Find all real solutions to the equation $4x^2 - 40 \lfloor x \rfloor + 51 = 0$.

The sum of the real values of $x$ satisfying the equality $|x+2|=2|x-2|$ is: $\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{2}{3} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 6\dfrac{1}{3} \qquad \textbf{(E)}\ 6\dfrac{2}{3} $

The inscribed circle of $\Delta ABC$ $(AC<BC)$ is tangent to $AC$ and $BC$ in points $X$ and $Y$ respectively. A line is constructed through the middle point $M$ of $AB$, parallel to $XY$, which intersects $BC$ in $N$. Let $L\in BC$ be such that $NL=AC$ and $L$ is between $C$ and $N$. The lines $ML$ and $AC$ intersect in point $K$. Prove that $BN=CK$.

An infinite set $B$ consisting of positive integers has the following property. For each $a,b \in B$ with $a>b$ the number $\frac{a-b}{(a,b)}$ belongs to $B$. Prove that $B$ contains all positive integers. Here, $(a,b)$ is the greatest common divisor of numbers $a$ and $b$.

In $ \triangle ABC$, $ AB \equal{} BC$, and $ BD$ is an altitude. Point $ E$ is on the extension of $ \overline{AC}$ such that $ BE \equal{} 10$. The values of $ \tan CBE$, $ \tan DBE$, and $ \tan ABE$ form a geometric progression, and the values of $ \cot DBE$, $ \cot CBE$, $ \cot DBC$ form an arithmetic progression. What is the area of $ \triangle ABC$? [asy]unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); pair D=(0,0), C=(3,0), A=(-3,0), B=(0, 8), Ep=(6,0); draw(A--B--Ep--cycle); draw(D--B--C); label("$A$",A,S); label("$D$",D,S); label("$C$",C,S); label("$E$",Ep,S); label("$B$",B,N);[/asy]$ \textbf{(A)}\ 16 \qquad \textbf{(B)}\ \frac {50}{3} \qquad \textbf{(C)}\ 10\sqrt3 \qquad \textbf{(D)}\ 8\sqrt5 \qquad \textbf{(E)}\ 18$

Let $n\geq 3$ be an integer. Find the largest real number $M$ such that for any positive real numbers $x_1,x_2,\cdots,x_n$, there exists an arrangement $y_1,y_2,\cdots,y_n$ of real numbers satisfying \[\sum_{i=1}^n \frac{y_i^2}{y_{i+1}^2-y_{i+1}y_{i+2}+y_{i+2}^2}\geq M,\] where $y_{n+1}=y_1,y_{n+2}=y_2$.

In a country, mathematicians chose an $\alpha> 2$ and issued coins in denominations of 1 ruble, as well as $\alpha ^k$ rubles for each positive integer k. $\alpha$ was chosen so that the value of each coins, except the smallest, was irrational. Is it possible that any natural number of rubles can be formed with at most 6 of each denomination of coins?

Let $ABC$ be an acute triangle and $A_1, B_1$ and $C_1$, points on the sides $BC, CA$ and $AB$, respectively, such that $CB_1 = A_1B_1$ and $BC_1 = A_1C_1$. Let $D$ be the symmetric of $A_1$ with respect to $B_1C_1, O$ and $O_1$ are the circumcenters of triangles $ABC$ and $A_1B_1C_1$, respectively. If $A \ne D, O \ne O_1$ and $AD$ is perpendicular to $OO_1$, prove that $AB = AC$.

Find the product $ \cos a \cdot \cos 2a\cdot \cos 3a \cdots \cos 1006a$ where $a=\frac{2\pi}{2013}$.

An integer partition, is a way of writing n as a sum of positive integers. Two sums that differ only in the order of their summands are considered the same partition. [quote]For example, 4 can be partitioned in five distinct ways: 4 3 + 1 2 + 2 2 + 1 + 1 1 + 1 + 1 + 1[/quote] The number of partitions of n is given by the partition function $p\left ( n \right )$. So $p\left ( 4 \right ) = 5$ . Determine all the positive integers so that $p\left ( n \right )+p\left ( n+4 \right )=p\left ( n+2 \right )+p\left ( n+3 \right )$.

A uniform circular ring of radius $R$ is fixed in place. A particle is placed on the axis of the ring at a distance much greater than $R$ and allowed to fall towards the ring under the influence of the ring’s gravity. The particle achieves a maximum speed $v$. The ring is replaced with one of the same (linear) mass density but radius $2R$, and the experiment is repeated. What is the new maximum speed of the particle? (a) $\frac{1}{2}v$ (b) $\frac{1}{\sqrt{2}}v$ (c) $v$ (d) $\sqrt{2}v$ (e) $2v$

Determine the positive integers expressible in the form $\frac{x^2+y}{xy+1}$, for at least $2$ pairs $(x,y)$ of positive integers

Consider the isosceles right triangle $ABC$, with $\angle A = 90^o$ and the point $M \in (BC)$ such that $\angle AMB = 75^o$. On the inner bisector of the angle $MAC$ take a point $F$ such that $BF = AB$. Prove that: a) the lines $AM$ and $BF$ are perpendicular; b) the triangle $CFM$ is isosceles.

