Alice plays the following game of solitaire on a $20 \times 20$ chessboard. She begins by placing $100$ pennies, $100$ nickels, $100$ dimes, and $100$ quarters on the board so that each of the $400$ squares contains exactly one coin. She then chooses $59$ of these coins and removes them from the board. After that, she removes coins, one at a time, subject to the following rules: - A penny may be removed only if there are four squares of the board adjacent to its square (up, down, left, and right) that are vacant (do not contain coins). Squares “off the board” do not count towards this four: for example, a non-corner square bordering the edge of the board has three adjacent squares, so a penny in such a square cannot be removed under this rule, even if all three adjacent squares are vacant. - A nickel may be removed only if there are at least three vacant squares adjacent to its square. (And again, “off the board” squares do not count.) - A dime may be removed only if there are at least two vacant squares adjacent to its square (“off the board” squares do not count). - A quarter may be removed only if there is at least one vacant square adjacent to its square (“off the board” squares do not count). Alice wins if she eventually succeeds in removing all the coins. Prove that it is impossiblefor her to win.

Let $ \Delta ABC$ be a triangle, with $ m(\angle A)=90^{\circ}$ and $ m(\angle B)=30^{\circ}.$ If $M$ is the middle of $[AB],$ $N$ is the middle of $[BC],$ and $P\in[BC],\ Q\in[MN],$ such that \[\frac{PB}{PC}=4\cdot\frac{QM}{QN}+3,\] prove that $ \Delta APQ$ is an equilateral triangle. [b]Author: MARIN BANCOȘ[/b] [b]Regional Mathematical Contest GRIGORE MOISIL, Romania, Baia Mare, 24.03.2012, 7th grade[/b]

For what pairs of positive real numbers $(a,b)$ does the improper integral $(1)$ converge? \begin{align}\int_{b}^{\infty}\left(\sqrt{\sqrt{x+a}-\sqrt{x}}-\sqrt{\sqrt{x}-\sqrt{x-b}}\right)\,\mathrm{d}x \end{align}

Let $x,y,z \in\mathbb{Q}$,such that $(x+y+z)^3=9(x^2y+y^2z+z^2x).$ Prove that $x=y=z$

$n$ people are in the plane, so that the closest person is unique and each one shoot this closest person with a squirt gun. If $n$ is odd, prove that there exists at least one person that nobody shot. If $n$ is even, will there always be a person who escape? Justify that.

Let $D,E,F$ be points on the sides $BC,CA,AB$ of a triangle $ABC$, respectively. Suppose that the inradii of the triangles $AEF,BFD,CDE$ are all equal to $r_1$. If $r_2$ and $r$ are the inradii of triangles $DEF$ and $ABC$ respectively, prove that $r_1 +r_2 =r$.

There are $n>1000$ people at a round table. Some of them are knights who always tell the truth, and the rest are liars who always tell lies. Each of those sitting said the phrase: “among the $20$ people sitting clockwise from where I sit there are as many knights as among the $20$ people seated counterclockwise from where I sit”. For what $n$ could this happen?

Let $P$ be a point outside of the triangle $ABC$ in the plane of $ABC$. Prove that by using reflections $S_{AB}$, $S_{AC}$, and $S_{BC}$ across the lines $AB$, $AC$, and $BC$ one can shift point $P$ inside the triangle $ABC$.

There is given a convex quadrilateral $ ABCD$. Prove that there exists a point $ P$ inside the quadrilateral such that \[ \angle PAB \plus{} \angle PDC \equal{} \angle PBC \plus{} \angle PAD \equal{} \angle PCD \plus{} \angle PBA \equal{} \angle PDA \plus{} \angle PCB = 90^{\circ} \] if and only if the diagonals $ AC$ and $ BD$ are perpendicular. [i]Proposed by Dusan Djukic, Serbia[/i]

Let $k$ be a positive integer. Consider $k$ not necessarily distinct prime numbers such that their product is ten times their sum. What are these primes and what is the value of $k$?

Let $\mathcal M_n(\mathbb R)$ denote the set of all real $n\times n$ matrices. Find all surjective functions $f:\mathcal M_n(\mathbb R)\to\{0,1,\ldots,n\}$ which satisfy $$f(XY)\le\min\{f(X),f(Y)\}$$for all $X,Y\in\mathcal M_n(\mathbb R)$.

The Nationwide Basketball Society (NBS) has $8001$ teams, numbered $2000$ through $10000$. For each $n$, team $n$ has $n+1$ players, and in a sheer coincidence, this year each player attempted $n$ shots and on team $n$, exactly one player made $0$ shots, one player made $1$ shot, . . ., one player made $n$ shots. A player's [i]field goal percentage[/i] is defined as the percentage of shots the player made, rounded to the nearest tenth of a percent (For instance, $32.45\%$ rounds to $32.5\%$). A player in the NBS is randomly selected among those whose field goal percentage is $66.6\%$. If this player plays for team $k$, the probability that $k\geq 6000$ can be expressed as $\tfrac{p}{q}$ for relatively prime positive integers $p$ and $q$. Find $p+q$.

