Sammy has a wooden board, shaped as a rectangle with length $2^{2014}$ and height $3^{2014}$. The board is divided into a grid of unit squares. A termite starts at either the left or bottom edge of the rectangle, and walks along the gridlines by moving either to the right or upwards, until it reaches an edge opposite the one from which the termite started. Depicted below are two possible paths of the termite. [img]https://cdn.artofproblemsolving.com/attachments/3/0/39f3b2aa9c61ff24ffc22b968790f4c61da6f9.png[/img] The termite’s path dissects the board into two parts. Sammy is surprised to find that he can still arrange the pieces to form a new rectangle not congruent to the original rectangle. This rectangle has perimeter $P$. How many possible values of $P$ are there?

In triangle $ABC$, point $M$ is the midpoint of $BC, P$ is the intersection point of the tangents at points $B$ and $C$ of the circumscribed circle, $N$ is the midpoint of the segment $MP$. The segment $AN$ intersects the circumscribed circle at point $Q$. Prove that $\angle PMQ = \angle MAQ$.

A domino is a $2$ by $1$ rectangle . For what integers $m$ and $n$ can we cover an $m*n$ rectangle with non-overlapping dominoes???

Let $ ABC$ be an isosceles triangle with $ AC \equal{} BC.$ Its incircle touches $ AB$ in $ D$ and $ BC$ in $ E.$ A line distinct of $ AE$ goes through $ A$ and intersects the incircle in $ F$ and $ G.$ Line $ AB$ intersects line $ EF$ and $ EG$ in $ K$ and $ L,$ respectively. Prove that $ DK \equal{} DL.$

Given a triangle $ABC$, let $P$ lie on the circumcircle of the triangle and be the midpoint of the arc $BC$ which does not contain $A$. Draw a straight line $l$ through $P$ so that $l$ is parallel to $AB$. Denote by $k$ the circle which passes through $B$, and is tangent to $l$ at the point $P$. Let $Q$ be the second point of intersection of $k$ and the line $AB$ (if there is no second point of intersection, choose $Q = B$). Prove that $AQ = AC$.

An acute triangle $ABC$ is given. Let $D$ be the foot of altitude dropped for $A$. Tangents from $D$ to circles with diameters $AB$ and $AC$ intersects with the said circles at $K$ and $L$, in respective. Point $S$ in the plane is given so that $\angle ABC + \angle ABS = \angle ACB + \angle ACS = 180^\circ$. Prove that $A, K, L$ and $S$ lie on a circle.

We are given convex quadrilateral $ABCD$. The midpoints of $BC$ and $DA$ are $M$ and $N$ respectively. The diagonal $AC$ divides $MN$ in half. Prove that the areas of triangles $ABC$ and $ACD$ are equal .

Let $ABCD$ be a parallelogram. The interior angle bisector of $\angle ADC$ intersects the line $BC$ in $E$, and the perpendicular bisector of the side $AD$ intersects the line $DE$ in $M$. Let $F= AM \cap BC$. Prove that: a) $DE=AF$; b) $AD\cdot AB = DE\cdot DM$. [i]Daniela and Marius Lobaza, Timisoara[/i]

At the bottom-left corner of a $2014\times 2014$ chessboard, there are some green worms and at the top-left corner of the same chessboard, there are some brown worms. Green worms can move only to right and up, and brown worms can move only to right and down. After a while, the worms make some moves and all of the unit squares of the chessboard become occupied at least once throughout this process. Find the minimum total number of the worms.

Let $ABCD$ be a convex quadrilateral. Is it possible to find a point $P$ such that the segments drawn between $P$ and the midpoints of the sides of $ABCD$ divide the quadrilateral into four sections of equal area? If $P$ exists, is it unique?

Let the inscribed circle of the triangle $\vartriangle ABC$ touch side $BC$ at $M$ , side $CA$ at $N$ and side $AB$ at $P$ . Let $D$ be a point from $\left[ NP \right]$ such that $\frac{DP}{DN}=\frac{BD}{CD}$ . Show that $DM \perp PN$ .

In an acute $\Delta ABC$, $AH_a$ and $BH_b$ are altitudes and $M$ is the middle point of $AB$. The circumscribed circles of $\Delta AMH_a$ and $\Delta BMH_b$ intersect for a second time in $P$. Prove that point $P$ lies on the circumscribed circle of $\Delta ABC$.

Justify the following construction of the bisector of an angle with an inaccessible vertex: [img]https://cdn.artofproblemsolving.com/attachments/9/d/be4f7799d58a28cab3b4c515633b0e021c1502.png[/img] $M \in a$ and $N \in b$ are taken, the $4$ bisectors of the $4$ internal angles formed by $MN$ are traced with $a$ and $ b$. Said bisectors intersect at $P$ and $Q$, then $PQ$ is the bisector sought.

