In the following diagram, let $ABCD$ be a square and let $M,N,P$ and $Q$ be the midpoints of its sides. Prove that \[S_{A'B'C'D'} = \frac 15 S_{ABCD}.\] [asy] import graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttzz = rgb(0,0.2,0.6); pen qqzzff = rgb(0,0.6,1); draw((0,4)--(4,4),qqttzz+linewidth(1.6pt)); draw((4,4)--(4,0),qqttzz+linewidth(1.6pt)); draw((4,0)--(0,0),qqttzz+linewidth(1.6pt)); draw((0,0)--(0,4),qqttzz+linewidth(1.6pt)); draw((0,4)--(2,0),qqzzff+linewidth(1.2pt)); draw((2,4)--(4,0),qqzzff+linewidth(1.2pt)); draw((0,2)--(4,4),qqzzff+linewidth(1.2pt)); draw((0,0)--(4,2),qqzzff+linewidth(1.2pt)); dot((0,4),ds); label("$A$", (0.07,4.12), NE*lsf); dot((0,0),ds); label("$D$", (-0.27,-0.37), NE*lsf); dot((4,0),ds); label("$C$", (4.14,-0.39), NE*lsf); dot((4,4),ds); label("$B$", (4.08,4.12), NE*lsf); dot((2,4),ds); label("$M$", (2.08,4.12), NE*lsf); dot((4,2),ds); label("$N$", (4.2,1.98), NE*lsf); dot((2,0),ds); label("$P$", (1.99,-0.49), NE*lsf); dot((0,2),ds); label("$Q$", (-0.48,1.9), NE*lsf); dot((0.8,2.4),ds); label("$A'$", (0.81,2.61), NE*lsf); dot((2.4,3.2),ds); label("$B'$", (2.46,3.47), NE*lsf); dot((3.2,1.6),ds); label("$C'$", (3.22,1.9), NE*lsf); dot((1.6,0.8),ds); label("$D'$", (1.14,0.79), NE*lsf); clip((-4.44,-11.2)--(-4.44,6.41)--(16.48,6.41)--(16.48,-11.2)--cycle); [/asy] [$S_{X}$ denotes area of the $X.$]

in a quadrilateral $ABCD$, $E$ and $F$ are on $BC$ and $AD$ respectively such that the area of triangles $AED$ and $BCF$ is $\frac{4}{7}$ of the area of $ABCD$. $R$ is the intersection point of digonals of $ABCD$. $\frac{AR}{RC}=\frac{3}{5}$ and $\frac{BR}{RD}=\frac{5}{6}$. a) in what ratio does $EF$ cut the digonals?(13 points) b) find $\frac{AF}{FD}$.(5 points)

Let's say a positive integer $ n$ is [i]atresvido[/i] if the set of its divisors (including 1 and $ n$) can be split in in 3 subsets such that the sum of the elements of each is the same. Determine the least number of divisors an atresvido number can have.

For each positive integer $ n$, let $ c(n)$ be the largest real number such that \[ c(n) \le \left| \frac {f(a) \minus{} f(b)}{a \minus{} b}\right|\] for all triples $ (f, a, b)$ such that --$ f$ is a polynomial of degree $ n$ taking integers to integers, and --$ a, b$ are integers with $ f(a) \neq f(b)$. Find $ c(n)$. [i]Shaunak Kishore.[/i]

For each positive integer, define a function \[ f(n)=\begin{cases}0, &\text{if n is the square of an integer}\\ \\ \left\lfloor\frac{1}{\{\sqrt{n}\}}\right\rfloor, &\text{if n is not the square of an integer}\end{cases}. \] Find the value of $\sum_{k=1}^{200} f(k)$.

All squares of a $ 20\times 20$ table are empty. Misha* and Sasha** in turn put chips in free squares (Misha* begins). The player after whose move there are four chips on the intersection of two rows and two columns wins. Which of the players has a winning strategy? [i]Proposed by A. Golovanov[/i] [b]US Name Conversions: [/b] [i]Misha*: Naoki Sasha**: Richard[/i]

Sam colors each tile in a 4 by 4 grid white or black. A coloring is called [i]rotationally symmetric[/i] if the grid can be rotated 90, 180, or 270 degrees to achieve the same pattern. Two colorings are called [i]rotationally distinct[/i] if neither can be rotated to match the other. How many rotationally distinct ways are there for Sam to color the grid such that the colorings are [i]not[/i] rotationally symmetric? [i]Proposed by Gabriel Wu[/i]

