.638449718549554)--(0.,0.)--(7.,0.)--cycle, aqaqaq); /* draw figures */ draw((1.1583842866003107,4.638449718549554)--(0.,0.), uququq); draw((0.,0.)--(7.,0.), uququq); draw((7.,0.)--(1.1583842866003107,4.638449718549554), uququq); draw((1.1583842866003107,4.638449718549554)--(1.623345080409327,6.500264738079558)); draw((0.,0.)--(-0.46496079380901606,-1.8618150195300045)); draw((-3.0803965232149757,0.)--(0.,0.)); draw((7.,0.)--(10.080396523214976,0.)); draw((1.1583842866003107,4.638449718549554)--(0.007284204967787214,5.552463941947242)); draw((7.,0.)--(8.151100081632526,-0.9140142233976905)); draw((-0.46496079380901606,-1.8618150195300045)--(8.151100081632526,-0.9140142233976905)); draw((-3.0803965232149757,0.)--(0.007284204967787214,5.552463941947242)); draw((10.080396523214976,0.)--(1.623345080409327,6.500264738079558)); draw((0.,0.)--(3.7376079411107392,4.8751985535596685)); draw((-0.7646359770779035,4.164347956460432)--(7.,0.)); draw((1.1583842866003107,4.638449718549554)--(5.997084862772141,-1.150964422430769)); draw((0.,0.)--(7.966133662513563,1.6250661845198895)); draw((-2.308476341169285,1.3881159854868106)--(7.,0.)); draw((1.1583842866003107,4.638449718549554)--(1.6890544250513695,-1.624864820496926)); draw((2.0395968109217,2.660375186246903)--(2.9561195753832448,0.6030390855677443), linetype("2 2")); draw((3.4388364046369224,1.909931693481981)--(1.4816619768719694,0.8229159040072803), linetype("2 2")); draw((1.3969966570225139,1.8221911417546572)--(4.301698851378541,0.8775330211014288), linetype("2 2")); /* dots and labels */ dot((1.1583842866003107,4.638449718549554),linewidth(3.pt) + dotstyle); label("$A$", (0.6263408942608304,4.2), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.44658827292841696,0.04763072114368767), NE * labelscalefactor); dot((7.,0.),linewidth(3.pt) + dotstyle); label("$C$", (7.008893888822507,0.18518574257820614), NE * labelscalefactor); dot((1.623345080409327,6.500264738079558),linewidth(3.pt) + dotstyle); label("$B_1$", (1.7267810657369815,6.6777827542874775), NE * labelscalefactor); dot((-0.46496079380901606,-1.8618150195300045),linewidth(3.pt) + dotstyle); label("$A_1$", (-1.1068523758141076,-1.6305405403574376), NE * labelscalefactor); dot((10.080396523214976,0.),linewidth(3.pt) + dotstyle); label("$B_4$", (10.062615364668826,-0.612633381742001), NE * labelscalefactor); dot((-3.0803965232149757,0.),linewidth(3.pt) + dotstyle); label("$C_4$", (-3.3077327187664096,-0.612633381742001), NE * labelscalefactor); dot((0.007284204967787214,5.552463941947242),linewidth(3.pt) + dotstyle); label("$C_1$", (0.1036318128096586,5.714897604245849), NE * labelscalefactor); dot((8.151100081632526,-0.9140142233976905),linewidth(3.pt) + dotstyle); label("$A_4$", (8.521999124602214,-1.1903644717669786), NE * labelscalefactor); dot((-2.308476341169285,1.3881159854868106),linewidth(3.pt) + dotstyle); label("$C_3$", (-2.9776006673235647,1.7808239912186203), NE * labelscalefactor); dot((-0.7646359770779035,4.164347956460432),linewidth(3.pt) + dotstyle); label("$C_2$", (-1.1618743843879151,4.504413415622086), NE * labelscalefactor); dot((1.6890544250513695,-1.624864820496926),linewidth(3.pt) + dotstyle); label("$A_2$", (1.6167370485893664,-2.125738617521704), NE * labelscalefactor); dot((5.997084862772141,-1.150964422430769),linewidth(3.pt) + dotstyle); label("$A_3$", (6.211074764502297,-1.603029536070534), NE * labelscalefactor); dot((7.966133662513563,1.6250661845198895),linewidth(3.pt) + dotstyle); label("$B_3$", (8.081823056011753,1.7808239912186203), NE * labelscalefactor); dot((3.7376079411107392,4.8751985535596685),linewidth(3.pt) + dotstyle); label("$B_2$", (3.8451283958285725,5.027122497073257), NE * labelscalefactor); dot((2.0395968109217,2.660375186246903),linewidth(3.pt) + dotstyle); label("$D$", (1.7542920700238853,2.991308179842383), NE * labelscalefactor); dot((3.4388364046369224,1.909931693481981),linewidth(3.pt) + dotstyle); label("$E$", (3.542507348672631,2.083445038374561), NE * labelscalefactor); dot((4.301698851378541,0.8775330211014288),linewidth(3.pt) + dotstyle); label("$F$", (4.22,0.93), NE * labelscalefactor); dot((2.9561195753832448,0.6030390855677443),linewidth(3.pt) + dotstyle); label("$G$", (2.909754250073844,0.10265272971749505), NE * labelscalefactor); dot((1.4816619768719694,0.8229159040072803),linewidth(3.pt) + dotstyle); label("$H$", (0.9839839499905795,0.43278478116033936), NE * labelscalefactor); dot((1.3969966570225139,1.8221911417546572),linewidth(3.pt) + dotstyle); label("$I$", (0.9839839499905795,1.8908680083662353), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

