Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as shown. The ratio of the area of the inscribed square to the area of the large square is [asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((1,0)--(1,0.2)); draw((2,0)--(2,0.2)); draw((3,1)--(2.8,1)); draw((3,2)--(2.8,2)); draw((1,3)--(1,2.8)); draw((2,3)--(2,2.8)); draw((0,1)--(0.2,1)); draw((0,2)--(0.2,2)); draw((2,0)--(3,2)--(1,3)--(0,1)--cycle); [/asy] $\textbf{(A)}\ \dfrac{\sqrt{3}}{3} \qquad \textbf{(B)}\ \dfrac{5}{9} \qquad \textbf{(C)}\ \dfrac{2}{3} \qquad \textbf{(D)}\ \dfrac{\sqrt{5}}{3} \qquad \textbf{(E)}\ \dfrac{7}{9}$

(a) Find a polygon which can be cut by a straight line into two congruent parts so that one side of the polygon is divided in half while another side at a ratio of $1 : 2$. (b) Does there exist a convex polygon with this property?

In triangle $ABC$, $X$ and $Y$ are the tangency points of incircle (with center $I$) with sides $AB$ and $AC$ respectively. A tangent line to the circumcircle of triangle $ABC$ (with center $O$) at point $A$, intersects the extension of $BC$ at $D$. If $D,X$ and $Y$ are collinear then prove that $D,I$ and $O$ are also collinear. [i]proposed by Amirhossein Zabeti[/i]

Let $z$ be a complex number satisfying $(z+\tfrac{1}{z})(z+\tfrac{1}{z}+1)=1$. Evaluate $(3z^{100}+\tfrac{2}{z^{100}}+1)(z^{100}+\tfrac{2}{z^{100}}+3)$.

Let $A_1A_2A_3A_4A_5$ be a regular pentagon with side length 1. The sides of the pentagon are extended to form the 10-sided polygon shown in bold at right. Find the ratio of the area of quadrilateral $A_2A_5B_2B_5$ (shaded in the picture to the right) to the area of the entire 10-sided polygon. [asy] size(8cm); defaultpen(fontsize(10pt)); pair A_2=(-0.4382971011,5.15554989475), B_4=(-2.1182971011,-0.0149584477027), B_5=(-4.8365942022,8.3510997895), A_3=(0.6,8.3510997895), B_1=(2.28,13.521608132), A_4=(3.96,8.3510997895), B_2=(9.3965942022,8.3510997895), A_5=(4.9982971011,5.15554989475), B_3=(6.6782971011,-0.0149584477027), A_1=(2.28,3.18059144705); filldraw(A_2--A_5--B_2--B_5--cycle,rgb(.8,.8,.8)); draw(B_1--A_4^^A_4--B_2^^B_2--A_5^^A_5--B_3^^B_3--A_1^^A_1--B_4^^B_4--A_2^^A_2--B_5^^B_5--A_3^^A_3--B_1,linewidth(1.2)); draw(A_1--A_2--A_3--A_4--A_5--cycle); pair O = (A_1+A_2+A_3+A_4+A_5)/5; label("$A_1$",A_1, 2dir(A_1-O)); label("$A_2$",A_2, 2dir(A_2-O)); label("$A_3$",A_3, 2dir(A_3-O)); label("$A_4$",A_4, 2dir(A_4-O)); label("$A_5$",A_5, 2dir(A_5-O)); label("$B_1$",B_1, 2dir(B_1-O)); label("$B_2$",B_2, 2dir(B_2-O)); label("$B_3$",B_3, 2dir(B_3-O)); label("$B_4$",B_4, 2dir(B_4-O)); label("$B_5$",B_5, 2dir(B_5-O)); [/asy]

On circle $ O$, points $ C$ and $ D$ are on the same side of diameter $ \overline{AB}$, $ \angle AOC \equal{} 30^\circ$, and $ \angle DOB \equal{} 45^\circ$. What is the ratio of the area of the smaller sector $ COD$ to the area of the circle? [asy]unitsize(6mm); defaultpen(linewidth(0.7)+fontsize(8pt)); pair C = 3*dir (30); pair D = 3*dir (135); pair A = 3*dir (0); pair B = 3*dir(180); pair O = (0,0); draw (Circle ((0, 0), 3)); label ("$C$", C, NE); label ("$D$", D, NW); label ("$B$", B, W); label ("$A$", A, E); label ("$O$", O, S); label ("$45^\circ$", (-0.3,0.1), WNW); label ("$30^\circ$", (0.5,0.1), ENE); draw (A--B); draw (O--D); draw (O--C);[/asy]$ \textbf{(A)}\ \frac {2}{9} \qquad \textbf{(B)}\ \frac {1}{4} \qquad \textbf{(C)}\ \frac {5}{18} \qquad \textbf{(D)}\ \frac {7}{24} \qquad \textbf{(E)}\ \frac {3}{10}$

A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\tfrac{r-\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers, and $r+s+t<1000$. Find $r+s+t$.

