Let $ABC$ an acute triangle. (a) Find the locus of points that are centers of rectangles whose vertices lie on the sides of $ABC$; (b) Determine if exist some points that are centers of $3$ distinct rectangles whose vertices lie on the sides of $ABC$.

Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is $ 4 : 3$. The horizontal length of a “$ 27$-inch” television screen is closest, in inches, to which of the following? [asy]import math; unitsize(7mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((0,0)--(4,0)--(4,3)--(0,3)--(0,0)--(4,3)); fill((0,0)--(4,0)--(4,3)--cycle,mediumgray); label(rotate(aTan(3.0/4.0))*"Diagonal",(2,1.5),NW); label(rotate(90)*"Height",(4,1.5),E); label("Length",(2,0),S);[/asy]$ \textbf{(A)}\ 20 \qquad \textbf{(B)}\ 20.5 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 21.5 \qquad \textbf{(E)}\ 22$

A $10 \times 10$ table consists of $100$ unit cells. A [i]block[/i] is a $2 \times 2$ square consisting of $4$ unit cells of the table. A set $C$ of $n$ blocks covers the table (i.e. each cell of the table is covered by some block of $C$ ) but no $n -1$ blocks of $C$ cover the table. Find the largest possible value of $n$.

Square $ABCD$ has side length $4$. Points $E$ and $F$ are the midpoints of sides $AB$ and $CD$, respectively. Eight $1$ by $2$ rectangles are placed inside the square so that no two of the eight rectangles overlap (see diagram). If the arrangement of eight rectangles is chosen randomly, then there are relatively prime positive integers $m$ and $n$ so that $\tfrac{m}{n}$ is the probability that none of the rectangles crosses the line segment $EF$ (as in the arrangement on the right). Find $m + n$. [asy] size(200); defaultpen(linewidth(0.8)+fontsize(10pt)); real r = 7; path square=origin--(4,0)--(4,4)--(0,4)--cycle; draw(square^^shift((r,0))*square,linewidth(1)); draw((1,4)--(1,0)^^(3,4)--(3,0)^^(0,2)--(1,2)^^(1,3)--(3,3)^^(1,1)--(3,1)^^(2,3)--(2,1)^^(3,2)--(4,2)); draw(shift((r,0))*((2,4)--(2,0)^^(0,2)--(4,2)^^(0,1)--(4,1)^^(0,3)--(2,3)^^(3,4)--(3,2))); label("A",(4,4),NE); label("A",(4+r,4),NE); label("B",(0,4),NW); label("B",(r,4),NW); label("C",(0,0),SW); label("C",(r,0),SW); label("D",(4,0),SE); label("D",(4+r,0),SE); label("E",(2,4),N); label("E",(2+r,4),N); label("F",(2,0),S); label("F",(2+r,0),S); [/asy]

Consider a rectangle $ABCD$ with $AB = a$ and $AD = b.$ Let $l$ be a line through $O,$ the center of the rectangle, that cuts $AD$ in $E$ such that $AE/ED = 1/2$. Let $M$ be any point on $l,$ interior to the rectangle. Find the necessary and sufficient condition on $a$ and $b$ that the four distances from M to lines $AD, AB, DC, BC$ in this order form an arithmetic progression.

Let $ABIH$,$BDEC$ and $ACFG$ be arbitrary rectangles constructed (externally) on the sides of triangle $ABC$.Choose point $S$ outside rectangle $ABIH$ (on the opposite side as triangle $ABC$) such that $\angle SHI=\angle FAC$ and $\angle HIS=\angle EBC$.Prove that the lines $FI,EH$ and $CS$ are concurrent(i.e., the three lines intersect in one point).

Let $H$ be a rectangle with angle between two diagonal $\leq 45^{0}$. Rotation $H$ around the its center with angle $0^{0}\leq x\leq 360^{0}$ we have rectangle $H_{x}$. Find $x$ such that $[H\cap H_{x}]$ minimum, where $[S]$ is area of $S$.

