Given a scalene acute triangle $ABC$, let M be the midpoints of its side $BC$ and $N$ the midpoint of the arc $BAC$ of its circumcircle. Let $\omega$ be the circle with diameter $BC$ and $D$, $E$ its intersections with the bisector of angle $\angle BAC$. Points $D'$, $E'$ lie on $\omega$ such that $DED'E' $ is a rectangle. Prove that $D'$, $E'$, $M$, $N$ lie on a single circle. [i] (Patrik Bak)[/i]

Let $ABCD$ be a rectangle with $AB>BC$. Let points $E,F$ be on side $CD$ such that $CE=ED$ and $BC=CF$. Show that if $AC$ is prependicular to $BE$ then $AB=BF$.

$ABCD$ is a rectangular sheet of paper that has been folded so that corner $B$ is matched with point $B'$ on edge $AD$. The crease is $EF$, where $E$ is on $AB$ and $F$is on $CD$. The dimensions $AE=8$, $BE=17$, and $CF=3$ are given. The perimeter of rectangle $ABCD$ is $m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(25,0), C=(25,70/3), D=(0,70/3), E=(8,0), F=(22,70/3), Bp=reflect(E,F)*B, Cp=reflect(E,F)*C; draw(F--D--A--E); draw(E--B--C--F, linetype("4 4")); filldraw(E--F--Cp--Bp--cycle, white, black); pair point=( 12.5, 35/3 ); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$B^\prime$", Bp, dir(point--Bp)); label("$C^\prime$", Cp, dir(point--Cp));[/asy]

Of the following statements, the one that is incorrect is: $ \textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}$ $ \textbf{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}$ $ \textbf{(C)}\ \text{Doubling the radius of a given circle doubles the area.}$ $ \textbf{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}$ $ \textbf{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}$

Several knights are arranged on an infinite chessboard. No square is attacked by more than one knight (in particular, a square occupied by a knight can be attacked by one knight but not by two). Sasha outlined a $ 14\times 16$ rectangle. What maximum number of knights can this rectangle contain?

In the figure, $ ABCD$ is a $ 2\times 2$ square, $ E$ is the midpoint of $ \overline{AD}$, and $ F$ is on $ \overline{BE}$. If $ \overline{CF}$ is perpendicular to $ \overline{BE}$, then the area of quadrilateral $ CDEF$ is [asy]defaultpen(linewidth(.8pt)); dotfactor=4; pair A = (0,2); pair B = origin; pair C = (2,0); pair D = (2,2); pair E = midpoint(A--D); pair F = foot(C,B,E); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("$A$",A,N);label("$B$",B,S);label("$C$",C,S);label("$D$",D,N);label("$E$",E,N);label("$F$",F,NW); draw(A--B--C--D--cycle); draw(B--E); draw(C--F); draw(rightanglemark(B,F,C,4));[/asy]$ \textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \minus{} \frac {\sqrt {3}}{2}\qquad \textbf{(C)}\ \frac {11}{5}\qquad \textbf{(D)}\ \sqrt {5}\qquad \textbf{(E)}\ \frac {9}{4}$

Let $ABCD$ be a rectangle and $P$ a point outside of it such that $\angle{BPC} = 90^{\circ}$ and the area of the pentagon $ABPCD$ is equal to $AB^{2}$. Show that $ABPCD$ can be divided in 3 pieces with straight cuts in such a way that a square can be built using those 3 pieces, without leaving any holes or placing pieces on top of each other. Note: the pieces can be rotated and flipped over.

A convex, closed figure lies inside a given circle. The figure is seen from every point of the circumference at a right angle (that is, the two rays drawn from the point and supporting the convex figure are perpendicular). Prove that the center of the circle is a center of symmetry of the figure.

Let $ ABC$ be a triangle. Squares $ AB_{c}B_{a}C$, $ CA_{b}A_{c}B$ and $ BC_{a}C_{b}A$ are outside the triangle. Square $ B_{c}B_{c}'B_{a}'B_{a}$ with center $ P$ is outside square $ AB_{c}B_{a}C$. Prove that $ BP,C_{a}B_{a}$ and $ A_{c}B_{c}$ are concurrent.

On a chessboard ($8\times 8$ squares with sides of length $1$) two diagonally opposite corner squares are taken away. Can the board now be covered with nonoverlapping rectangles with sides of lengths $1$ and $2$?

Let $ABC$ be a triangle with $\angle B=90$. The incircle of $ABC$ touches the side $BC$ at $D$. The incenters of triangles $ABD$ and $ADC$ are $X$ and $Z$ , respectively. The lines $XZ$ and $AD$ are intersecting at the point $K$. $XZ$ and circumcircle of $ABC$ are intersecting at $U$ and $V$. Let $M$ be the midpoint of line segment $[UV]$ . $AD$ intersects the circumcircle of $ABC$ at $Y$ other than $A$. Prove that $|CY|=2|MK|$ .