Let $I$ be the center of the incircle of non-isosceles triangle $ABC,A_{1}=AI\cap BC$ and $B_{1}=BI\cap AC.$ Let $l_{a}$ be the line through $A_{1}$ which is parallel to $AC$ and $l_{b}$ be the line through $B_{1}$ parallel to $BC.$ Let $l_{a}\cap CI=A_{2}$ and $l_{b}\cap CI=B_{2}.$ Also $N=AA_{2}\cap BB_{2}$ and $M$ is the midpoint of $AB.$ If $CN\parallel IM$ find $\frac{CN}{IM}$.

In $\triangle ABC$, medians $\overline{AD}$ and $\overline{CE}$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC?$ [asy] unitsize(75); pathpen = black; pointpen=black; pair A = MP("A", D((0,0)), dir(200)); pair B = MP("B", D((2,0)), dir(-20)); pair C = MP("C", D((1/2,1)), dir(100)); pair D = MP("D", D(midpoint(B--C)), dir(30)); pair E = MP("E", D(midpoint(A--B)), dir(-90)); pair P = MP("P", D(IP(A--D, C--E)), dir(150)*2.013); draw(A--B--C--cycle); draw(A--D--E--C); [/asy] $\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 13.5 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 14.5 \qquad \textbf{(E)}\ 15 $

Acute triangle $ABP$, where $AB > BP$, has altitudes $BH$, $PQ$, and $AS$. Let $C$ denote the intersection of lines $QS$ and $AP$, and let $L$ denote the intersection of lines $HS$ and $BC$. If $HS = SL$ and $HL$ is perpendicular to $BC$, find the value of $\frac{SL}{SC}$.

Let $f(x)$ be a cubic polynomial with roots $x_1 , x_2$ and $x_3.$ Assume that $f(2x)$ is divisible by $f'(x)$ and compute the ratio $x_1 : x_2: x_3 .$

Points $ M,N$ are taken on sides $ BC,CD$ respectively of parallelogram $ ABCD$. Let $ E\equal{}BD\cap AM, F\equal{}BD\cap AN$. Diagonal $ BD$ cuts triangle $ AMN$ into two parts. Prove that these two parts have equal area if and only if the point $ K$ given by $ EK\parallel{}AD, FK\parallel{}AB$ lies on segment $ MN$.

The symbol of the Olympiad shows $5$ regular hexagons with side $a$, located inside a regular hexagon with side $b$. Find ratio $\frac{a}{b}$. [img]https://1.bp.blogspot.com/-OwyAl75LwiM/YIsThl3SG6I/AAAAAAAANS0/LwHEsAfyZMcqVIS8h_jr_n46OcMJaSTgQCLcBGAsYHQ/s0/2019%2BUkraine%2Bcorrespondence%2B5-12%2Bp8.png[/img]

A triangle $ABC$ is given in the plane with $AB = \sqrt{7},$ $BC = \sqrt{13}$ and $CA = \sqrt{19},$ circles are drawn with centers at $A,B$ and $C$ and radii $\frac{1}{3},$ $\frac{2}{3}$ and $1,$ respectively. Prove that there are points $A',B',C'$ on these three circles respectively such that triangle $ABC$ is congruent to triangle $A'B'C'.$

$ABCD$ is a parallelogram, line $DF$ is drawn bisecting $BC$ at $E$ and meeting $AB$ (extended) at $F$ from vertex $C$. Line $CH$ is drawn bisecting side $AD$ at $G$ and meeting $AB$ (extended) at $H$. Lines $DF$ and $CH$ intersect at $I$. If the area of parallelogram $ABCD$ is $x$, find the area of triangle $HFI$ in terms of $x$.

