A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper? [asy]size(270pt); defaultpen(fontsize(10pt)); filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); dot((-3,3)); label("$A$",(-3,3),NW); draw((1,3)--(-3,-1),dashed+linewidth(.5)); draw((-1,3)--(3,-1),dashed+linewidth(.5)); draw((-1,-3)--(3,1),dashed+linewidth(.5)); draw((1,-3)--(-3,1),dashed+linewidth(.5)); draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5)); draw((0,3)--(0,-3),linetype("2.5 2.5")+linewidth(.5)); draw((3,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5)); label('$w$',(-1,-1),SW); label('$w$',(1,-1),SE); draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle); draw((4.5,0)--(8.5,0)); draw((6.5,2)--(6.5,-2)); label("$A$",(6.5,0),NW); dot((6.5,0)); [/asy] $\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h $

If the operation $x * y$ is defined by $x * y = (x+1)(y+1) - 1$, then which one of the following is FALSE? $\textbf{(A)} \ x * y = y *x$ for all real $x$ and $y$. $\textbf{(B)} \ x * (y + z) = ( x * y ) + (x * z)$ for all real $x,y,$ and $z$ $\textbf{(C)} \ (x-1) * (x+1) = (x * x) - 1$ for all real $x$. $\textbf{(D)} \ x * 0 = x$ for all real $x$. $\textbf{(E)} \ x * (y * z) = (x * y) * z$for all real $x,y,$ and $z$.

If a positive sequence $\{a_n\}_{n\geq 1}$ satisfies $\int_0^{a_n} x^{n}\ dx=2$, then find $\lim_{n\to\infty} a_n.$

On a $ 10 \times 10$ chessboard, some $ 4n$ unit squares are chosen to form a region $ \mathcal{R}.$ This region $ \mathcal{R}$ can be tiled by $ n$ $ 2 \times 2$ squares. This region $ \mathcal{R}$ can also be tiled by a combination of $ n$ pieces of the following types of shapes ([i]see below[/i], with rotations allowed). Determine the value of $ n.$

Let $a,b,c,x,y,$ and $z$ be complex numbers such that \[a=\dfrac{b+c}{x-2},\qquad b=\dfrac{c+a}{y-2},\qquad c=\dfrac{a+b}{z-2}.\] If $xy+yz+xz=67$ and $x+y+z=2010$, find the value of $xyz$.

Suppose $x,y$ and $z$ are integers that satisfy the system of equations \[x^2y+y^2z+z^2x=2186\] \[xy^2+yz^2+zx^2=2188.\] Evaluate $x^2+y^2+z^2.$

Each member of the sequence $112002, 11210, 1121, 117, 46, 34,\ldots$ is obtained by adding five times the rightmost digit to the number formed by omitting that digit. Determine the billionth ($10^9$th) member of this sequence.

Show that each integer $n$ can be written as the sum of five perfect cubes (not necessarily positive).

Let $ABC$ be a triangle and let $I$ and $O$ denote its incentre and circumcentre respectively. Let $\omega_A$ be the circle through $B$ and $C$ which is tangent to the incircle of the triangle $ABC$; the circles $\omega_B$ and $\omega_C$ are defined similarly. The circles $\omega_B$ and $\omega_C$ meet at a point $A'$ distinct from $A$; the points $B'$ and $C'$ are defined similarly. Prove that the lines $AA',BB'$ and $CC'$ are concurrent at a point on the line $IO$. [i](Russia) Fedor Ivlev[/i]

Let $f$ be a polynomial. We say that a complex number $p$ is a double attractor if there exists a polynomial $h(x)$ so that $f(x)-f(p) = h(x)(x-p)^2$ for all x \in R. Now, consider the polynomial $$f(x) = 12x^5 - 15x^4 - 40x^3 + 540x^2 - 2160x + 1,$$ and suppose that it’s double attractors are $a_1, a_2, ... , a_n$. If the sum $\sum^{n}_{i=1}|a_i|$ can be written as $\sqrt{a} +\sqrt{b}$, where $a, b$ are positive integers, find $a + b$.

In the adjoining figure $ TP$ and $ T'Q$ are parallel tangents to a circle of radius $ r$, with $ T$ and $ T'$ the points of tangency. $ PT''Q$ is a third tangent with $ T''$ as point of tangency. If $ TP\equal{}4$ and $ T'Q\equal{}9$ then $ r$ is [asy]unitsize(45); pair O = (0,0); pair T = dir(90); pair T1 = dir(270); pair T2 = dir(25); pair P = (.61,1); pair Q = (1.61, -1); draw(unitcircle); dot(O); label("O",O,W); label("T",T,N); label("T'",T1,S); label("T''",T2,NE); label("P",P,NE); label("Q",Q,S); draw(O--T2); label("$r$",midpoint(O--T2),NW); draw(T--P); label("4",midpoint(T--P),N); draw(T1--Q); label("9",midpoint(T1--Q),S); draw(P--Q);[/asy] $ \textbf{(A)}\ 25/6 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 25/4 \\ \qquad \textbf{(D)}\ \text{a number other than }25/6, 6, 25/4 \\ \qquad \textbf{(E)}\ \text{not determinable from the given information}$

Let $k<<n$ denote that $k<n$ and $k\mid n$. Let $f:\{1,2,...,2013\}\rightarrow \{1,2,...,M\}$ be such that, if $n\leq 2013$ and $k<<n$, then $f(k)<<f(n)$. What’s the least possible value of $M$?