In an acute, scalene triangle $\triangle ABC$, let $O$ be the circumcenter. Let $M$ be the midpoint of $AC$. Let the perpendicular from $A$ to $BC$ be $D$. Let the circumcircle of $\triangle OAM$ hit $DM$ at $P(\not= M)$. Prove that $B, O, P$ are colinear.

Note that $k\ge 1$ for an odd natural number $$k! ! = k \cdot (k - 2) \cdot ... \cdot 1.$$ Prove that $2^n$ divides $(2^n -1)!! -1$ for all $n \ge 3$.

[asy]defaultpen(linewidth(.8pt)); pair A = (0,11); pair B = (2,0); pair D = (4,0); pair E = (7,0); pair C = (13,0); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,S); label("$E$",E,S); label("$2$",midpoint(B--D),N); label("$3$",midpoint(D--E),NW); label("$6$",midpoint(E--C),NW); draw(A--B--C--cycle); draw(A--D); draw(A--E);[/asy] In triangle $ ABC$ in the adjoining figure, $ AD$ and $ AE$ trisect $ \angle BAC$. The lengths of $ BD$, $ DE$ and $ EC$ are $ 2$, $ 3$, and $ 6$, respectively. The length of the shortest side of $ \triangle ABC$ is $ \textbf{(A)}\ 2\sqrt{10}\qquad \textbf{(B)}\ 11\qquad \textbf{(C)}\ 6\sqrt{6}\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ \text{not uniquely determined by the given information}$

In triangle $ABC, AB=\sqrt{30}, AC=\sqrt{6},$ and $BC=\sqrt{15}.$ There is a point $D$ for which $\overline{AD}$ bisects $\overline{BC}$ and $\angle ADB$ is a right angle. The ratio \[ \frac{\text{Area}(\triangle ADB)}{\text{Area}(\triangle ABC)} \] can be written in the form $m/n,$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

[b]p1.[/b] (a) Put the numbers $1$ to $6$ on the circle in such way that for any five consecutive numbers the sum of first three (clockwise) is larger than the sum of remaining two. (b) Can you arrange these numbers so it works both clockwise and counterclockwise. [b]p2.[/b] A girl has a box with $1000$ candies. Outside the box there is an infinite number of chocolates and muffins. A girl may replace: $\bullet$ two candies in the box with one chocolate bar, $\bullet$ two muffins in the box with one chocolate bar, $\bullet$ two chocolate bars in the box with one candy and one muffin, $\bullet$ one candy and one chocolate bar in the box with one muffin, $\bullet$ one muffin and one chocolate bar in the box with one candy. Is it possible that after some time it remains only one object in the box? [b]p3.[/b] Find any integer solution of the puzzle: $WE+ST+RO+NG=128$ (different letters mean different digits between $1$ and $9$). [b]p4.[/b] Two consecutive three‐digit positive integer numbers are written one after the other one. Show that the six‐digit number that is obtained is not divisible by $1001$. [b]p5.[/b] There are $9$ straight lines drawn in the plane. Some of them are parallel some of them intersect each other. No three lines do intersect at one point. Is it possible to have exactly $17$ intersection points? PS. You should use hide for answers. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].

Let $a,b,c$ be positive real numbers. Prove that \[\frac{a}b+\frac{b}c+\frac{c}a \geq \frac{c+a}{c+b}+\frac{a+b}{a+c}+\frac{b+c}{b+a}\]

Find all functions $f : \mathbb R\to\mathbb R $ such that \[f(x+yf(x^2))=f(x)+xf(xy)\] for all real numbers $x$ and $y$.

We write $1$ or $-1$ on each unit square of a $2007 \times 2007$ board. Find the number of writings such that for every square on the board the absolute value of the sum of numbers on the square is less then or equal to $1$.

In quadrilateral $ABCD$, $AC = BD$ and $\measuredangle B = 60^\circ$. Denote by $M$ and $N$ the midpoints of $\overline{AB}$ and $\overline{CD}$, respectively. If $MN = 12$ and the area of quadrilateral $ABCD$ is 420, then compute $AC$. [i]Proposed by Aaron Lin[/i]

Evaluate the following sum $$\frac{1}{\log_2{\frac{1}{7}}}+\frac{1}{\log_3{\frac{1}{7}}}+\frac{1}{\log_4{\frac{1}{7}}}+\frac{1}{\log_5{\frac{1}{7}}}+\frac{1}{\log_6{\frac{1}{7}}}-\frac{1}{\log_7{\frac{1}{7}}}-\frac{1}{\log_8{\frac{1}{7}}}-\frac{1}{\log_9{\frac{1}{7}}}-\frac{1}{\log_{10}{\frac{1}{7}}}$$

The tangents to the circumcircle of triangle $ABC$ at $A$ and $B$ meet at point $D$. The circle passing through the projections of $D$ to $BC, CA, AB$, meet $AB$ for the second time at point $C'$. Points $A', B'$ are defined similarly. Prove that $AA', BB', CC'$ concur.

Congruent unit circles intersect in such a way that the center of each circle lies on the circumference of the other. Let $R$ be the region in which two circles overlap. Determine the perimeter of $R.$