Let $AH_A, BH_B, CH_C$ be the altitudes of triangle $ABC$. Prove that if $\frac{H_BC}{AC} = \frac{H_CA}{AB}$, then the line symmetric to $BC$ with respect to line $H_BH_C$ is tangent to the circumscribed circle of triangle $H_BH_CA$. [i](Proposed by Mykhailo Bondarenko)[/i]

In a trapezoid $ABCD$ whose parallel sides $AB$ and $CD$ are in ratio $\frac{AB}{CD}=\frac32$, the points $ N$ and $M$ are marked on the sides $BC$ and $AB$ respectively, in such a way that $BN = 3NC$ and $AM = 2MB$ and segments $AN$ and $DM$ are drawn that intersect at point $P$, find the ratio between the areas of triangle $APM$ and trapezoid $ABCD$. [img]https://cdn.artofproblemsolving.com/attachments/7/8/21d59ca995d638dfcb76f9508e439fd93a5468.png[/img]

Let $PAB$ and $PBC$ be two similar right-angled triangles (in the same plane) with $\angle PAB = \angle PBC = 90^\circ$ such that $A$ and $C$ lie on opposite sides of the line $PB$. If $PC = AC$, calculate the ratio $\frac{PA}{AB}$.

In square $ABCE$, $AF=2FE$ and $CD=2DE$. What is the ratio of the area of $\triangle BFD$ to the area of square $ABCE$? [asy] size((100)); draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((3,0)--(9,9)--(0,3)--cycle); dot((3,0)); dot((0,3)); dot((9,9)); dot((0,0)); dot((9,0)); dot((0,9)); label("$A$", (0,9), NW); label("$B$", (9,9), NE); label("$C$", (9,0), SE); label("$D$", (3,0), S); label("$E$", (0,0), SW); label("$F$", (0,3), W); [/asy] $ \textbf{(A)}\ \frac{1}{6}\qquad\textbf{(B)}\ \frac{2}{9}\qquad\textbf{(C)}\ \frac{5}{18}\qquad\textbf{(D)}\ \frac{1}{3}\qquad\textbf{(E)}\ \frac{7}{20} $

Let $\gamma,\Gamma$ be two concentric circles with radii $r,R$ with $r<R$. Let $ABCD$ be a cyclic quadrilateral inscribed in $\gamma$. If $\overrightarrow{AB}$ denotes the Ray starting from $A$ and extending indefinitely in $B's$ direction then Let $\overrightarrow{AB}, \overrightarrow{BC}, \overrightarrow{CD} , \overrightarrow{DA}$ meet $\Gamma$ at the points $C_1,D_1,A_1,B_1$ respectively. Prove that \[\frac{[A_1B_1C_1D_1]}{[ABCD]} \ge \frac{R^2}{r^2}\] where $[.]$ denotes area.

Let $C_1$ and $C_2$ be tangent circles internally at point $A$, with $C_2$ inside of $C_1$. Let $BC$ be a chord of $C_1$ that is tangent to $C_2$. Prove that the ratio between the length $BC$ and the perimeter of the triangle $ABC$ is constant, that is, it does not depend of the selection of the chord $BC$ that is chosen to construct the trangle.

As shown in the figure, triangle $ABC$ is divided into six smaller triangles by lines drawn from the vertices through a common interior point. The areas of four of these triangles are as indicated. Find the area of triangle $ABC$. [asy] size(200); pair A=origin, B=(14,0), C=(9,12), D=foot(A, B,C), E=foot(B, A, C), F=foot(C, A, B), H=orthocenter(A, B, C); draw(F--C--A--B--C^^A--D^^B--E); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("84", centroid(H, C, E), fontsize(9.5)); label("35", centroid(H, B, D), fontsize(9.5)); label("30", centroid(H, F, B), fontsize(9.5)); label("40", centroid(H, A, F), fontsize(9.5));[/asy]

Points $M$ and $N$ are considered in the interior of triangle $ABC$ such that $\angle MAB = \angle NAC$ and $\angle MBA = \angle NBC$. Prove that $$\frac{AM \cdot AN}{AB \cdot AC}+ \frac{BM\cdot BN}{BA \cdot BC}+ \frac{CM \cdot CN }{CA \cdot CB}=1$$

On the sides $BC,CA$ and $AB$ of a triangle $ABC$ of area $S$ are taken points $A' ,B' ,C'$ respectively such that $AC' /AB = BA' /BC = CB' /CA = p$, where $0 < p < 1$ is variable. (a) Find the area of triangle $A' B' C'$ in terms of $ p$. (b) Find the value of $p$ which minimizes this area. (c) Find the locus of the intersection point $P$ of the lines through $A' $ and $C'$ parallel to $AB$ and $AC$ respectively.

Given a triangle $ABC$ inscribed in $(O)$ with $2$ symmedians $AD, CF(D,F$ are on the sides $BC, AB,$ respectively$).$ The ray $DF$ intersects $(O)$ at $P.$ The line passing through $P$ and perpendicular to $OA$ intersects $AB,AC$ at $Q,R,$ respectively$.$ Compute the ratio $\dfrac{PR}{PQ}.$