Find the sum of all $5$-digit positive integers that they have only the digits $1, 2$, and $5$, none repeated more than three consecutive times.

Let $(a_n)_{n\geq 1}$ be a sequence of positive real numbers with the property that $$(a_{n+1})^2 + a_na_{n+2} \leq a_n + a_{n+2}$$ for all positive integers $n$. Show that $a_{2022}\leq 1$.

Consider a function $f: \mathbb{Z} \rightarrow \mathbb{Z}$. We call an integer $a$ [i]spanning[/i] if for all integers $b \neq a$, there exists a positive integer $k$ with $f^k(a)=b$. Find, with proof, the maximum possible number of [i]spanning[/i] numbers of $f$. Note: $\mathbb{Z}$ represents the set of all integers, so $f$ is a function from the set of integers to itself. $f^k(a)$ is defined as $f$ applied $k$ times to $a$.

Let $\sigma(\cdot)$ denote the divisor sum function and $d(\cdot)$ denote the divisor counting function. Find all positve integers $n$ such that $\sigma(d(n))=n.$ [i]Andrei Bâra[/i]

For positive integers $n,$ let the numbers $c(n)$ be determined by the rules $c(1)=1,c(2n)=c(n),$ and $c(2n+1)=(-1)^nc(n).$ Find the value of \[\sum_{n=1}^{2013}c(n)c(n+2).\]

Suppose that $n\ge2$ and $a_1,a_2,...,a_n$ are natural numbers that $ (a_1,a_2,...,a_n)=1$. Find all strictly increasing function $f: \mathbb{Z} \to \mathbb{R} $ that: $$ \forall x_1,x_2,...,x_n \in \mathbb{Z} : f(\sum_{i=1}^{n} {x_ia_i}) = \sum_{i=1}^{n} {f(x_ia_i})$$ [i]Proposed by Navid Safaei and Ali Mirzaei [/i]

Let $ABCD$ be a square. The line segment $AB$ is divided internally at $H$ so that $|AB|\cdot |BH|=|AH|^2$. Let $E$ be the midpoints of $AD$ and $X$ be the midpoint of $AH$. Let $Y$ be a point on $EB$ such that $XY$ is perpendicular to $BE$. Prove that $|XY|=|XH|$.

In triangle $ABC$ ($AB > BC$), $K$ and $M$ are the midpoints of sides $AB$ and $AC$, $O$ is the point of intersection of the angle bisectors. Let $P$ be the intersection point of lines $KM$ and $CO$, and the point $Q$ is such that $QP \perp KM$ and $QM \parallel BO$. Prove that $QO \perp AC$.

The $A{}$-excircle $\omega_A{}$ of the triangle $ABC$ touches the side of the $BC$ at point $A_1$ and the extensions of the sides $AB$ and $AC$ are at points $C_1$ and $B_1$ respectively. Let $P{}$ be the middle of the segment $B_1C_1$. The line $A_1P$ intersects $\omega_A{}$ a second time at point $X{}$. The tangents to the circumcircle of the triangle $ABC$ at point $A{}$ and to $\omega_A{}$ at point $X{}$ intersect at point $R$. Prove that $RP = RX$.

Let $M$ be the midpoint of side $AC$ of the triangle $ABC$. Let $P$ be a point on the side $BC$. If $O$ is the point of intersection of $AP$ and $BM$ and $BO = BP$, determine the ratio $\frac{OM}{PC}$ .

What is the value of $$3 + \frac{1}{3+\frac{1}{3+\frac{1}{3}}}?$$ $\textbf{(A) } \frac{31}{10} \qquad \textbf{(B) } \frac{49}{15} \qquad \textbf{(C) } \frac{33}{10} \qquad \textbf{(D) } \frac{109}{33} \qquad \textbf{(E) } \frac{15}{4}$

Two magicians are about to show the next trick. A circle is drawn on the board with one semicircle marked. Viewers mark 100 points on this circle, then the first magician erases one of them. After this, the second one for the first time looks at the drawing and determines from the remaining 99 points whether the erased point was lying on the marked semicircle. Prove that such a trick will not always succeed.

Let $\mathbb{N}$ denote the set of positive integers. Find all functions $f:\mathbb{N}\longrightarrow\mathbb{N}$ such that \[n+f(m)\mid f(n)+nf(m)\] for all $m,n\in \mathbb{N}$ [i]Proposed by Dorlir Ahmeti, Albania[/i]