Ali puts $5$ points on the plane such that no three of them are collinear. Ramtin adds a sixth point that is not collinear with any two of the former points.Ali wants to eventually construct two triangles from the six points such that one can be placed inside another. Can Ali put the 5 points in such a manner so that he would always be able to construct the desired triangles? (We say that triangle $T_1$ can be placed inside triangle $T_2$ if $T_1$ is congruent to a triangle that is located completely inside $T_2$.)

In a trapezoid $ABCD$ with bases $AD$ and $BC$, the bisector of the angle $\angle DAB$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $P$ and $S$, respectively, and the bisector of the angle $\angle BCD$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $Q$ and $R$, respectively. Prove that if $PS\parallel RQ$, then $AB = CD$.

Some towns in a country are connected by two–way roads, so that for any two towns there is a unique path along the roads connecting them. It is known that there is exactly 100 towns which are directly connected to only one town. Prove that we can construct 50 new roads in order to obtain a net in which every two towns will be connected even if one road gets closed.

For each non-negative integer $n$, let $u_n = \left( 2+\sqrt{5} \right)^n + \left( 2-\sqrt{5} \right)^n$. a) Prove that $u_n$ is a positive integer for all $n \geq 0$. When $n$ changes, what is the largest possible remainder when $u_n$ is divided by $24$? b) Find all pairs of positive integers $(a, b)$ such that $a, b < 500$ and for all odd positive integers $n$, $u_n \equiv a^n - b^n \pmod {1111}$.

The Fibonacci numbers are defined by $F_1=1,$ $F_2=1,$ and $F_n=F_{n-1}+F_{n-2}$ for $n\geq 3.$ What is $$\dfrac{F_2}{F_1}+\dfrac{F_4}{F_2}+\dfrac{F_6}{F_3}+\cdots+\dfrac{F_{20}}{F_{10}}?$$ $\textbf{(A) }318 \qquad\textbf{(B) }319\qquad\textbf{(C) }320\qquad\textbf{(D) }321\qquad\textbf{(E) }322$

Dan has a fair $6$-sided die with faces labeled $1,2,3,4,+,$ and $-.$ In order to complete the equation \[ \underline{\qquad} \ \underline{\qquad} \ \underline{\qquad}=\underline{\qquad}, \] Dan repeatedly rolls his die and fills in a blank with the character he obtained, starting with the leftmost blank and progressing rightward. The probability that, when all blanks are filled, Dan forms a true equation, is $\frac{p}{q},$ where $p$ and $q$ are relatively prime integers. Find $p+q.$

Find all positive integers $k$ for which there is an irrational $\alpha>1$ and a positive integer $N$ such that $\left\lfloor\alpha^{n}\right\rfloor$ is of the form $m^2-k$ com $m \in \mathbb{Z}$ for every integer $n>N$.

The incircle of a triangle $ABC$ is centered at $I$ and touches $AC,AB$ and $BC$ at points $K,L,M$, respectively. The median $BB_1$ of $\triangle ABC$ intersects $MN$ at $D$. Prove that the points $I,D,K$ are collinear.

The number of postive integers less than $1000$ divisible by neither $5$ nor $7$ is: $\text{(A)}\ 688 \qquad \text{(B)}\ 686\qquad \text{(C)}\ 684 \qquad \text{(D)}\ 658\qquad \text{(E)}\ 630$

Let $a,b$ be integers with $0<a<b$. A set $\{x,y,z\}$ of non-negative integers is [i]olympic[/i] if $x<y<z$ and if $\{z-y,y-x\}=\{a,b\}$. Show that the set of all non-negative integers is the union of pairwise disjoint olympic sets.

What is the value of \[\frac{2^{2014}+2^{2012}}{2^{2014}-2^{2012}}?\] $ \textbf{(A)}\ -1\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ \frac{5}{3}\qquad\textbf{(D)}\ 2013\qquad\textbf{(E)}\ 2^{4024} $

The game of circulate is played with a deck of $kn$ cards each with a number in $1,2,\ldots,n$ such that there are $k$ cards with each number. First, $n$ piles numbered $1,2,\ldots,n$ of $k$ cards each are dealt out face down. The player then flips over a card from pile $1$, places that card face up at the bottom of the pile, then next flips over a card from the pile whose number matches the number on the card just flipped. The player repeats this until he reaches a pile in which every card has already been flipped and wins if at that point every card has been flipped. Hamster has grown tired of losing every time, so he decides to cheat. He looks at the piles beforehand and rearranges the $k$ cards in each pile as he pleases. When can Hamster perform this procedure such that he will win the game? [i]Brian Hamrick.[/i]

Consider a non-zero real number $a$ such that $\{a\} + \left\{\frac{1}{a}\right\}=1$, where $\{x\}$ denotes the fractional part of $x$. Prove that for any positive integer $n$, $\{a^n\} + \left\{\frac{1}{a^n}\right\}= 1$.

