In the plane we are given $3 \cdot n$ points ($n>$1) no three collinear, and the distance between any two of them is $\leq 1$. Prove that we can construct $n$ pairwise disjoint triangles such that: The vertex set of these triangles are exactly the given 3n points and the sum of the area of these triangles $< 1/2$.

$\text{ABCD}$ is a rectangle, $\text{D}$ is the center of the circle, and $\text{B}$ is on the circle. If $\text{AD}=4$ and $\text{CD}=3$, then the area of the shaded region is between [asy] pair A,B,C,D; A=(0,4); B=(3,4); C=(3,0); D=origin; draw(circle(D,5)); fill((0,5)..(1.5,4.7697)..B--A--cycle,black); fill(B..(4,3)..(5,0)--C--cycle,black); draw((0,5)--D--(5,0)); label("A",A,NW); label("B",B,NE); label("C",C,S); label("D",D,SW); [/asy] $\text{(A)}\ 4\text{ and }5 \qquad \text{(B)}\ 5\text{ and }6 \qquad \text{(C)}\ 6\text{ and }7 \qquad \text{(D)}\ 7\text{ and }8 \qquad \text{(E)}\ 8\text{ and }9$

Let $ABC$ a triangle and $P$ a point such that the triangles $PAB, PBC, PCA$ have the same area and the same perimeter. Prove that if: a) $P$ is in the interior of the triangle $ABC$ then $ABC$ is equilateral. b) $P$ is in the exterior of the triangle $ABC$ then $ABC$ is right angled triangle.

One hundred and one of the squares of an $n\times n$ table are colored blue. It is known that there exists a unique way to cut the table to rectangles along boundaries of its squares with the following property: every rectangle contains exactly one blue square. Find the smallest possible $n$.

In parallelogram $ ABCD$, point $ M$ is on $ \overline{AB}$ so that $ \frac{AM}{AB} \equal{} \frac{17}{1000}$ and point $ N$ is on $ \overline{AD}$ so that $ \frac{AN}{AD} \equal{} \frac{17}{2009}$. Let $ P$ be the point of intersection of $ \overline{AC}$ and $ \overline{MN}$. Find $ \frac{AC}{AP}$.

$ ABCD$ is a rectangle of area 2. $ P$ is a point on side $ CD$ and $ Q$ is the point where the incircle of $ \triangle PAB$ touches the side $ AB$. The product $ PA \cdot PB$ varies as $ ABCD$ and $ P$ vary. When $ PA \cdot PB$ attains its minimum value, a) Prove that $ AB \geq 2BC$, b) Find the value of $ AQ \cdot BQ$.

Square $ABCD$ is divided into four rectangles by $EF$ and $GH$. $EF$ is parallel to $AB$ and $GH$ parallel to $BC$. $\angle BAF = 18^\circ$. $EF$ and $GH$ meet at point $P$. The area of rectangle $PFCH$ is twice that of rectangle $AGPE$. Given that the value of $\angle FAH$ in degrees is $x$, find the nearest integer to $x$. [asy] size(100); defaultpen(linewidth(0.7)+fontsize(10)); pair D2(pair P) { dot(P,linewidth(3)); return P; } // NOTE: I've tampered with the angles to make the diagram not-to-scale. The correct numbers should be 72 instead of 76, and 45 instead of 55. pair A=(0,1), B=(0,0), C=(1,0), D=(1,1), F=intersectionpoints(A--A+2*dir(-76),B--C)[0], H=intersectionpoints(A--A+2*dir(-76+55),D--C)[0], E=F+(0,1), G=H-(1,0), P=intersectionpoints(E--F,G--H)[0]; draw(A--B--C--D--cycle); draw(F--A--H); draw(E--F); draw(G--H); label("$A$",D2(A),NW); label("$B$",D2(B),SW); label("$C$",D2(C),SE); label("$D$",D2(D),NE); label("$E$",D2(E),plain.N); label("$F$",D2(F),S); label("$G$",D2(G),W); label("$H$",D2(H),plain.E); label("$P$",D2(P),SE); [/asy]

A point $P$ in the interior of a convex polyhedron in Euclidean space is called a [i]pivot point[/i] of the polyhedron if every line through $P$ contains exactly $0$ or $2$ vertices of the polyhedron. Determine, with proof, the maximum number of pivot points that a polyhedron can contain.

Let $A_1A_2$ and $B_1B_2$ be the diagonals of a rectangle, and let $O$ be its center. Find and construct the set of all points $P$ that satisfy simultaneously the four inequaliies: $$A_1P > OP , \\A_2P > OP, \ \ B_1P > OP , \ \ B_2P > OP.$$

A rectangle with perimeter $ 176$ is divided into five congruent rectangles as shown in the diagram. What is the perimeter of one of the five congruent rectangles? [asy]defaultpen(linewidth(.8pt)); draw(origin--(0,3)--(4,3)--(4,0)--cycle); draw((0,1)--(4,1)); draw((2,0)--midpoint((0,1)--(4,1))); real r = 4/3; draw((r,3)--foot((r,3),(0,1),(4,1))); draw((2r,3)--foot((2r,3),(0,1),(4,1)));[/asy]$ \textbf{(A)}\ 35.2\qquad \textbf{(B)}\ 76\qquad \textbf{(C)}\ 80\qquad \textbf{(D)}\ 84\qquad \textbf{(E)}\ 86$

The points of intersection of $ xy \equal{} 12$ and $ x^2 \plus{} y^2 \equal{} 25$ are joined in succession. The resulting figure is: $ \textbf{(A)}\ \text{a straight line} \qquad\textbf{(B)}\ \text{an equilateral triangle} \qquad\textbf{(C)}\ \text{a parallelogram}$ $ \textbf{(D)}\ \text{a rectangle} \qquad\textbf{(E)}\ \text{a square}$