Tom, Dick and Harry started out on a $ 100$-mile journey. Tom and Harry went by automobile at the rate of $ 25$ mph, while Dick walked at the rate of $ 5$ mph. After a certain distance, Harry got off and walked on at $ 5$ mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was: $ \textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 7 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ \text{none of these answers}$

Let $ABC$ be a triangle and let the tangent at $B{}$ to its circumcircle meet the internal bisector of the angle $A{}$ at $P{}$. The line through $P{}$ parallel to $AC$ meets $AB$ at $Q{}$. Assume that $Q{}$ lies in the interior of segment $AB$ and let the line through $Q{}$ parallel to $BC$ meet $AC$ at $X{}$ and $PC$ at $Y{}$. Prove that $PX$ is tangent to the circumcircle of the triangle $XYC$.

Let $ABC$ be a triangle with $\angle CAB > 45^o$ and $\angle CBA > 45^o$. Construct an isosceles right angled triangle $RAB$ with $AB$ as its hypotenuse and $R$ inside $ABC$. Also construct isosceles right angled triangles $ACQ$ and $BCP$ having $AC$ and $BC$ respectively as their hypotenuses and lying entirely outside $ABC$. Show that $CQRP$ is a parallelogram.

A circle $K_1$ of radius $r_1 = 1\slash 2$ is inscribed in a semi-circle $H$ with diameter $AB$ and radius $1.$ A sequence of different circles $K_2, K_3, \ldots$ with radii $r_2, r_3, \ldots$ respectively are drawn so that for each $n\geq 1$, the circle $K_{n+1}$ is tangent to $H$, $K_n$ and $AB.$ Prove that $a_n = 1\slash r_n$ is an integer for each $n$, and that it is a perfect square for $n$ even and twice a perfect square for $n$ odd.

An isosceles right-angled triangle $ABC$ is given. On the extensions of sides $AB$ and $AC$, behind vertices $B$ and $C$ equal segments $BK$ and $CL$ were laid. $E$ and F are the points of intersection of the segment $KL$ and the lines perpendicular to the $KC$ , passing through the points $B$ and $A$, respectively. Prove that $EF = FL$.

The convex pentagon $ABC DE$ is given, such that $AB,BC,CD$ and $DE$ are parallel to one of the diagonals. Prove that this also applies to $EA$.

Consider $D$ the midpoint of the base $[BC]$ of the isosceles triangle ABC in which $\angle BAC < 90^o$. On the perpendicular from $B$ on the line $BC$ consider the point $E$ such that $\angle EAB= \angle BAC$, and on the line passing though $C$ parallel to the line $AB$ we consider the point $F$ such that $F$ and $D$ are on different side of the line $AC$ and $\angle FAC = \angle CAD$. Prove that $AE = CF$ and $BF = EF$

Let $I, O$ be the incenter and the circumcenter of triangle $ABC$, and $D,E,F$ be the circumcenters of triangle $ BIC, CIA, AIB$. Let $ P, Q, R$ be the midpoints of segments $ DI, EI, FI $. Prove that the circumcenter of triangle $PQR $, $M$, is the midpoint of segment $IO$.

On the diagonal $AC$ of the convex quadrilateral $ABCD$ points $P$,$Q$ are chosen such that triangles $ABD$,$PCD$ and $QBD$ are similar to each other in this order. Prove that $AQ=PC$ [i]M. Zorka[/i]

In the figure, $ \angle EAB$ and $ \angle ABC$ are right angles. $ AB \equal{} 4, BC \equal{} 6, AE \equal{} 8$, and $ \overline{AC}$ and $ \overline{BE}$ intersect at $ D$. What is the difference between the areas of $ \triangle ADE$ and $ \triangle BDC$? [asy]unitsize(4mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair A=(0,0), B=(4,0), C=(4,6), Ep=(0,8); pair D=extension(A,C,Ep,B); draw(A--C--B--A--Ep--B); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$E$",Ep,N); label("$D$",D,2.5*N); label("$4$",midpoint(A--B),S); label("$6$",midpoint(B--C),E); label("$8$",(0,3),W);[/asy]$ \textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 9$

The base of a triangle is of length $b$, and the latitude is of length $h$. A rectangle of height $x$ is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is: $ \textbf{(A)}\ \frac{bx}{h}(h-x)\qquad\textbf{(B)}\ \frac{hx}{b}(b-x)\qquad\textbf{(C)}\ \frac{bx}{h}(h-2x)\qquad$ $\textbf{(D)}\ x(b-x)\qquad\textbf{(E)}\ x(h-x) $

Let $ABCD$ be a convex quadrilateral with these properties: $\angle ADC = 135^o$ and $\angle ADB - \angle ABD = 2\angle DAB = 4\angle CBD$. If $BC = \sqrt2 CD$ , prove that $AB = BC + AD$. by Mahdi Etesami Fard