Prove that if the four faces of a tetrahedron are of the same area they are equal.

5 blocks of volume 1 cm$^3$, 1 cm$^3$, 1 cm$^3$, 1 cm$^3$ and 4 cm$^3$ are placed one above another to form a structure as shown in the figure. Suppose sum of surface areas of upper face of each is 48 cm$^2$ . Determine the minimum possible height of the whole structure.

A pair of positive integers $m, n$ is called [i]guerrera[/i], if there exists positive integers $a, b, c, d$ such that $m=ab$, $n=cd$ and $a+b=c+d$. For example the pair $8, 9$ is [i]guerrera[/i] cause $8= 4 \cdot 2$, $9= 3 \cdot 3$ and $4+2=3+3$. We paint the positive integers if the following order: We start painting the numbers $3$ and $5$. If a positive integer $x$ is not painted and a positive $y$ is painted such that the pair $x, y$ is [i]guerrera[/i], we paint $x$. Find all positive integers $x$ that can be painted.

We have $n>2$ non-zero integers such that each one of them is divisible by the sum of the other $n-1$ numbers. Prove that the sum of all the given numbers is zero.

The lines with equations $ax-2y=c$ and $2x+by=-c$ are perpendicular and intersect at $(1, -5)$. What is $c$? $\textbf{(A) } -13\qquad \textbf{(B) } -8\qquad \textbf{(C) } 2\qquad \textbf{(D) } 8\qquad \textbf{(E) } 13$

A board that has some of its squares painted black is called [i]acceptable [/i] if there are no four black squares that form a $2 \times 2$ subboard. Find the largest real number $\lambda$ such that for every positive integer $n$ the following proposition holds: mercy: if an $n \times n$ board is acceptable and has fewer than $\lambda n^2$ dark squares, then an additional square black can be painted so that the board is still acceptable.

On the last afternoon of the Kubik family vacation, Michael puts down a copy of $\textit{Mathematical Olympiad Challenges}$ and goes out to play tennis. Wendy notices the book and decides to see if there are a few problems in it that she can solve. She decides to focus her energy on one problem in particular: \[\begin{array}{l}\text{Given 69 distinct positive integers not exceeding 100, prove that one can}\\\text{choose four of them }a,b,c,d\text{ such that }a<b<c\text{ and } a+b+c=d. \text{ Is this}\\\text{statement true for 68 numbers?}\end{array}\] After some time working on the problem, Wendy finally feels like she has a grip on the solution. When Michael returns, she explains her solutions to him. "Well done!" he tells her. "Now, see if you can solve this generalization. Consider the set \[S=\{1,2,3,\ldots,2007,2008\}.\] Find the smallest value of $t$ such that given any subset $T$ of $S$ where $|T|=t$, then there are necessarily distinct $a,b,c,d\in T$ for which $a+b+c=d$." Find the answer to Michael's generalization.

Given collinear points $A,B,C$ such that $AB = BC$. How can you construct a point $D$ on $AB$ such that $AD = 2DB$, using only a straightedge? (You are not allowed to measure distances)

At the theater children get in for half price. The price for $5$ adult tickets and $4$ child tickets is $\$24.50$. How much would $8$ adult tickets and $6$ child tickets cost? $\textbf{(A) }\$35\qquad \textbf{(B) }\$38.50\qquad \textbf{(C) }\$40\qquad \textbf{(D) }\$42\qquad \textbf{(E) }\$42.50$

Points $X$ and $Y$ are the midpoints of arcs $AB$ and $BC$ of the circumscribed circle of triangle $ABC$. Point $T$ lies on side $AC$. It turned out that the bisectors of the angles $ATB$ and $BTC$ pass through points $X$ and $Y$ respectively. What angle $B$ can be in triangle $ABC$?

Prove that any polynomial of the form $1+a_nx^n + a_{n+1}x^{n+1} + \cdots + a_kx^k$ ($k\ge n$) has at least $n-2$ non-real roots (counting multiplicity), where the $a_i$ ($n\le i\le k$) are real and $a_k\ne 0$. [i]David Yang.[/i]

1. Let $ABC$ be an acute-angled triangle with $AB\neq AC$. The circle with diameter $BC$ intersects the sides $AB$ and $AC$ at $M$ and $N$ respectively. Denote by $O$ the midpoint of the side $BC$. The bisectors of the angles $\angle BAC$ and $\angle MON$ intersect at $R$. Prove that the circumcircles of the triangles $BMR$ and $CNR$ have a common point lying on the side $BC$.