Let $n \ge 3$ be an integer, and consider a $n \times n$-board, divided into $n^2$ unit squares. For all $m \ge 1$, arbitrarily many $1\times m$-rectangles (type I) and arbitrarily many $m\times 1$-rectangles (type II) are available. We cover the board with $N$ such rectangles, without overlaps, and such that every rectangle lies entirely inside the board. We require that the number of type I rectangles used is equal to the number of type II rectangles used.(Note that a $1 \times 1$-rectangle has both types.) What is the minimal value of $N$ for which this is possible?

Let $k$ and $n$ be positive integers. Consider an array of $2\left(2^n-1\right)$ rows by $k$ columns. A $2$-coloring of the elements of the array is said to be [i]acceptable[/i] if any two columns agree on less than $2^n-1$ entries on the same row. Given $n$, determine the maximum value of $k$ for an acceptable $2$-coloring to exist.

The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle? $\textbf{(A)} \text{ } \frac{\sqrt{3}-1}{2} \qquad \textbf{(B)} \text{ } \frac{1}{2} \qquad \textbf{(C)} \text{ } \frac{\sqrt{5}-1}{2} \qquad \textbf{(D)} \text{ } \frac{\sqrt{2}}{2} \qquad \textbf{(E)} \text{ } \frac{\sqrt{6}-1}{2}$

Fold the next seven corners into a rectangle. [img]https://cdn.artofproblemsolving.com/attachments/b/b/2b8b9d6d4b72024996a66d41f865afb91bb9b7.png[/img]

Find the geometric locus of the center of a rectangle whose vertices lie on the perimeter of a given triangle.

A cubic block with dimensions $n$ by $n$ by $n$ is made up of a collection of $1$ by $1$ by $1$ unit cubes. What is the smallest value of $n$ so that if the outer layer of unit cubes are removed from the block, more than half the original unit cubes will still remain?

Let $A_n$ be the number of ways to tile a $4 \times n$ rectangle using $2 \times 1$ tiles. Prove that $A_n$ is divisible by 2 if and only if $A_n$ is divisible by 3.

A rectangle is said to be inscribed in a parallelogram if its vertices lie one on each side of the parallelogram. On the larger side $AB$ of the parallelogram $ABCD$, find all those points $K$ that are the vertices of the rectangles inscribed in $ABCD$.

In the diagram below, square $ABCD$ with side length 23 is cut into nine rectangles by two lines parallel to $\overline{AB}$ and two lines parallel to $\overline{BC}$. The areas of four of these rectangles are indicated in the diagram. Compute the largest possible value for the area of the central rectangle. [asy] size(250); defaultpen (linewidth (0.7) + fontsize (10)); draw ((0,0)--(23,0)--(23,23)--(0,23)--cycle); label("$A$", (0,23), NW); label("$B$", (23, 23), NE); label("$C$", (23,0), SE); label("$D$", (0,0), SW); draw((0,6)--(23,6)); draw((0,19)--(23,19)); draw((5,0)--(5,23)); draw((12,0)--(12,23)); label("13", (17/2, 21)); label("111",(35/2,25/2)); label("37",(17/2,3)); label("123",(2.5,12.5));[/asy] [i]Proposed by Lewis Chen[/i]

The graph of the equation $9x+223y=2007$ is drawn on graph paper with each square representing one unit in each direction. How many of the $1$ by $1$ graph paper squares have interiors lying entirely below the graph and entirely in the first quadrant?

We have an $n \times n$ board, and in every square there is an integer. The sum of all integers on the board is $0$. We define an action on a square where the integer in the square is decreased by the number of neighbouring squares, and the number inside each of the neighbouring squares is increased by 1. Determine if there exists $n\geq 2$ such that we can turn all the integers into zeros in a finite number of actions.

In a $2^n \times 2^n$ square with $n$ positive integer is covered with at least two non-overlapping rectangle pieces with integer dimensions and a power of two as surface. Prove that two rectangles of the covering have the same dimensions (Two rectangles have the same dimensions as they have the same width and the same height, wherein they, not allowed to be rotated.)

Into each box of a $ 2012 \times 2012 $ square grid, a real number greater than or equal to $ 0 $ and less than or equal to $ 1 $ is inserted. Consider splitting the grid into $2$ non-empty rectangles consisting of boxes of the grid by drawing a line parallel either to the horizontal or the vertical side of the grid. Suppose that for at least one of the resulting rectangles the sum of the numbers in the boxes within the rectangle is less than or equal to $ 1 $, no matter how the grid is split into $2$ such rectangles. Determine the maximum possible value for the sum of all the $ 2012 \times 2012 $ numbers inserted into the boxes.

In the diagram, congruent rectangles $ABCD$ and $DEFG$ have a common vertex $D$. Sides $BC$ and $EF$ meet at $H$. Given that $DA = DE = 8$, $AB = EF = 12$, and $BH = 7$. Find the area of $ABHED$. [img]https://cdn.artofproblemsolving.com/attachments/f/b/7225fa89097e7b20ea246b3aa920d2464080a5.png[/img]