Punches in the buses of a certain bus company always cut exactly six holes into the ticket. The possible locations of the holes form a $3 \times 3$ table as shown in the figure. Mr. Freerider wants to put together a collection of tickets such that, for any combination of punch holes, he would have a ticket with the same combination in his collection. The ticket can be viewed both from the front and from the back. Find the smallest number of tickets in such a collection. [img]https://cdn.artofproblemsolving.com/attachments/b/b/de5f09317a9a109fbecccecdc033de18217806.png[/img]

Prove that there exists $m \in \mathbb{N}$ such that there exists an integral sequence $\lbrace a_n \rbrace$ which satisfies: [b]I.[/b] $a_0 = 1, a_1 = 337$; [b]II.[/b] $(a_{n + 1} a_{n - 1} - a_n^2) + \frac{3}{4}(a_{n + 1} + a_{n - 1} - 2a_n) = m, \forall$ $n \geq 1$; [b]III. [/b]$\frac{1}{6}(a_n + 1)(2a_n + 1)$ is a perfect square $\forall$ $n \geq 1$.

Determine all positive integers$ n$ such that $f_n(x,y,z) = x^{2n} + y^{2n} + z^{2n} - xy - yz - zx$ divides $g_n(x,y, z) = (x - y)^{5n} + (y -z)^{5n} + (z - x)^{5n}$, as polynomials in $x, y, z$ with integer coefficients.

$a,b$ are real numbers (neither is $0$).Given two conditions: A: $a>0$. B: $a>b$ and $a^{-1}>b^{-1}$. Then, which one of the followings are true? $(\text{A})$A is sufficient but unnecessary condition of B. $(\text{B})$A is necessary but insufficient condition of B. $(\text{C})$A is sufficient and necessary condition of B. $(\text{D})$A is insufficient and unnecessary condition of B.

Find all natural $ x $ for which $ 3x+1 $ and $ 6x-2 $ are perfect squares, and the number $ 6x^2-1 $ is prime.

In rhombus $ABCD$, point $P$ lies on segment $\overline{AD}$ such that $BP\perp AD$, $AP = 3$, and $PD = 2$. What is the area of $ABCD$? [asy] import olympiad; size(180); real r = 3, s = 5, t = sqrt(r*r+s*s); defaultpen(linewidth(0.6) + fontsize(10)); pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0); draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D)); label("$A$",A,SW); label("$B$", B, NW); label("$C$",C,NE); label("$D$",D,SE); label("$P$",P,S); [/asy] $\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$

The set $M$ consists of $n$ points on the plane and satisfies the conditions: $\bullet$ there are $7$ points in the set $M$, which are vertices of a convex heptagon, $\bullet$ for arbitrary five points with $M$, which are vertices of a convex pentagon, there is a point that also belongs to $M$ and lies inside this pentagon. Find the smallest possible value that $n$ can take .

Let $ABC$ be an equilateral triangle. Let $A_1,B_1,C_1$ be interior points of $ABC$ such that $BA_1=A_1C$, $CB_1=B_1A$, $AC_1=C_1B$, and $$\angle BA_1C+\angle CB_1A+\angle AC_1B=480^\circ$$ Let $BC_1$ and $CB_1$ meet at $A_2,$ let $CA_1$ and $AC_1$ meet at $B_2,$ and let $AB_1$ and $BA_1$ meet at $C_2.$ Prove that if triangle $A_1B_1C_1$ is scalene, then the three circumcircles of triangles $AA_1A_2, BB_1B_2$ and $CC_1C_2$ all pass through two common points. (Note: a scalene triangle is one where no two sides have equal length.) [i]Proposed by Ankan Bhattacharya, USA[/i]

Let $f : [0, 1] \to [0, 1]$ satisfy $f(0) = 0, f(1) = 1$ and \[f(x + y) - f(x) = f(x) - f(x - y)\] for all $x, y \geq 0$ with $x - y, x + y \in [0, 1].$ Prove that $f(x) = x$ for all $x \in [0, 1].$

Let $n$ be a positive integer, with prime factorization $$n = p_1^{e_1}p_2^{e_2} \cdots p_r^{e_r}$$ for distinct primes $p_1, \ldots, p_r$ and $e_i$ positive integers. Define $$rad(n) = p_1p_2\cdots p_r,$$ the product of all distinct prime factors of $n$. Find all polynomials $P(x)$ with rational coefficients such that there exists infinitely many positive integers $n$ with $P(n) = rad(n)$.

Let $ABC$ be an equilateral triangle with side length $1$. Points $D$ and $E$ lie on $\overline{BC}$ and $\overline{AC}$ respectively such that $\triangle BDE$ is right isosceles, while points $F$ and $G$ lie on $\overline{BC}$ and $\overline{AB}$ respectively such that $\triangle CFG$ is right isosceles. Find the area of the intersection of $\triangle BDE$ and $\triangle CFG$. [i]Proposed by Ishin Shah[/i]

In an exotic country, the National Bank issues coins that can take any value in the interval $[0, 1]$. Find the smallest constant $c > 0$ such that the following holds, no matter the situation in that country: [i]Any citizen of the exotic country that has a finite number of coins, with a total value of no more than $1000$, can split those coins into $100$ boxes, such that the total value inside each box is at most $c$.[/i]

Let f be a function such that $f(0) = 0, f(1) = 1$, and $f(n) = 2f(n-1)- f(n- 2) + (-1)^n(2n - 4)$ for all integers $n \ge 2$. Find f(n) in terms of $n$.