A 2 x 3 rectangle and a 3 x 4 rectangle are contained within a square without overlapping at any interior point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square? $ \textbf{(A) } 16 \qquad \textbf{(B) } 25 \qquad \textbf{(C) } 36 \qquad \textbf{(D) } 49 \qquad \textbf{(E) } 64$

Each unit square of a 3-by-3 unit-square grid is to be colored either blue or red. For each square, either color is equally likely to be used. The probability of obtaining a grid that does not have a 2-by-2 red square is $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Let $ABC$ be a triangle such that $CA \neq CB$, the points $A_{1}$ and $B_{1}$ are tangency points for the ex-circles relative to sides $CB$ and $CA$, respectively, and $I$ the incircle. The line $CI$ intersects the cincumcircle of the triangle $ABC$ in the point $P$. The line that trough $P$ that is perpendicular to $CP$, intersects the line $AB$ in $Q$. Prove that the lines $QI$ and $A_{1}B_{1}$ are parallels.

A rectangle intersects a circle as shown: $AB=4$, $BC=5$, and $DE=3$. Then $EF$ equals: [asy]size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F); draw(D--G--H--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, SW); label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); [/asy] $\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$

Let $ABCD$ be a fixed convex quadrilateral. Say a point $K$ is [i]pastanaga[/i] if there's a rectangle $PQRS$ centered at $K$ such that $A\in PQ, B\in QR, C\in RS, D\in SP$. Prove there exists a circle $\omega$ depending only on $ABCD$ that contains all pastanaga points.

A rectangle is divided into $n^2$ smaller rectangle by $n - 1$ horizontal lines and $n - 1$ vertical lines. Between those rectangles there are exactly $5660$ which are not congruent. For what minimum value of $n$ is this possible?

I couldn't solve this problem and the only solution I was able to find was very unnatural (it was an official solution, I think) and I couldn't be satisfied with it, so I ask you if you can find some different solutions. The problem is really great one! If $M$ is the centroid of a triangle $ABC$, prove that the following inequality holds: \[\sin\angle CAM+\sin\angle CBM\leq\frac{2}{\sqrt3}.\] The equality occurs in a very strange case, I don't remember it.

Consider a $1 \times n$ rectangle and some tiles of size $1 \times 1$ of four different colours. The rectangle is tiled in such a way that no two neighboring square tiles have the same colour. a) Find the number of distinct symmetrical tilings. b) Find the number of tilings such that any consecutive square tiles have distinct colours.

Two congruent rectangles are located as shown in the figure. Find the area of the shaded part. [img]https://cdn.artofproblemsolving.com/attachments/2/e/10b164535ab5b3a3b98ce1a0b84892cd11d76f.png[/img]

A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even. [i]Proposed by Jeck Lim, Singapore[/i]

Consider on the coordinate plane all rectangles whose (i) vertices have integer coordinates; (ii) edges are parallel to coordinate axes; (iii) area is $2^k$, where $k = 0,1,2....$ Is it possible to color all points with integer coordinates in two colors so that no such rectangle has all its vertices of the same color?

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.

A square $ (n \minus{} 1) \times (n \minus{} 1)$ is divided into $ (n \minus{} 1)^2$ unit squares in the usual manner. Each of the $ n^2$ vertices of these squares is to be coloured red or blue. Find the number of different colourings such that each unit square has exactly two red vertices. (Two colouring schemse are regarded as different if at least one vertex is coloured differently in the two schemes.)

Consider a rectangle $ABCD$ with $AB = 2$ and $BC = \sqrt3$. The point $M$ lies on the side $AD$ so that $MD = 2 AM$ and the point $N$ is the midpoint of the segment $AB$. On the plane of the rectangle rises the perpendicular MP and we choose the point $Q$ on the segment $MP$ such that the measure of the angle between the planes $(MPC)$ and $(NPC)$ shall be $45^o$, and the measure of the angle between the planes $(MPC)$ and $(QNC)$ shall be $60^o$. a) Show that the lines $DN$ and $CM$ are perpendicular. b) Show that the point $Q$ is the midpoint of the segment $MP$.

A rectangular piece of paper $ABCD$ has sides of lengths $AB = 1$, $BC = 2$. The rectangle is folded in half such that $AD$ coincides with $BC$ and $EF$ is the folding line. Then fold the paper along a line $BM$ such that the corner $A$ falls on line $EF$. How large, in degrees, is $\angle ABM$? [asy] size(180); pathpen = rgb(0,0,0.6)+linewidth(1); pointpen = black+linewidth(3); pointfontpen = fontsize(10); pen dd = rgb(0,0,0.6) + linewidth(0.7) + linetype("4 4"), dr = rgb(0.8,0,0), dg = rgb(0,0.6,0), db = rgb(0,0,0.6)+linewidth(1); pair A=(0,1), B=(0,0), C=(2,0), D=(2,1), E=A/2, F=(2,.5), M=(1/3^.5,1), N=reflect(B,M)*A; D(B--M--D("N",N,NE)--B--D("C",C,SE)--D("D",D,NE)--M); D(D("M",M,plain.N)--D("A",A,NW)--D("B",B,SW),dd); D(D("E",E,W)--D("F",F,plain.E),dd); [/asy]