In $\triangle ABC$, $H$ is the orthocenter, and $AD,BE$ are arbitrary cevians. Let $\omega_1, \omega_2$ denote the circles with diameters $AD$ and $BE$, respectively. $HD,HE$ meet $\omega_1,\omega_2$ again at $F,G$. $DE$ meets $\omega_1,\omega_2$ again at $P_1,P_2$ respectively. $FG$ meets $\omega_1,\omega_2$ again $Q_1,Q_2$ respectively. $P_1H,Q_1H$ meet $\omega_1$ at $R_1,S_1$ respectively. $P_2H,Q_2H$ meet $\omega_2$ at $R_2,S_2$ respectively. Let $P_1Q_1\cap P_2Q_2 = X$, and $R_1S_1\cap R_2S_2=Y$. Prove that $X,Y,H$ are collinear. [i]Ray Li.[/i]

Let $D$ and $E$ be points on respectively $[AB]$ and $[AC]$ of $\triangle ABC$ where $|AB|=|AC|$, $m(\widehat{BAC})=40^\circ$. Let $F$ be a point on $BC$ such that $C$ is between $B$ and $F$. If $|BE|=|CF|$, $|AD|=|AE|$, and $m(\widehat{BEC})=60^\circ$, then what is $m(\widehat{DFB})$ ? $ \textbf{(A)}\ 45^\circ \qquad\textbf{(B)}\ 40^\circ \qquad\textbf{(C)}\ 35^\circ \qquad\textbf{(D)}\ 30^\circ \qquad\textbf{(E)}\ 25^\circ $

Let $C$ be the curve expressed by $x=\sin t,\ y=\sin 2t\ \left(0\leq t\leq \frac{\pi}{2}\right).$ (1) Express $y$ in terms of $x$. (2) Find the area of the figure $D$ enclosed by the $x$-axis and $C$. (3) Find the volume of the solid generated by a rotation of $D$ about the $y$-axis.

Prove no three lattice points in the plane form an equilateral triangle.

There are eight rooks on a chessboard, no two attacking each other. Prove that some two of the pairwise distances between the rooks are equal. (The distance between two rooks is the distance between the centers of their cell.)

Let $ABCDEF$ be an inscribed hexagon with $$AB.CD.EF=BC.DE.FA$$ Let $B_1$ be the reflection point of $B$ with respect to $AC$ and $D_1$ be the reflection point of $D$ with respect to $CE,$ and finally let $F_1$ be the reflection point of $F$ with respect to $AE.$ Prove that $\triangle B_1D_1F_1\sim BDF.$

Let $\triangle ABC$ be an equilateral triangle and let $P$ be a point on $\left[AB\right]$. $Q$ is the point on $BC$ such that $PQ$ is perpendicular to $AB$. $R$ is the point on $AC$ such that $QR$ is perpendicular to $BC$. And $S$ is the point on $AB$ such that $RS$ is perpendicular to $AC$. $Q'$ is the point on $BC$ such that $PQ'$ is perpendicular to $BC$. $R'$ is the point on $AC$ such that $Q'R'$ is perpendicular to $AC$. And $S'$ is the point on $AB$ such that $R'S'$ is perpendicular to $AB$. Determine $\frac{|PB|}{|AB|}$ if $S=S'$.

Is it possible to divide the plane into polygons so that each polygon is transformed into itself under some rotation by $360/7$ degrees about some point? All sides of these polygons must be greater than $1$ cm. (A polygon is the part of a plane bounded by one non-self-intersect-ing closed broken line, not necessarily convex.) (A. Andjans, Riga)

In an acute triangle $\bigtriangleup ABC$, $AB<AC<BC$, and $A_1,B_1,C_1$ are the projections of $A,B,C$ to the corresponding sides. Let the reflection of $B_1$ wrt $CC_1$ be $Q$, and the reflection of $C_1$ wrt $BB_1$ be $P$. Prove that the circumcirle of $A_1PQ$ passes through the midpoint of $BC$.

Let $ABC$ be an acute triangle, $H$ its orthocentre, $D$ a point on the side $[BC]$, and $P$ a point such that $ADPH$ is a parallelogram. Show that $\angle BPC > \angle BAC$.