$\triangle ABC$ is a triangle and $D,E,F$ are points on $BC$, $CA$, $AB$ respectively. It is given that $BF=BD$, $CD=CE$ and $\angle BAC=48^{\circ}$. Find the angle $\angle EDF$ $ \textbf{(A) }64^{\circ}\qquad\textbf{(B) }66^{\circ}\qquad\textbf{(C) }68^{\circ}\qquad\textbf{(D) }70^{\circ}\qquad\textbf{(E) }72^{\circ} $

$F$* is the multiplicative group of the field $F$. $F$* is of finitely generated. Is it true that $F$* is cyclic? Additional question: (wasn’t at the olympiad) $K$* is the multiplicative group of the field $K$. $L \subseteq $$K$* is a finitely generated subgroup. Is it true that $L$ is cyclic?

An acute isosceles triangle, $ABC$ is inscribed in a circle. Through $B$ and $C$, tangents to the circle are drawn, meeting at point $D$. If $\angle ABC=\angle ACB=2\angle D$ and $x$ is the radian measure of $\angle A$, then $x=$ [asy] defaultpen(linewidth(0.7)+fontsize(10)); real x=180/7; pair D=origin, B=dir(3x), C=dir(4x), A=intersectionpoint(C--C+dir(2x), B--B+dir(5x)), O=circumcenter(A,B,C); markscalefactor=0.015; draw(B--C--D--B--A--C^^Circle(O, abs(O-C))^^anglemark(C,A,B)); dot(A^^B^^C^^D); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$x$", A+0.1*dir(270), S);[/asy] $\text{(A)} \ \frac37\pi \qquad \text{(B)} \ \frac49\pi \qquad \text{(C)} \ \frac5{11}\pi \qquad \text{(D)} \ \frac6{13}\pi \qquad \text{(E)} \ \frac7{15}\pi$

The number of digits in the number $N=2^{12}\times 5^8$ is $\textbf{(A) }9\qquad\textbf{(B) }10\qquad\textbf{(C) }11\qquad\textbf{(D) }12\qquad \textbf{(E) }20$

Supppose five points in a plane are situated so that no two of the straight lines joining them are parallel, perpendicular, or coincident. From each point perpendiculars are drawn to all the lines joining the other four points. Determine the maxium number of intersections that these perpendiculars can have.

Consider a rectangular sheet of paper $ABCD$ such that the lengths of $AB$ and $AD$ are respectively $7$ and $3$ centimetres. Suppose that $B'$ and $D'$ are two points on $AB$ and $AD$ respectively such that if the paper is folded along $B'D'$ then $A$ falls on $A'$ on the side $DC$. Determine the maximum possible area of the triangle $AB'D'$.

There are $n$ people at a party. Prove that there are two people such that, of the remaining $n-2$ people, there are at least $\left\lfloor\frac{n}{2}\right\rfloor-1$ of them, each of whom either knows both or else knows neither of the two. Assume that knowing is a symmetric relation, and that $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.

Let $ABCD$ be a square with diagonals $AC$ and $BD$, and $P$ a point in one of the sides of the square. Show that the sum of the distances from P to the diagonals is constant.

Given a string of at least one character in which each character is either A or B, Kathryn is allowed to make these moves: [list] [*] she can choose an appearance of A, erase it, and replace it with BB, or [*] she can choose an appearance of B, erase it, and replace it with AA. [/list] Kathryn starts with the string A. Let $a_n$ be the number of strings of length $n$ that Kathryn can reach using a sequence of zero or more moves. (For example, $a_1=1$, as the only string of length 1 that Kathryn can reach is A.) Then $\sum_{n=1}^{\infty} \frac{a_n}{5^n} = \frac{m}{n}$, where $m$ and $n$ are positive integers with $\gcd(m,n)=1$. Compute $100m+n$. [i]Proposed by Luke Robitaille[/i]

Let $O$ be an inner point of an acute-angled triangle $ABC$. Let $A_1, B_1$ and $C_1$ be the projections of $O$ on the sides $BC, AC$ and $AB$ respectively . Let $P$ be the intersection of the perpendiculars on $B_1C_1$ and $A_1C_1$ from points$ A$ and $B$ respectilvey. Let $H$ be the projection of $P$ on $AB$. Show that points $A_1, B_1, C_1$ and $H$ lie on a circle.

A model used to estimate the time it will take to hike to the top of the mountain on a trail is of the form $T = aL + bG,$ where $a$ and $b$ are constants, $T$ is the time in minutes, $L$ is the length of the trail in miles, and $G$ is the altitude gain in feet. The model estimates that it will take $69$ minutes to hike to the top if a trail is $1.5$ miles long and ascends $800$ feet, as well as if a trail is $1.2$ miles long and ascends $1100$ feet. How many minutes does the model estimate it will take to hike to the top if the trail is $4.2$ miles long and ascends $4000$ feet? $\textbf{(A) } 240 \qquad \textbf{(B) } 246 \qquad \textbf{(C) } 252 \qquad \textbf{(D) } 258 \qquad \textbf{(E) } 264$

On side $BC$ of parallelogram $ABCD$ ($A$ is acute) lies point $T$ so that triangle $ATD$ is an acute triangle. Let $O_1$, $O_2$, and $O_3$ be the circumcenters of triangles $ABT$, $DAT$, and $CDT$ respectively. Prove that the orthocenter of triangle $O_1O_2O_3$ lies on line $AD$.