The sum of the distances from one vertex of a square with sides of length two to the midpoints of each of the sides of the square is $\textbf{(A) }2\sqrt{5}\qquad\textbf{(B) }2+\sqrt{3}\qquad\textbf{(C) }2+2\sqrt{3}\qquad\textbf{(D) }2+\sqrt{5}\qquad \textbf{(E) }2+2\sqrt{5}$

Here’s a screenshot of the problem. If someone could LaTEX a diagram, that would be great!

Let $S$ be the square one of whose diagonals has endpoints $(0.1,0.7)$ and $(-0.1,-0.7)$. A point $v=(x,y)$ is chosen uniformly at random over all pairs of real numbers $x$ and $y$ such that $0\le x \le 2012$ and $0 \le y \le 2012$. Let $T(v)$ be a translated copy of $S$ centered at $v$. What is the probability that the square region determined by $T(v)$ contains exactly two points with integer coordinates in its interior? $ \textbf{(A)}\ 0.125\qquad\textbf{(B)}\ 0.14\qquad\textbf{(C)}\ 0.16\qquad\textbf{(D)}\ 0.25\qquad\textbf{(E)}\ 0.32 $

Circles $ \omega_1$ and $ \omega_2$ meet at $ P$ and $ Q$. Segments $ AC$ and $ BD$ are chords of $ \omega_1$ and $ \omega_2$ respectively, such that segment $ AB$ and ray $ CD$ meet at $ P$. Ray $ BD$ and segment $ AC$ meet at $ X$. Point $ Y$ lies on $ \omega_1$ such that $ PY \parallel BD$. Point $ Z$ lies on $ \omega_2$ such that $ PZ \parallel AC$. Prove that points $ Q,X,Y,Z$ are collinear.

Let $ ABC$ be a triangle. Point $ D$ lies on its sideline $ BC$ such that $ \angle CAD \equal{} \angle CBA.$ Circle $ (O)$ passing through $ B,D$ intersects $ AB,AD$ at $ E,F$, respectively. $ BF$ meets $ DE$ at $ G$.Denote by$ M$ the midpoint of $ AG.$ Show that $ CM\perp AO.$

Pentagon $ABCDE$ consists of a square $ACDE$ and an equilateral triangle $ABC$ that share the side $\overline{AC}$. A circle centered at $C$ has area 24. The intersection of the circle and the pentagon has half the area of the pentagon. Find the area of the pentagon. [asy]/* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(4.26cm); 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 = -1.52, xmax = 2.74, ymin = -2.18, ymax = 6.72; /* image dimensions */ draw((0,1)--(2,1)--(2,3)--(0,3)--cycle); draw((0,3)--(2,3)--(1,4.73)--cycle); /* draw figures */ draw((0,1)--(2,1)); draw((2,1)--(2,3)); draw((2,3)--(0,3)); draw((0,3)--(0,1)); draw((0,3)--(2,3)); draw((2,3)--(1,4.73)); draw((1,4.73)--(0,3)); draw(circle((0,3), 1.44)); label("$C$",(-0.4,3.14),SE*labelscalefactor); label("$A$",(2.1,3.1),SE*labelscalefactor); label("$B$",(0.86,5.18),SE*labelscalefactor); label("$D$",(-0.28,0.88),SE*labelscalefactor); label("$E$",(2.1,0.8),SE*labelscalefactor); /* dots and labels */ clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

In a triangle $ABC$, let $I$ denote its incenter. Points $D, E, F$ are chosen on the segments $BC, CA, AB$, respectively, such that $BD + BF = AC$ and $CD + CE = AB$. The circumcircles of triangles $AEF, BFD, CDE$ intersect lines $AI, BI, CI$, respectively, at points $K, L, M$ (different from $A, B, C$), respectively. Prove that $K, L, M, I$ are concyclic.