Let $ABC$ be a triangle with orthocenter $H$ and circumcenter $O$. Let $P$ be a point in the plane such that $AP \perp BC$. Let $Q$ and $R$ be the reflections of $P$ in the lines $CA$ and $AB$, respectively. Let $Y$ be the orthogonal projection of $R$ onto $CA$. Let $Z$ be the orthogonal projection of $Q$ onto $AB$. Assume that $H \neq O$ and $Y \neq Z$. Prove that $YZ \perp HO$. [asy] import olympiad; unitsize(30); pair A,B,C,H,O,P,Q,R,Y,Z,Q2,R2,P2; A = (-14.8, -6.6); B = (-10.9, 0.3); C = (-3.1, -7.1); O = circumcenter(A,B,C); H = orthocenter(A,B,C); P = 1.2 * H - 0.2 * A; Q = reflect(A, C) * P; R = reflect(A, B) * P; Y = foot(R, C, A); Z = foot(Q, A, B); P2 = foot(A, B, C); Q2 = foot(P, C, A); R2 = foot(P, A, B); draw(B--(1.6*A-0.6*B)); draw(B--C--A); draw(P--R, blue); draw(R--Y, red); draw(P--Q, blue); draw(Q--Z, red); draw(A--P2, blue); draw(O--H, darkgreen+linewidth(1.2)); draw((1.4*Z-0.4*Y)--(4.6*Y-3.6*Z), red+linewidth(1.2)); draw(rightanglemark(R,Y,A,10), red); draw(rightanglemark(Q,Z,B,10), red); draw(rightanglemark(C,Q2,P,10), blue); draw(rightanglemark(A,R2,P,10), blue); draw(rightanglemark(B,P2,H,10), blue); label("$\textcolor{blue}{H}$",H,NW); label("$\textcolor{blue}{P}$",P,N); label("$A$",A,W); label("$B$",B,N); label("$C$",C,S); label("$O$",O,S); label("$\textcolor{blue}{Q}$",Q,E); label("$\textcolor{blue}{R}$",R,W); label("$\textcolor{red}{Y}$",Y,S); label("$\textcolor{red}{Z}$",Z,NW); dot(A, filltype=FillDraw(black)); dot(B, filltype=FillDraw(black)); dot(C, filltype=FillDraw(black)); dot(H, filltype=FillDraw(blue)); dot(P, filltype=FillDraw(blue)); dot(Q, filltype=FillDraw(blue)); dot(R, filltype=FillDraw(blue)); dot(Y, filltype=FillDraw(red)); dot(Z, filltype=FillDraw(red)); dot(O, filltype=FillDraw(black)); [/asy]

Each square of $1\times 1$, of a $n\times n$ grid is colored using red or blue, in such way that between all the $2\times 2$ subgrids, there are all the possible colorations of a $2\times 2$ grid using red or blue, (colorations that can be obtained by using rotation or symmetry, are said to be different, so there are 16 possibilities). Find: a) The minimum value of $n$. b) For that value, find the least possible number of red squares.

By the time a party is over, $ 28$ handshakes have occurred. If everyone shook everyone else's hand once, how many people attended the party?

Point $P$ is taken in the interior of the triangle $ABC$, so that \[\angle PAB = \angle PCB = \dfrac {1} {4} (\angle A + \angle C).\] Let $L$ be the foot of the angle bisector of $\angle B$. The line $PL$ meets the circumcircle of $\triangle APC$ at point $Q$. Prove that $QB$ is the angle bisector of $\angle AQC$. [i]Proposed by S. Berlov[/i]

Let $ABCD$ be a quadrilateral with $AD = 20$ and $BC = 13$. The area of $\triangle ABC$ is $338$ and the area of $\triangle DBC$ is $212$. Compute the smallest possible perimeter of $ABCD$. [i]Proposed by Evan Chen[/i]

Let $ \Gamma(I,r)$ and $ \Gamma(O,R)$ denote the incircle and circumcircle, respectively, of a triangle $ ABC$. Consider all the triangels $ A_iB_iC_i$ which are simultaneously inscribed in $ \Gamma(O,R)$ and circumscribed to $ \Gamma(I,r)$. Prove that the centroids of these triangles are concyclic.

$ M$ is the midpoint of base $ BC$ in a trapezoid $ ABCD$. A point $ P$ is chosen on the base $ AD$. The line $ PM$ meets the line $ CD$ at a point $ Q$ such that $ C$ lies between $ Q$ and $ D$. The perpendicular to the bases drawn through $ P$ meets the line $ BQ$ at $ K$. Prove that $ \angle QBC \equal{} \angle KDA$. [i]Proposed by S. Berlov[/i]

Let $ABC$ be a triangle. Prove that there exists a unique point $P$ for which one can find points $D$, $E$ and $F$ such that the quadrilaterals $APBF$, $BPCD$, $CPAE$, $EPFA$, $FPDB$, and $DPEC$ are all parallelograms. [i]Proposed by Lewis Chen[/i]