Determine all natural numbers $x$, $y$ and $z$, such that $x\leq y\leq z$ and \[\left(1+\frac1x\right)\left(1+\frac1y\right)\left(1+\frac1z\right) = 3\text{.}\]

For every positive integer $n$ let $$f(n) = \frac{n^4+n^3+n^2-n+1}{n^6-1}$$ Given $$\sum_{n = 2}^{20} f(n) = \frac{a}{b}$$ for relatively prime positive integers $a$ and $b$, find the sum of the prime factors of $b$. [i]Proposed by Harry Kim[/i]

Let $ n > 1$ be an integer, and $ n$ can divide $ 2^{\phi(n)} \plus{} 3^{\phi(n)} \plus{} \cdots \plus{} n^{\phi(n)},$ let $ p_{1},p_{2},\cdots,p_{k}$ be all distinct prime divisors of $ n$. Show that $ \frac {1}{p_{1}} \plus{} \frac {1}{p_{2}} \plus{} \cdots \plus{} \frac {1}{p_{k}} \plus{} \frac {1}{p_{1}p_{2}\cdots p_{k}}$ is an integer. ( where $ \phi(n)$ is defined as the number of positive integers $ \leq n$ that are relatively prime to $ n$.)

Find all functions $f:\mathbb Q\to\mathbb Q$ such that for all $x,y,z,w\in\mathbb Q$, $$f(f(xyzw)+x+y)+f(z)+f(w)=f(f(xyzw)+z+w)+f(x)+f(y).$$

Let $P$ and $Q$ be the midpoints of sides $AB$ and $BC,$ respectively, of $\triangle ABC.$ Suppose $\angle A = 30^{\circ}$ and $\angle PQC = 110^{\circ}.$ Find $\angle B$ in degrees.

At Durant University, an A grade corresponds to raw scores between $90$ and $100$, and a B grade corresponds to raw scores between $80$ and $90$. Travis has $3$ equally weighted exams in his math class. Given that Travis earned an A on his first exam and a B on his second (but doesn't know his raw score for either), what is the minimum score he needs to have a $90\%$ chance of getting an A in the class? Note that scores on exams do not necessarily have to be integers.

In an acute angled triangle the feet of the altitudes are joined to form a new triangle. In this new triangle it is known that two sides are parallel to sides of the original triangle . Prove that the third side is also parallel to one of the sides of the original triangle .

In a right triangle rectangle $ABC$ such that $AB = AC$, $M$ is the midpoint of $BC$. Let $P$ be a point on the perpendicular bisector of $AC$, lying in the semi-plane determined by $BC$ that does not contain $A$. Lines $CP$ and $AM$ intersect at $Q$. Calculate the angles that form the lines $AP$ and $BQ$.

We examine the following two sequences: The Fibonacci sequence: $F_{0}= 0, F_{1}= 1, F_{n}= F_{n-1}+F_{n-2 }$ for $n \geq 2$; The Lucas sequence: $L_{0}= 2, L_{1}= 1, L_{n}= L_{n-1}+L_{n-2}$ for $n \geq 2$. It is known that for all $n \geq 0$ \[F_{n}=\frac{\alpha^{n}-\beta^{n}}{\sqrt{5}},L_{n}=\alpha^{n}+\beta^{n}, \] where $\alpha=\frac{1+\sqrt{5}}{2},\beta=\frac{1-\sqrt{5}}{2}$. These formulae can be used without proof. Show that $L_{2n+1}+(-1)^{n+1}(n \geq 1)$ can be written as a product of three (not necessarily distinct) Fibonacci numbers.