[asy] import olympiad; import cse5; size(5cm); pointpen = black; pair A = Drawing((10,17.32)); pair B = Drawing((0,0)); pair C = Drawing((20,0)); draw(A--B--C--cycle); pair X = 0.85*A + 0.15*B; pair Y = 0.82*A + 0.18*C; pair W = (-11,0) + X; pair Z = (19, 9); draw(W--X, EndArrow); draw(X--Y, EndArrow); draw(Y--Z, EndArrow); anglepen=black; anglefontpen=black; MarkAngle("\theta", C,Y,Z, 3); [/asy] The cross-section of a prism with index of refraction $1.5$ is an equilateral triangle, as shown above. A ray of light comes in horizontally from air into the prism, and has the opportunity to leave the prism, at an angle $\theta$ with respect to the surface of the triangle. Find $\theta$ in degrees and round to the nearest whole number. [i](Ahaan Rungta, 5 points)[/i]

Let $ABC$ be a right angled triangle at $A$. Denote $D$ the foot of the altitude through $A$ and $O_1, O_2$ the incentres of triangles $ADB$ and $ADC$. The circle with centre $A$ and radius $AD$ cuts $AB$ in $K$ and $AC$ in $L$. Show that $O_1, O_2, K$ and $L$ are on a line.

How many rectangles are in this figure? [asy] pair A,B,C,D,E,F,G,H,I,J,K,L; A=(0,0); B=(20,0); C=(20,20); D=(0,20); draw(A--B--C--D--cycle); E=(-10,-5); F=(13,-5); G=(13,5); H=(-10,5); draw(E--F--G--H--cycle); I=(10,-20); J=(18,-20); K=(18,13); L=(10,13); draw(I--J--K--L--cycle);[/asy] $ \textbf{(A)}\ 8\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 12 $

The letter F shown below is rotated $90^\circ$ clockwise around the origin, then reflected in the $y$-axis, and then rotated a half turn around the origin. What is the final image? [asy] import cse5;pathpen=black;pointpen=black; size(2cm); D((0,-2)--MP("y",(0,7),N)); D((-3,0)--MP("x",(5,0),E)); D((1,0)--(1,2)--(2,2)--(2,3)--(1,3)--(1,4)--(3,4)--(3,5)--(0,5)); [/asy][asy] import cse5;pathpen=black;pointpen=black; unitsize(0.2cm); D((0,-2)--MP("y",(0,7),N)); D(MP("\textbf{(A) }",(-3,0),W)--MP("x",(5,0),E)); D((1,0)--(1,2)--(2,2)--(2,3)--(1,3)--(1,4)--(3,4)--(3,5)--(0,5)); // D((18,-2)--MP("y",(18,7),N)); D(MP("\textbf{(B) }",(13,0),W)--MP("x",(21,0),E)); D((17,0)--(17,2)--(16,2)--(16,3)--(17,3)--(17,4)--(15,4)--(15,5)--(18,5)); // D((36,-2)--MP("y",(36,7),N)); D(MP("\textbf{(C) }",(29,0),W)--MP("x",(38,0),E)); D((31,0)--(31,1)--(33,1)--(33,2)--(34,2)--(34,1)--(35,1)--(35,3)--(36,3)); // D((0,-17)--MP("y",(0,-8),N)); D(MP("\textbf{(D) }",(-3,-15),W)--MP("x",(5,-15),E)); D((3,-15)--(3,-14)--(1,-14)--(1,-13)--(2,-13)--(2,-12)--(1,-12)--(1,-10)--(0,-10)); // D((15,-17)--MP("y",(15,-8),N)); D(MP("\textbf{(E) }",(13,-15),W)--MP("x",(22,-15),E)); D((15,-14)--(17,-14)--(17,-13)--(18,-13)--(18,-14)--(19,-14)--(19,-12)--(20,-12)--(20,-15)); [/asy]

Triangle $ABC$ is equilateral with $AB=1$. Points $E$ and $G$ are on $\overline{AC}$ and points $D$ and $F$ are on $\overline{AB}$ such that both $\overline{DE}$ and $\overline{FG}$ are parallel to $\overline{BC}$. Furthermore, triangle $ADE$ and trapezoids $DFGE$ and $FBCG$ all have the same perimeter. What is $DE+FG$? [asy] size(180); pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); real s=1/2,m=5/6,l=1; pair A=origin,B=(l,0),C=rotate(60)*l,D=(s,0),E=rotate(60)*s,F=m,G=rotate(60)*m; draw(A--B--C--cycle^^D--E^^F--G); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,S); label("$E$",E,NW); label("$F$",F,S); label("$G$",G,NW); [/asy] $\textbf{(A) }1\qquad \textbf{(B) }\dfrac{3}{2}\qquad \textbf{(C) }\dfrac{21}{13}\qquad \textbf{(D) }\dfrac{13}{8}\qquad \textbf{(E) }\dfrac{5}{3}\qquad$