The bisector of the angle $BAC$of the acute triangle $ABC$ ( $AC \ne AB$) intersects its circumscribed circle for the second time at the point $W$. Let $O$ be the center of the circumscribed circle $\Delta ABC$. The line $AW$ intersects for the second time the circumcribed circles of triangles $OWB$ and $OWC$ at the points $N$ and $M$, respectively. Prove that $BN + MC = AW$. (Mitrofanov V., Hilko D.)

Let $\triangle ABC$ be an acute-angled triangle with $AB<AC$. Let $M$ and $N$ be the midpoints of $AB$ and $AC$, respectively; let $AD$ be an altitude in this triangle. A point $K$ is chosen on the segment $MN$ so that $BK=CK$. The ray $KD$ meets the circumcircle $\Omega$ of $ABC$ at $Q$. Prove that $C, N, K, Q$ are concyclic.

Suppose that $(a_1,\ldots,a_{20})$ and $(b_1,\ldots,b_{20})$ are two sequences of integers such that the sequence $(a_1,\ldots,a_{20},b_1,\ldots,b_{20})$ contains each of the numbers $1,\ldots,40$ exactly once. What is the maximum possible value of the sum \[\sum_{i=1}^{20}\sum_{j=1}^{20}\min(a_i,b_j)?\]

Find the greatest positive integer $A$ with the following property: For every permutation of $\{1001,1002,...,2000\}$ , the sum of some ten consecutive terms is great than or equal to $A$.

In the bottom-left corner of a chessboard (with 8 rows and 8 columns), there is a king. Marius and Alexandru play a game, with Alexandru going first. On their turn, each player moves the king either one square to the right, one square up, or one square diagonally up-right. The player who moves the king to the top-right corner square wins. Who will win if both players play optimally?

Tom can run to Beth's house in $63$ minutes. Beth can run to Tom's house in $84$ minutes. At noon Tom starts running from his house toward Beth's house while at the same time Beth starts running from her house toward Tom's house. When they meet, they both run at Beth's speed back to Beth's house. At how many minutes after noon will they arrive at Beth's house?

For how many positive integers $n$, $1\leq n\leq 2008$, can the set \[\{1,2,3,\ldots,4n\}\] be divided into $n$ disjoint $4$-element subsets such that every one of the $n$ subsets contains the element which is the arithmetic mean of all the elements in that subset?

An integer lattice point in the Cartesian plane is a point $(x,y)$ where $x$ and $y$ are both integers. Suppose nine integer lattice points are chosen such that no three of them lie on the same line. Out of all 36 possible line segments between pairs of those nine points, some line segments may contain integer lattice points besides the original nine points. What is the minimum number of line segments that must contain an integer lattice point besides the original nine points? Prove your answer.

Consider finitely many points in the plane with no three points on a line. All these points can be coloured red or green such that any triangle with vertices of the same colour contains at least one point of the other colour in its interior. What is the maximal possible number of points with this property?

The numbers from $1$ to $n^2$ are written in order in a grid of $n \times n$, one number in each square, in such a way that the first row contains the numbers from $1$ to $n$ from left to right; the second row contains the numbers $n + 1$ to $2n$ from left to right, and so on and so forth. An allowed move on the grid consists in choosing any two adjacent squares (i.e. two squares that share a side), and add (or subtract) the same integer to both of the numbers that appear on those squares. Find all values of $n$ for which it is possible to make every squares to display $0$ after making any number of moves as necessary and, for those cases in which it is possible, find the minimum number of moves that are necessary to do this.

