Given circles $\omega_1$ and $\omega_2$ with two intersection points $A$ and $B$. Points $X$ on $\omega_1$ and $Y$ on $\omega_2$ are chosen such that $XY$ is tangent to both circles and $XY$ is closer to $B$ than $A$. If $C$ and $D$ are the reflections of $X$ and $Y$ with respect to $B$, prove that $\angle CAD \lt 90^\circ$.
Consider an acute scalene triangle $ABC$ with circumcircle $\Gamma$. The external angle bisector of $\angle BAC$ meets $BC$ at $X$. Lines $\ell_b$ and $\ell_c$ are tangent lines from $B$ and $C$ to $\Gamma$. A line passes through $X$ and intersects $\ell_b$ and $\ell_c$ at points $Y$ and $Z$, such that $X, Y$ and $Z$ lie on this order. Suppose that the circumcircle of triangles $AYB$ and $AZC$ intersect at $N$, for the second time. If $D$ is the intersection point of $\ell_b$ and $\ell_c$, show that $ND$ is the angle bisector of $\angle YNZ$.
Let $ABCD$ be a rhombus and let $\omega$ be its incircle. Let $M$ be the midpoint of $AB$ and $K$ be a point inside $ABCD$ such that $MK$ is tangent to $\omega$. Prove that $CDKM$ is cyclic.
Determine for which $n \geq 3$ it is possible to find $n$ positive integers such that two by two have at least one factor in common other than 1, but three to three are first between them. If you add the condition that all integers are less than 5000, what is the maximum value of $n$ possible?
Two wizards perform in the following issue. At first the first wizard locks the second wizard in a cabin where he can neither see nor hear anything. To start the game, the first wizard invites Daniel, a member of the public, to place a white or black pawn on every box of a chessboard $n \times n$. Then he asks Daniele to indicate a box $C$ of his choice. At this point, the first wizard chooses a box $D$ (not necessarily different from $C$) and replaces the token found on $D$ with one of the other color (white with black or black with white). The cabin in which the second wizard was then opened is then opened. Observing the chessboard, the second wizard manages to guess what the $C$ box is. For which $n$ can the two wizards implement such a strategy that their number always succeeds?
Let $N$ be the number of ordered sixtuples of integers $(a, b, c, d, e, f)$ such as $a^{3}+b^{3}+c^{3}+d^{3}+e^{3}+f^{3}=168$ and $-202120212021^{9}\lt a b c d e f\lt 202120212021^{9}$, where $abcdef$ is the product of you are whole. Which of the following is the rest of $N$ in division by 6?
Let $h$ be a positive integer and let $a_{n}$ the sequence defined for recurrence as follows: $$
\begin{aligned}
a_{0} & =1 \\
a_{n+1} & = \begin{cases}\frac{a_{n}}{2} & \text{ if } a_{n} \text{ is even } \\
a_{n}+h & \text{ if } a_{n} \text{ It's odd.}\end{cases}
\end{aligned}
$$ (for example, if $h=27$ we have: $a_{1}=28, a_{2}=14, a_{3}=7, a_{4}=34, a_{5}=17, a_{6}=44, \ldots$) For what values of $h$ does $n\gt 0$ exist for which $a_{n}=1$?
Let $ABCD$ be a convex quadrilateral such that $\widehat{CAB} = \widehat{CDA}$ and $\widehat{BCA} = \widehat{ACD}$. Said $M$ the midpoint of $AB$ proves that $\widehat{BCM} = \widehat{DBA}$.
Determine all the $\{a, b\}$ pairs of positive integers with the following property: however positive integers color with two colors $A$ and $B$, there are always two positive color integers $A$ with difference $a$ or two positive integers of the $B$ color with difference $b$.
In each of the boxes of a square table $4 \times 4$ the figure is written $1$ or the figure $2$. It is known that the sum of $9$ figures contained in each of the $4$ squares $3 \times 3$ contained in the table is multiple of $4$, while the sum of all $16$ digits is not multiple of $4$. determine the maximum and minimum possible value for the sum of all $16$ figures.
Let $x_{0}, x_{1}, x_{2}, \ldots$ be a sequence of rational numbers defined by recurrence as follows: $x_{0}$ is any rational number, and, for $n \geq 0$, $$
x_{n+1}= \begin{cases}\left|\frac{x_{n}}{2}-1\right| & \text { se il numeratore di } x_{n} \text { è pari }, \\ \left|\frac{1}{x_{n}}-1\right| & \text { se il numeratore di } x_{n} \text { è dispari },\end{cases}
$$ where for numerator of a rational number means that of the fraction reduced to a minimum. Prove that for each value of $x_{0}$:
(a) the sequence contains only a finite number of distinct terms;
(b) The sequence contains exactly one of the numbers 0 and $2 / 3$ (i.and.: either there is an index $k$ such that $x_{k}=0$, or there is an index $m$ such that $x_{m}=2 / 3$, but they do not exist both).
Let $ABC$ be an isosceles triangle with $AB = AC$ and $\widehat{BAC} \lt 60^\circ$. Let $D$ be the point on $AC$ such that $\widehat{DBC} = \widehat{BAC}$. Let $E$ be the intersection of the perpendicular bisector of $BD$ with the line through $A$ parallel to $BC$. Let $F$ lie on the line $AC$, on the same side of $C$ as $A$, with $FA$ twice $AC$. Finally, let $r$ be the line through $F$ perpendicular to $AB$, $s$ the line through $E$ perpendicular to $AC$, and $t$ the line $BD$. Prove:
(a) the lines $EB$ and $AC$ are parallel;
(b) the lines $r$, $s$ and $t$ are concurrent.
Let $n$ be an integer greater than or equal to 2. There are $n$ people in single file, each of which is either a villain (and always lies) or a knight (and always tells the truth). Each person, except the first, indicates one of the people in front of her and declares "this person is a villain" or "this person is a knight". Knowing that there are closely more villains than knights, prove that by attending the statements it is possible to determine for each of the people whether it is a villain or a knight.
A mysterious machine contains a secret combination consisting of 2016 integers $x_{1}, \ldots, x_{2016}$. We know that all combination numbers are equal except one. It is possible to query the machine by subjecting it a sequence of 2016 integers $y_{1}, \ldots, y_{2016}$. The machine responds by revealing the value of the $$
x_{1} y_{1}+\cdots+x_{2016} y_{2016} \text{.}
$$ sum after answering the first question, the machine accepts a second question, then a third, and so on. How many questions are needed to determine the combination:
(a) knowing that the different number is equal to zero?
(b) not knowing what the number is different?
two circles $\gamma_{1}$ and $\gamma_{2}$ of $A$ and $B$ centers are given in the plane, and intersecting in two points $C$ and $D$. Assume that the through circumference for $A$, $B$ and $C$ intersects further $\gamma_{1}$ and $\gamma_{2}$ in $E$ and $F$ respectively, and that the arc $E F$ not containing $C$ lie out of the two circles delimited by $\gamma_{1}$ and $\gamma_{2}$. Prove that the $E F$ arc not containing $C$ is bisected by the line $C D$.
Let $ABC$ be a right triangle, with hypotenuse $AC$, and let $H$ the foot of the height conducted by $B$ ad $AC$. Knowing that the lengths $AB$, $BC$ and $BH$ constitute the sides of a new right triangle, determine the possible values of $\frac{AH}{CH}$.