Quadrilateral $ABCD (\overline{AD} < \overline{BC})$ is inscribed in a circle, and $H(\neq A, B)$ is a point on segment $AB.$ The circumcircle of triangle $BCH$ meets $BD$ at $E(\neq B)$ and line $HE$ meets $AD$ at $F$. The circle passes through $C$ and tangent to line $BD$ at $E$ meets $EF$ at $G(\neq E).$ Prove that $\angle DFG = \angle FCG.$

$a,b$ are natural numbers such that $\frac{a+1}{b}$ and $\frac{b+1}{a}$ are integers.Let $d=GCD(a;b)$.Prove that $d^2\le a+b$

A chord $AB$ is drawn in a circle. On its extensions beyond points $A$ and $B$, points $P$ and $Q$ respectively are taken such that $AP = BQ$. Through $P$ and $Q$ two tangents to the circle are drawn, intersecting at point $M$. Find the locus of points $M$ ($P$ and $Q$ move along a straight line and for any $P$ and $Q$ all possible pairs of tangents are taken, which determine four points from the desired locus of points) .

For a positive integer $n$, $S(n)$ denotes the sum of its digits and $U(n)$ its unit digit. Determine all positive integers $n$ with the property that \[n = S(n) + U(n)^2.\]

Let $ABC$ be a triangle,$O$ its circumcenter and $R$ the radius of its circumcircle.Denote by $O_{1}$ the symmetric of $O$ with respect to $BC$,$O_{2}$ the symmetric of $O$ with respect to $AC$ and by $O_{3}$ the symmetric of $O$ with respect to $AB$. (a)Prove that the circles $C_{1}(O_{1},R)$, $C_{2}(O_{2},R)$, $C_{3}(O_{3},R)$ have a common point. (b)Denote by $T$ this point.Let $l$ be an arbitary line passing through $T$ which intersects $C_{1}$ at $L$, $C_{2}$ at $M$ and $C_{3}$ at $K$.From $K,L,M$ drop perpendiculars to $AB,BC,AC$ respectively.Prove that these perpendiculars pass through a point.

Rank the [i]magnitudes[/i] of the [i]distance[/i] traveled during the ten second interval. (A) $\text{I} > \text{II} > \text{III}$ (B) $\text{II} > \text{I} > \text{III}$ (C) $\text{III} > \text{II} > \text{I}$ (D) $\text{I} > \text{II = III}$ (E) $\text{I = II = III}$

Let $n$ and $k$ be positive integers, where $n > 1$ is odd. Suppose $n$ voters are to elect one of the $k$ cadidates from a set $A$ according to the rule of "majoritarian compromise" described below. After each voter ranks the candidates in a column according to his/her preferences, these columns are concatenated to form a $k$ x $n$ voting matrix. We denote the number of ccurences of $a \in A$ in the $i$-th row of the voting matrix by $a_{i}$ . Let $l_{a}$ stand for the minimum integer $l$ for which $\sum^{l}_{i=1}{a_{i}}> \frac{n}{2}$. Setting $l'= min \{l_{a} | a \in A\}$, we will regard the voting matrices which make the set $\{a \in A | l_{a} = l' \}$ as admissible. For each such matrix, the single candidate in this set will get elected according to majoritarian compromise. Moreover, if $w_{1} \geq w_{2} \geq ... \geq
w_{k} \geq 0$ are given, for each admissible voting matrix, $\sum^{k}_{i=1}{w_{i}a_{i}}$ is called the total weighted score of $a \in A$. We will say that the system $(w_{1},w_{2}, . . . , w_{k})$ of weights represents majoritarian compromise if the total score of the elected candidate is maximum among the scores of all candidates. (a) Determine whether there is a system of weights representing majoritarian compromise if $k = 3$. (b) Show that such a system of weights does not exist for $k > 3$.

Let $r$ and $s$ be positive real numbers such that $(r+s-rs)(r+s+rs)=rs$. Find the minimum value of $r+s-rs$ and $r+s+rs$

In the triangle $ABC$ it is known that $BC = 5, AC - AB = 3$. Prove that $r <2 <r_a$ . (here $r$ is the radius of the circle inscribed in the triangle $ABC$, $r_a$ is the radius of an exscribed circle that touches the sides of $BC$). (Mykola Moroz)