A triangle $ABC$ is inscribed in a rectangle $APQR$ so that points $B$ and $C$ lie on segments $PQ$ and $QR$, respectively. If $\alpha,\beta,\gamma$ are the angles of the triangle, prove that $$\cot\alpha\cdot S_{BCQ}=\cot\beta\cdot S_{ACR}+\cot\gamma\cdot S_{ABP}.$$

Given an oriented line $\Delta$ and a fixed point $A$ on it, consider all trapezoids $ABCD$ one of whose bases $AB$ lies on $\Delta$, in the positive direction. Let $E,F$ be the midpoints of $AB$ and $CD$ respectively. Find the loci of vertices $B,C,D$ of trapezoids that satisfy the following: [i](i) [/i] $|AB| \leq a$ ($a$ fixed); [i](ii) [/i] $|EF| = l$ ($l$ fixed); [i](iii)[/i] the sum of squares of the nonparallel sides of the trapezoid is constant. [hide="Remark"] [b]Remark.[/b] The constants are chosen so that such trapezoids exist.[/hide]

We consider the ways to divide a $1$ by $1$ square into rectangles (of which the sides are parallel to those of the square). All rectangles must have the same circumference, but not necessarily the same shape. a) Is it possible to divide the square into 20 rectangles, each having a circumference of $2:5$? b) Is it possible to divide the square into 30 rectangles, each having a circumference of $2$?

If $R$ and $S$ are two rectangles with integer sides such that the perimeter of $R$ equals the area of $S$ and the perimeter of $S$ equals the area of $R$, then we call $R$ and $S$ a friendly pair of rectangles. Find all friendly pairs of rectangles.

In a billiard with shape of a rectangle $ABCD$ with $AB=2013$ and $AD=1000$, a ball is launched along the line of the bisector of $\angle BAD$. Supposing that the ball is reflected on the sides with the same angle at the impact point as the angle shot , examine if it shall ever reach at vertex B.

In $ xy \minus{}$plane, there are $ b$ blue and $ r$ red rectangles whose sides are parallel to the axis. Any parallel line to the axis can intersect at most one rectangle with same color. For any two rectangle with different colors, there is a line which is parallel to the axis and which intersects only these two rectangles. $ (b,r)$ cannot be ? $\textbf{(A)}\ (1,7) \qquad\textbf{(B)}\ (2,6) \qquad\textbf{(C)}\ (3,4) \qquad\textbf{(D)}\ (3,3) \qquad\textbf{(E)}\ \text{None}$

Cut the $9 \times 10$ grid rectangle along the grid lines into several squares so that there are exactly two of them with odd sidelengths.

Every natural number is coloured in one of the $ k$ colors. Prove that there exist four distinct natural numbers $ a, b, c, d$, all coloured in the same colour, such that $ ad \equal{} bc$, $ \displaystyle \frac b a$ is power of 2 and $ \displaystyle \frac c a$ is power of 3.

Rectangle $ABCD$ is divided into four parts of equal area by five segments as shown in the figure, where $XY = YB + BC + CZ = ZW = WD + DA + AX$, and $PQ$ is parallel to $AB$. Find the length of $AB$ (in cm) if $BC = 19$ cm and $PQ = 87$ cm. [asy] size(250); pair A=origin, B=(96,0), C=(96,22), D=(0,22), W=(16,22), X=(20,0), Y=(80,0), Z=(76,22), P=(24,11), Q=(72,11); draw(P--X--A--D--W--P--Q--Y--B--C--Z--Q^^W--Z^^X--Y); dot(A^^B^^C^^D^^P^^Q^^W^^X^^Y^^Z); pair point=(48,11); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$P$", P, dir(point--P)); label("$Q$", Q, dir(point--Q)); label("$W$", W, N); label("$X$", X, S); label("$Y$", Y, S); label("$Z$", Z, N);[/asy]

Let $n$ and $t$ be positive integers. What is the number of ways to place $t$ dominoes $(1\times 2$ or $2\times 1$ rectangles) in a $2\times n$ table so that there is no $2\times 2$ square formed by $2$ dominoes and each $2\times 3$ rectangle either does not have a horizontal domino in the middle and last cell in the first row or does not have a horizontal domino in the first and middle cell in the second row (or both)?

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 $m$ and $n$ be positive integers. Mr. Fat has a set $S$ containing every rectangular tile with integer side lengths and area of a power of $2$. Mr. Fat also has a rectangle $R$ with dimensions $2^m \times 2^n$ and a $1 \times 1$ square removed from one of the corners. Mr. Fat wants to choose $m + n$ rectangles from $S$, with respective areas $2^0, 2^1, \ldots, 2^{m + n - 1}$, and then tile $R$ with the chosen rectangles. Prove that this can be done in at most $(m + n)!$ ways. [i]Palmer Mebane.[/i]

Determine all positive integers $m$ satisfying the condition that there exists a unique positive integer $n$ such that there exists a rectangle which can be decomposed into $n$ congruent squares and can also be decomposed into $m+n$ congruent squares.

$ABCD$ is a cyclic quadrilateral. The lines $AD$ and $BC$ meet at X, and the lines $AB$ and $CD$ meet at $Y$ . The line joining the midpoints $M$ and $N$ of the diagonals $AC$ and $BD$, respectively, meets the internal bisector of angle $AXB$ at $P$ and the external bisector of angle $BYC$ at $Q$. Prove that $PXQY$ is a rectangle