Points $A_1,A_2,...,A_n$ are selected from the equilateral triangle with a side that is equal to $1$. Denote by $d_k$ the least distance from $A_k$ to all other selected points. Prove that $d_1^2+...+d_n^2 \le 3,5$.

Let $h$ and $k$ be positive integers. Prove that for every $\varepsilon >0,$ there are positive integers $m$ and $n$ such that \[\varepsilon < \left|h\sqrt{m}-k\sqrt{n}\right|<2\varepsilon.\]

Let $g$ be a straight line and $n$ a given positive integer. Prove that there are always n different points on g to choose as well as a point not lying on g in such a way that the distance between each two of these $n + 1$ points is an integer.

Find all positive real solutions to the equation $x+\left\lfloor\frac x3\right\rfloor=\left\lfloor\frac{2x}3\right\rfloor+\left\lfloor\frac{3x}5\right\rfloor$

The average of the five numbers in a list is $54$. The average of the first two numbers is $48$. What is the average of the last three numbers? $\textbf{(A)}\ 55\qquad \textbf{(B)}\ 56\qquad \textbf{(C)}\ 57\qquad \textbf{(D)}\ 58 \qquad \textbf{(E)}\ 59$

Given that $a_1, a_2, a_3, . . . , a_{99}$ is a permutation of $1, 2, 3, . . . , 99,$ find the maximum possible value of $$|a_1 - 1| + |a_2 - 2| + |a_3 - 3| + \dots + |a_{99} - 99|.$$

A closed subset $S$ of $\mathbb{R}^{2}$ lies in $a<x<b$. Show that its projection on the $y$-axis is closed.

What is $10 \cdot \left(\tfrac{1}{2} + \tfrac{1}{5} + \tfrac{1}{10}\right)^{-1}?$ ${ \textbf{(A)}\ 3\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ \frac{25}{2}\qquad\textbf{(D)}}\ \frac{170}{3}\qquad\textbf{(E)}\ 170$

Let $n$ be a positive integer. Ana and Beto play a game on a $2 \times n$ board (with 2 rows and $n$ columns). First, Ana writes a digit from 1 to 9 in each cell of the board such that in each column the two written digits are different. Then, Beto erases a digit from each column. Reading from left to right, a number with $n$ digits is formed. Beto wins if this number is a multiple of $n$; otherwise, Ana wins. Determine which of the two players has a winning strategy in the following cases: $\bullet$ (a) $n = 1001$. $\bullet$ (b) $n = 1003$.

The Fox and Pinocchio have grown a tree on the Field of Miracles with 11 golden coins. It is known that exactly 4 of them are counterfeit. All the real coins weigh the same, the counterfeit coins also weigh the same but are lighter. The Fox and Pinocchio have collected the coins and wish to divide them. The Fox is going to give 4 coins to Pinocchio, but Pinocchio wants to check whether they all are real. Can he check this using two weighings on a balance scale with no weights?

If $x_1 + x_2 + \cdots + x_n = \sum_{i=1}^{n} x_i = \frac{1}{2}$ and $x_i > 0$ ; then prove that: $ \frac{1-x_1}{1+x_1} \cdot \frac{1-x_2}{1+x_2} \cdots \frac{1-x_n}{1+x_n} = \prod_{i=1}^{n} \frac{1-x_i}{1+x_i} \geq \frac{1}{3}$

Leta,b,c are postive real numbers,proof that $ \frac{a}{b\plus{}2c}\plus{}\frac{b}{c\plus{}2a}\plus{}\frac{c}{a\plus{}2b}\geq1$

Let $a_1, a_2, \dots, a_{2^{2016}}$ be positive integers not bigger than $2016$. We know that for each $n \leq 2^{2016}$, $a_1a_2 \dots a_{n} +1 $ is a perfect square. Prove that for some $i $ , $a_i=1$.

On the sides $AB,AD$ of the rhombus $ABCD$ are the points $E,F$ such that $AE=DF$. The lines $BC,DE$ intersect at $P$ and $CD,BF$ intersect at $Q$. Prove that: (a) $\frac{PE}{PD} + \frac{QF}{QB} = 1$; (b) $P,A,Q$ are collinear. [i]Virginia Tica, Vasile Tica[/i]