In $\triangle ABC$, $AB<AC$. $D$ and $P$ are the feet of the internal and external angle bisectors of $\angle BAC$, respectively. $M$ is the midpoint of segment $BC$, and $\omega$ is the circumcircle of $\triangle APD$. Suppose $Q$ is on the minor arc $AD$ of $\omega$ such that $MQ$ is tangent to $\omega$. $QB$ meets $\omega$ again at $R$, and the line through $R$ perpendicular to $BC$ meets $PQ$ at $S$. Prove $SD$ is tangent to the circumcircle of $\triangle QDM$. [i]Proposed by Ray Li[/i]

Let $ABC$ and $PQR$ be two triangles such that [list] [b](a)[/b] $P$ is the mid-point of $BC$ and $A$ is the midpoint of $QR$. [b](b)[/b] $QR$ bisects $\angle BAC$ and $BC$ bisects $\angle QPR$ [/list] Prove that $AB+AC=PQ+PR$.

Find $\frac{area(CDF)}{area(CEF)}$ in the figure. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(5.75cm); 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 = -2, xmax = 21, ymin = -2, ymax = 16; /* image dimensions */ /* draw figures */ draw((0,0)--(20,0)); draw((13.48,14.62)--(7,0)); draw((0,0)--(15.93,9.12)); draw((13.48,14.62)--(20,0)); draw((13.48,14.62)--(0,0)); label("6",(15.16,12.72),SE*labelscalefactor); label("10",(18.56,5.1),SE*labelscalefactor); label("7",(3.26,-0.6),SE*labelscalefactor); label("13",(13.18,-0.71),SE*labelscalefactor); label("20",(5.07,8.33),SE*labelscalefactor); /* dots and labels */ dot((0,0),dotstyle); label("$B$", (-1.23,-1.48), NE * labelscalefactor); dot((20,0),dotstyle); label("$C$", (19.71,-1.59), NE * labelscalefactor); dot((7,0),dotstyle); label("$D$", (6.77,-1.64), NE * labelscalefactor); dot((13.48,14.62),dotstyle); label("$A$", (12.36,14.91), NE * labelscalefactor); dot((15.93,9.12),dotstyle); label("$E$", (16.42,9.21), NE * labelscalefactor); dot((9.38,5.37),dotstyle); label("$F$", (9.68,4.5), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle AFE?$ [asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy] $\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$

See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url] Triangle $ABC$ and straight line $l$ are given at the plane. Construct using a compass and a ruler the straightline which is parallel to $l$ and bisects the area of triangle $ABC.$

Points $E$ and $F$ lie inside rectangle $ABCD$ with $AE=DE=BF=CF=EF$. If $AB=11$ and $BC=8$, find the area of the quadrilateral $AEFB$.

[asy] import cse5; size(180); real a=4, b=3; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); //Credit to chezbgone2 for the diagram[/asy] In $\triangle ABC$, $AB = 10~ AC = 8$ and $BC = 6$. Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB$. Let $Q$ and $R$ be the points of intersection, distinct from $C$ , of circle $P$ with sides $AC$ and $BC$, respectively. The length of segment $QR$ is $\textbf{(A) }4.75\qquad\textbf{(B) }4.8\qquad\textbf{(C) }5\qquad\textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }3\sqrt{3}$

