Let $A,B$ and $C$ be three points on a circle of radius $1$. (a) Show that the area of the triangle $ABC$ equals \[\frac12(\sin(2\angle ABC)+\sin(2\angle BCA)+\sin(2\angle CAB))\] (b) Suppose that the magnitude of $\angle ABC$ is fixed. Then show that the area of the triangle $ABC$ is maximized when $\angle BCA=\angle CAB$ (c) Hence or otherwise, show that the area of the triangle $ABC$ is maximum when the triangle is equilateral.

Let $m$ and $n$ be positive integers such that $$(m^3 -27)(n^3- 27) = 27(m^2n^2 + 27):$$ Find the maximum possible value of $m^3 + n^3$.

$x_1, x_2,\cdots,x_{n+1}$ are posive real numbers satisfying the equation $\frac{1}{1+x_1} + \frac{1}{1+x_2} + \cdots + \frac{1}{1+x_{n+1}} =1$ Prove that $x_1x_2 \cdots x_{n+1} \geq n^{n+1}$.

The last four digits of a perfect square are equal. Prove that all of them are zeros.

A polygon $A$ that doesn't intersect itself and has perimeter $p$ is called [b]Rotund[/b] if for each two points $x,y$ on the sides of this polygon which their distance on the plane is less than $1$ their distance on the polygon is at most $\frac{p}{4}$. (Distance on the polygon is the length of smaller path between two points on the polygon) Now we shall prove that we can fit a circle with radius $\frac{1}{4}$ in any rotund polygon. The mathematicians of two planets earth and Tarator have two different approaches to prove the statement. In both approaches by "inner chord" we mean a segment with both endpoints on the polygon, and "diagonal" is an inner chord with vertices of the polygon as the endpoints. [b]Earth approach: Maximal Chord[/b] We know the fact that for every polygon, there exists an inner chord $xy$ with a length of at most 1 such that for any inner chord $x'y'$ with length of at most 1 the distance on the polygon of $x,y$ is more than the distance on the polygon of $x',y'$. This chord is called the [b]maximal chord[/b]. On the rotund polygon $A_0$ there's two different situations for maximal chord: (a) Prove that if the length of the maximal chord is exactly $1$, then a semicircle with diameter maximal chord fits completely inside $A_0$, so we can fit a circle with radius $\frac{1}{4}$ in $A_0$. (b) Prove that if the length of the maximal chord is less than one we still can fit a circle with radius $\frac{1}{4}$ in $A_0$. [b]Tarator approach: Triangulation[/b] Statement 1: For any polygon that the length of all sides is less than one and no circle with radius $\frac{1}{4}$ fits completely inside it, there exists a triangulation of it using diagonals such that no diagonal with length more than $1$ appears in the triangulation. Statement 2: For any polygon that no circle with radius $\frac{1}{4}$ fits completely inside it, can be divided into triangles that their sides are inner chords with length of at most 1. The mathematicians of planet Tarator proved that if the second statement is true, for each rotund polygon there exists a circle with radius $\frac{1}{4}$ that fits completely inside it. (c) Prove that if the second statement is true, then for each rotund polygon there exists a circle with radius $\frac{1}{4}$ that fits completely inside it. They found out that if the first statement is true then the second statement is also true, so they put a bounty of a doogh on proving the first statement. A young earth mathematician named J.N., found a counterexample for statement 1, thus receiving the bounty. (d) Find a 1392-gon that is counterexample for statement 1. But the Tarators are not disappointed and they are still trying to prove the second statement. (e) (Extra points) Prove or disprove the second statement. Time allowed for this problem was 150 minutes.

The [i]base factorial[/i] number system is a unique representation for positive integers where the $n$th digit from the right ranges from $0$ to $n$ inclusive and has place value $n!$ for all $n \ge 1.$ For instance, $71$ can be written in base factorial as $2321_{!} = 2 \cdot 4! + 3 \cdot 3! + 2 \cdot 2! + 1 \cdot 1!.$ Let $S_{!}(n)$ be the base $10$ sum of the digits of $n$ when $n$ is written in base factorial. Compute $\sum_{n=1}^{700} S_{!}(n)$ (expressed in base $10$).

For $a,b,c>0$, prove that (i) $\frac{a+b+c}{2}-\frac{ab}{a+b}-\frac{bc}{b+c}-\frac{ca}{c+a}\ge 0$ (ii) $a(1+b)+b(1+c)+c(1+a)\ge 6\sqrt{abc}$

Given a finite sequence of integers $a_{1},$ $a_{2},$ $...,$ $a_{n}$ for $n\geq 2.$ Show that there exists a subsequence $a_{k_{1}},$ $a_{k_{2}},$ $...,$ $a_{k_{m}},$ where $1\leq k_{1}\leq k_{2}\leq...\leq k_{m}\leq n,$ such that the number $a_{k_{1}}^{2}+a_{k_{2}}^{2}+...+a_{k_{m}}^{2}$ is divisible by $n.$ [b]Note by Darij:[/b] Of course, the $1\leq k_{1}\leq k_{2}\leq ...\leq k_{m}\leq n$ should be understood as $1\leq k_{1}<k_{2}<...<k_{m}\leq n;$ else, we could take $m=n$ and $k_{1}=k_{2}=...=k_{m},$ so that the number $a_{k_{1}}^{2}+a_{k_{2}}^{2}+...+a_{k_{m}}^{2}=n^{2}a_{k_{1}}^{2}$ will surely be divisible by $n.$

A game about passing barriers rules that in the $n$th barrier, you need to throw a dice for $n$ times. If the sum of points you get is larger than $2^n$, then you can pass this barrier. [b](a)[/b] How many barriers can you pass at most? [b](b)[/b] Find the probablity of passing the first three barriers.