Let $ABC$ be an acute scalene triangle with orthocenter $H$ and circumcenter $O$. Let the line through $A$ tangent to the circumcircle of triangle $AHO$ intersect the circumcircle of triangle $ABC$ at $A$ and $P \neq A$. Let the circumcircles of triangles $AOP$ and $BHP$ intersect at $P$ and $Q \neq P$. Let line $PQ$ intersect segment $BO$ at $X$. Suppose that $BX=2$, $OX=1$, and $BC=5$. Then $AB \cdot AC = \sqrt{k}+m\sqrt{n}$ for positive integers $k$, $m$, and $n$, where neither $k$ nor $n$ is divisible by the square of any integer greater than $1$. Compute $100k+10m+n$. [i]Proposed by Luke Robitaille[/i]

Three semicircles of radius $ 1$ are constructed on diameter $ AB$ of a semicircle of radius $ 2$. The centers of the small semicircles divide $ \overline{AB}$ into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles? [asy]import graph; unitsize(14mm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dashed=linetype("4 4"); dotfactor=3; pair A=(-2,0), B=(2,0); fill(Arc((0,0),2,0,180)--cycle,mediumgray); fill(Arc((-1,0),1,0,180)--cycle,white); fill(Arc((0,0),1,0,180)--cycle,white); fill(Arc((1,0),1,0,180)--cycle,white); draw(Arc((-1,0),1,60,180)); draw(Arc((0,0),1,0,60),dashed); draw(Arc((0,0),1,60,120)); draw(Arc((0,0),1,120,180),dashed); draw(Arc((1,0),1,0,120)); draw(Arc((0,0),2,0,180)--cycle); dot((0,0)); dot((-1,0)); dot((1,0)); draw((-2,-0.1)--(-2,-0.3),gray); draw((-1,-0.1)--(-1,-0.3),gray); draw((1,-0.1)--(1,-0.3),gray); draw((2,-0.1)--(2,-0.3),gray); label("$A$",A,W); label("$B$",B,E); label("1",(-1.5,-0.1),S); label("2",(0,-0.1),S); label("1",(1.5,-0.1),S);[/asy]$ \textbf{(A)}\ \pi\minus{}\sqrt3 \qquad \textbf{(B)}\ \pi\minus{}\sqrt2 \qquad \textbf{(C)}\ \frac{\pi\plus{}\sqrt2}{2} \qquad \textbf{(D)}\ \frac{\pi\plus{}\sqrt3}{2}$ $ \textbf{(E)}\ \frac{7}{6}\pi\minus{}\frac{\sqrt3}{2}$

The country of HMMTLand has $8$ cities. Its government decides to construct several two-way roads between pairs of distinct cities. After they finish construction, it turns out that each city can reach exactly $3$ other cities via a single road, and from any pair of distinct cities, either exactly $0$ or $2$ other cities can be reached from both cities by a single road. Compute the number of ways HMMTLand could have constructed the roads.

Triangle $ABC$ is right angled such that $\angle ACB=90^{\circ}$ and $\frac {AC}{BC} = 2$. Let the line parallel to side $AC$ intersects line segments $AB$ and $BC$ in $M$ and $N$ such that $\frac {CN}{BN} = 2$. Let $O$ be the intersection point of lines $CM$ and $AN$. On segment $ON$ lies point $K$ such that $OM+OK=KN$. Let $T$ be the intersection point of angle bisector of $\angle ABC$ and line from $K$ perpendicular to $AN$. Determine value of $\angle MTB$.

Find all positive integers $k$ such that there exist two positive integers $m$ and $n$ satisfying \[m(m + k) = n(n + 1).\]

Consider the equation $ax^{2}+by^{2}=1$, where $a,b$ are fixed rational numbers. Prove that either such an equation has no solutions in rational numbers, or it has infinitely many solutions.

There are $900$ students in an International School. There are $59$ international boys and $59$ international girls. The Students are partitioned into $30$ classrooms (each classrooms have equal number of student) and in each of the classrooms, the student will labelled number from $1$ to $30$. The Partition must satisfy at least one follow condition: [list=i] [*] Any Two international boys in same classroom can't have consecutive numbers. [*] For every classroom, the student who is labelled $1$ must be a boy. [/list] Prove that there are $2$ classrooms, each of which has $2$ international boys with their labels difference equal.