The diagram below shows four regular hexagons each with side length $1$ meter attached to the sides of a square. This figure is drawn onto a thin sheet of metal and cut out. The hexagons are then bent upward along the sides of the square so that $A_1$ meets $A_2$, $B_1$ meets $B_2$, $C_1$ meets $C_2$, and $D_1$ meets $D_2$. If the resulting dish is filled with water, the water will rise to the height of the corner where the $A_1$ and $A_2$ meet. there are relatively prime positive integers $m$ and $n$ so that the number of cubic meters of water the dish will hold is $\sqrt{\frac{m}{n}}$. Find $m+n$. [asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -4.3, xmax = 14.52, ymin = -8.3, ymax = 6.3; /* image dimensions */ draw((0,1)--(0,0)--(1,0)--(1,1)--cycle); draw((1,1)--(1,0)--(1.87,-0.5)--(2.73,0)--(2.73,1)--(1.87,1.5)--cycle); draw((0,1)--(1,1)--(1.5,1.87)--(1,2.73)--(0,2.73)--(-0.5,1.87)--cycle); draw((0,0)--(1,0)--(1.5,-0.87)--(1,-1.73)--(0,-1.73)--(-0.5,-0.87)--cycle); draw((0,1)--(0,0)--(-0.87,-0.5)--(-1.73,0)--(-1.73,1)--(-0.87,1.5)--cycle); /* draw figures */ draw((0,1)--(0,0)); draw((0,0)--(1,0)); draw((1,0)--(1,1)); draw((1,1)--(0,1)); draw((1,1)--(1,0)); draw((1,0)--(1.87,-0.5)); draw((1.87,-0.5)--(2.73,0)); draw((2.73,0)--(2.73,1)); draw((2.73,1)--(1.87,1.5)); draw((1.87,1.5)--(1,1)); draw((0,1)--(1,1)); draw((1,1)--(1.5,1.87)); draw((1.5,1.87)--(1,2.73)); draw((1,2.73)--(0,2.73)); draw((0,2.73)--(-0.5,1.87)); draw((-0.5,1.87)--(0,1)); /* dots and labels */ dot((1.87,-0.5),dotstyle); label("$C_1$", (1.72,-0.1), NE * labelscalefactor); dot((1.87,1.5),dotstyle); label("$B_2$", (1.76,1.04), NE * labelscalefactor); dot((1.5,1.87),dotstyle); label("$B_1$", (0.96,1.8), NE * labelscalefactor); dot((-0.5,1.87),dotstyle); label("$A_2$", (-0.26,1.78), NE * labelscalefactor); dot((-0.87,1.5),dotstyle); label("$A_1$", (-0.96,1.08), NE * labelscalefactor); dot((-0.87,-0.5),dotstyle); label("$D_2$", (-1.02,-0.18), NE * labelscalefactor); dot((-0.5,-0.87),dotstyle); label("$D_1$", (-0.22,-0.96), NE * labelscalefactor); dot((1.5,-0.87),dotstyle); label("$C_2$", (0.9,-0.94), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]

Let $ABC$ be a triangle, where $AB = AC$ and $BC = 12$. Let $D$ be the midpoint of $BC$. Let $E$ be a point in $AC$ such that $DE \perp AC$. Let $F$ be a point in $AB$ such that $EF \parallel BC$. If $EC = 4$, determine the length of $EF$.

Let $ABC$ be a triangle. Find all points $P$ on segment $BC$ satisfying the following property: If $X$ and $Y$ are the intersections of line $PA$ with the common external tangent lines of the circumcircles of triangles $PAB$ and $PAC$, then \[\left(\frac{PA}{XY}\right)^2+\frac{PB\cdot PC}{AB\cdot AC}=1.\]

$BB_{1}$ is a bisector of an acute triangle $ABC$. A perpendicular from $B_{1}$ to $BC$ meets a smaller arc $BC$ of a circumcircle of $ABC$ in a point $K$. A perpendicular from $B$ to $AK$ meets $AC$ in a point $L$. $BB_{1}$ meets arc $AC$ in $T$. Prove that $K$, $L$, $T$ are collinear. [i]V. Astakhov[/i]

Let $ABC$ be a triangle with symmedian point $K$. Select a point $A_1$ on line $BC$ such that the lines $AB$, $AC$, $A_1K$ and $BC$ are the sides of a cyclic quadrilateral. Define $B_1$ and $C_1$ similarly. Prove that $A_1$, $B_1$, and $C_1$ are collinear. [i]Proposed by Sammy Luo[/i]

Diagonals $AC$ and $BD$ of a trapezoid $ABCD$ meet at $P$. The circumcircles of triangles $ABP$ and $CDP$ intersect the line $AD$ for the second time at points $X$ and $Y$ respectively. Let $M$ be the midpoint of segment $XY$. Prove that $BM = CM$.