$22$ mathematicians are meeting together. Each mathematician has at least $3$ friends (friends are mutual). And each mathematician can pass his or her information to any mathematician through the transfer between friends. Is it possible to divide these $22$ mathematicians into $2$-person groups (that is, two people in each group, a total of $11$ groups), so that the mathematicians in each group are friends? [hide=original wording in Chinese]仃22位数学家一起开会.每位数学家都至少有3个朋友(朋友是相互的).而且每 位数学家都可以通过朋友之间的传递.将门已的资料传给任意一位数学家.问：是否一定可 以将这22位数学家两两分组(即每组两人,共11组),使得每组的数学家都是朋友？[/hide]

If $2137^{753}$ is multiplied out, the units' digit in the final product in the final product is: ${{ \textbf{(A)}\ 1\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 7}\qquad\textbf{(E)}\ 9} $

The difference between the cubes of two consecutive positive integers is equal to $n^2$ for a positive integer $n$. Show that $n$ is the sum of two squares.

Let $O$ be the circumcenter of triangle $ABC$. Choose a point $X$ on the circumcircle $\odot (ABC)$ such that $OX\parallel BC$. Assume that $\odot(AXO)$ intersects $AB, AC$ at $E, F$, respectively, and $OE, OF$ intersect $BC$ at $P, Q$, respectively. Furthermore, assume that $\odot(XP Q)$ and $\odot (ABC)$ intersect at $R$. Prove that $OR$ and $\odot (XP Q)$ are tangent to each other. (ltf0501)

If $\tan x+\tan y=25$ and $\cot x + \cot y=30$, what is $\tan(x+y)$?

Let $p \equiv 2 \pmod 3$ be a prime, $k$ a positive integer and $P(x) = 3x^{\frac{2p-1}{3}}+3x^{\frac{p+1}{3}}+x+1$. For any integer $n$, let $R(n)$ denote the remainder when $n$ is divided by $p$ and let $S = \{0,1,\cdots,p-1\}$. At each step, you can either (a) replaced every element $i$ of $S$ with $R(P(i))$ or (b) replaced every element $i$ of $S$ with $R(i^k)$. Determine all $k$ such that there exists a finite sequence of steps that reduces $S$ to $\{0\}$. [i]Proposed by fattypiggy123[/i]

Let $x$ satisfy $x^4+x^3+x^2+x+1=0$. Compute the value of $(5x+x^2)(5x^2+x^4)(5x^3+x^6)(5x^4+x^8)$. [i]Proposed by Andrew Zhao[/i]

Solve the equation $~$ $x^4+2y^4+4z^4+8t^4=16xyzt$ $~$ in the set of integer numbers.

Evaluate $ \int_0^1 \frac{x}{\{(2x\minus{}1)\sqrt{x^2\plus{}x\plus{}1}\plus{}(2x\plus{}1)\sqrt{x^2\minus{}x\plus{}1}\}\sqrt{x^4\plus{}x^2\plus{}1}}\ dx$.

Is there a polyhedron whose area ratio of any two faces is at least $2$ ?

Joseph rolls a fair 6-sided dice repeatedly until he gets 3 of the same side in a row. What is the expected value of the number of times he rolls? [i]Lightning 3.1[/i]

Let $ABCD$ be a convex quadrilateral with $\angle{DAC}= \angle{BDC}= 36^\circ$ , $\angle{CBD}= 18^\circ$ and $\angle{BAC}= 72^\circ$. The diagonals and intersect at point $P$ . Determine the measure of $\angle{APD}$.

It is known that every student of the class for Sunday once visited the rink, and every boy met there with every girl. Prove that there was a point in time when all the boys, or all the girls of the class were simultaneously on the rink.

Two natural numbers have the property that the product of their positive divisors are equal. Does this imply that they are equal? [i]Proposed by Belarus for the 1999th IMO[/i]

Let $n,m$ be positive integers such that $n<m$ and $a_1, a_2, ..., a_m$ be different real numbers. (a) Find all polynomials $P$ with real coefficients and degree at most $n$ such that: $|P(a_i)-P(a_j)|=|a_i-a_j|$ for all $i,j=\{1, 2, ..., m\}$ such that $i<j$. (b) If $n,m\ge 2$ does there exist a polynomial $Q$ with real coefficients and degree $n$ such that: $|Q(a_i)-Q(a_j)|<|a_i-a_j|$ for all $i,j=\{1, 2, ..., m\}$ such that $i<j$ Edit: See #3

Given $8$ pairwise distinct positive integers $a_1,a_2,…,a_8$ such that the greatest common divisor of any three of them is equal to $1$. Show that there exists positive integer $n\geq 8$ and $n$ pairwise distinct positive integers $m_1,m_2,…,m_n$ with the greatest common divisor of all $n$ numbers equal to $1$ such that for any positive integers $1\leq p<q<r\leq n$, there exists positive integers $1\leq i<j\leq 8$ that $a_ia_j\mid m_p+m_q+m_r$.