Figure $ABCD$ is a trapezoid with $AB || DC, AB = 5, BC = 3 \sqrt 2, \measuredangle BCD = 45^\circ$, and $\measuredangle CDA = 60^\circ$. The length of $DC$ is $\textbf{(A) }7 + \frac{2}{3} \sqrt{3}\qquad \textbf{(B) }8\qquad \textbf{(C) }9 \frac{1}{2}\qquad \textbf{(D) }8 + \sqrt 3\qquad \textbf{(E) }8 + 3 \sqrt 3$

We define a [i]chessboard polygon[/i] to be a polygon whose sides are situated along lines of the form $ x \equal{} a$ or $ y \equal{} b$, where $ a$ and $ b$ are integers. These lines divide the interior into unit squares, which are shaded alternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping $ 1 \times 2$ rectangles. Finally, a [i]tasteful tiling[/i] is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a $ 3 \times 4$ rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner. [asy]size(300); pathpen = linewidth(2.5); void chessboard(int a, int b, pair P){ for(int i = 0; i < a; ++i) for(int j = 0; j < b; ++j) if((i+j) % 2 == 1) fill(shift(P.x+i,P.y+j)*unitsquare,rgb(0.6,0.6,0.6)); D(P--P+(a,0)--P+(a,b)--P+(0,b)--cycle); } chessboard(2,2,(2.5,0));fill(unitsquare,rgb(0.6,0.6,0.6));fill(shift(1,1)*unitsquare,rgb(0.6,0.6,0.6)); chessboard(4,3,(6,0)); chessboard(4,3,(11,0)); MP("\mathrm{Distasteful\ tilings}",(2.25,3),fontsize(12)); /* draw lines */ D((0,0)--(2,0)--(2,2)--(0,2)--cycle); D((1,0)--(1,2)); D((2.5,1)--(4.5,1)); D((7,0)--(7,2)--(6,2)--(10,2)--(9,2)--(9,0)--(9,1)--(7,1)); D((8,2)--(8,3)); D((12,0)--(12,2)--(11,2)--(13,2)); D((13,1)--(15,1)--(14,1)--(14,3)); D((13,0)--(13,3));[/asy] a) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully. b) Prove that such a tasteful tiling is unique.

Given a natural number $n$, calculate the number of rectangles in the plane, the coordinates of whose vertices are integers in the range $0$ to $n$, and whose sides are parallel to the axes.

Show that it is possible to cut any triangle into several pieces, so that a rectangle is formed when they are joined together.

Let $ABCD$ be a quadrilateral inscribed in circle $\omega$. Define $E = AA \cap CD$, $F = AA \cap BC$, $G = BE \cap \omega$, $H = BE \cap AD$, $I = DF \cap \omega$, and $J = DF \cap AB$. Prove that $GI$, $HJ$, and the $B$-symmedian are concurrent. [i]Proposed by Robin Park[/i]

Let $O, H$ be circumcenter and orthogonal center of triangle $ABC$. $M,N$ are midpoints of $BH$ and $CH$. $BB'$ is diagonal of circumcircle. If $HONM$ is a cyclic quadrilateral, prove that $B'N=\frac12AC$.

Let $a,b,c,$ and $d$ be the lengths of sides $MN,NP,PQ,$ and $QM$, respectively, of quadrilateral $MNPQ$. If $A$ is the area of $MNPQ$, then $\textbf{(A) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is convex}$ $\textbf{(B) }A=\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$ $\textbf{(C) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a rectangle}$ $\textbf{(D) }A\le\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$ $\textbf{(E) }A\ge\left(\frac{a+c}{2}\right)\left(\frac{b+d}{2}\right)\text{ if and only if }MNPQ\text{ is a parallelogram}$

On a large chessboard $2n$ of its $1 \times 1$ squares have been marked such thar the rook (which moves only horizontally or vertically) can visit all the marked squares without jumpin over any unmarked ones. Prove that the figure consisting of all the marked squares can be cut into rectangles. (A Shapovalov)