Find the largest prime that divides $1\cdot 2\cdot 3+2\cdot 3\cdot 4+\cdots +44\cdot 45\cdot 46$

Several zeros, ones and twos are written on the blackboard. An anonymous clean in turn pairs of different numbers, writing, instead of cleaned, the number not equal to each. ($0$ instead of pair $\{1,2\}, 1$ instead of $\{0,2\}, 2$ instead of $\{0,1\}$). Prove that if there remains one number only, it does not depend on the processing order.

A triangle $ABC$ is inscribed in a circle $\omega$. A variable line $\ell$ chosen parallel to $BC$ meets segments $AB$, $AC$ at points $D$, $E$ respectively, and meets $\omega$ at points $K$, $L$ (where $D$ lies between $K$ and $E$). Circle $\gamma_1$ is tangent to the segments $KD$ and $BD$ and also tangent to $\omega$, while circle $\gamma_2$ is tangent to the segments $LE$ and $CE$ and also tangent to $\omega$. Determine the locus, as $\ell$ varies, of the meeting point of the common inner tangents to $\gamma_1$ and $\gamma_2$. [i](Russia) Vasily Mokin and Fedor Ivlev[/i]

Prove the following statement: If a polynomial $f(x)$ with real coefficients takes only nonnegative values, then there exists a positive integer $n$ and polynomials $g_1(x), g_2(x),\cdots, g_n(x)$ such that \[f(x) = g_1(x)^2 + g_2(x)^2 +\cdots+ g_n(x)^2\]

Find all positive integers $n$ such that the equation $\frac{1}{x} + \frac{1}{y} = \frac{1}{n}$ has exactly $2011$ positive integer solutions $(x,y)$ where $x \leq y$.

Let $p_n(k)$ be the number of permutations of the set $\{1,2,3,\ldots,n\}$ which have exactly $k$ fixed points. Prove that $\sum_{k=0}^nk p_n(k)=n!$.[i](IMO Problem 1)[/i] [b][i]Original formulation [/i][/b] Let $S$ be a set of $n$ elements. We denote the number of all permutations of $S$ that have exactly $k$ fixed points by $p_n(k).$ Prove: (a) $\sum_{k=0}^{n} kp_n(k)=n! \ ;$ (b) $\sum_{k=0}^{n} (k-1)^2 p_n(k) =n! $ [i]Proposed by Germany, FR[/i]

The captain and his three sailors get $2009$ golden coins with the same value . The four people decided to divide these coins by the following rules : sailor $1$,sailor $2$,sailor $3$ everyone write down an integer $b_1,b_2,b_3$ , satisfies $b_1\ge b_2\ge b_3$ , and ${b_1+b_2+b_3=2009}$; the captain dosen't know what the numbers the sailors have written . He divides $2009$ coins into $3$ piles , with number of coins: $a_1,a_2,a_3$ , and $a_1\ge a_2\ge a_3$ . For sailor $k$ ($k=1,2,3$) , if $b_k<a_k$ , then he can take $b_k$ coins from the $k$th pile ; if $b_k\ge a_k$ , then he can't take any coins away . At last , the captain own the rest of the coins .If no matter what the numbers the sailors write , the captain can make sure that he always gets $n$ coins . Find the largest possible value of $n$ and prove your conclusion .

How many ways are there to fill in the $ 8$ spots in the picture with letters $A, B, C$ and $D$, using two copies of each letter, such that the spots with identical letters can be connected with a continuous line that stays within the box, without these four lines crossing each other or going through other spots? The lines do not have to be straight. [img]https://cdn.artofproblemsolving.com/attachments/f/f/66c30eaf6fa3b42c5197d0e3a3d59e9160bb8e.png[/img]

Evaluate $$\frac{1}{1!21!} + \frac{1}{3!19!} + \frac{1}{5!16!} + ... + \frac{1}{21!1!}$$

Find all the integers $n \ge 5$ such that the residue of $n$ when divided by each prime number smaller than $\frac{n}{2}$ is odd.

If $60^a = 3$ and $60^b = 5$, then $12^{[(1-a-b)/2(1-b)]}$ is $\text{(A)} \ \sqrt{3} \qquad \text{(B)} \ 2 \qquad \text{(C)} \ \sqrt{5} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \sqrt{12}$

Given a square $ABCD$ of side length $6$, the point $E$ is drawn on the line $AB$ such that the distance $EA$ is less than $EB$ and the triangle $\vartriangle BCE$ has the same area as $ABCD$. Compute the shaded area. [img]https://cdn.artofproblemsolving.com/attachments/a/8/5d945a593aee58af3af94f4e8e967eeaeefa6a.png[/img]

The digit in unit place of $1!+2!+\ldots+99!$ is $\textbf{(A)}~3$ $\textbf{(B)}~0$ $\textbf{(C)}~1$ $\textbf{(D)}~7$