Let $P$ be a point on the median $AM$ of a triangle $ABC$. Suppose that the tangents to the circumcircles of $ABP$ and $ACP$ at $B$ and $C$ respectively meet at $Q$. Show that $\angle PAB = \angle CAQ$.

Let $a_1, a_2, a_3, \ldots$ be a sequence of positive real numbers, and $s$ be a positive integer, such that \[a_n = \max \{ a_k + a_{n-k} \mid 1 \leq k \leq n-1 \} \ \textrm{ for all } \ n > s.\] Prove there exist positive integers $\ell \leq s$ and $N$, such that \[a_n = a_{\ell} + a_{n - \ell} \ \textrm{ for all } \ n \geq N.\] [i]Proposed by Morteza Saghafiyan, Iran[/i]

Let $n \geq 1$ be an integer. What is the maximum number of disjoint pairs of elements of the set $\{ 1,2,\ldots , n \}$ such that the sums of the different pairs are different integers not exceeding $n$?

Points $ A,B,C$ and $ D$ lie on a line, in that order, with $ AB\equal{}CD$ and $ BC\equal{}12$. Point $ E$ is not on the line, and $ BE\equal{}CE\equal{}10$. The perimeter of $ \triangle AED$ is twice the perimeter of $ \triangle BEC$. Find $ AB$. $ \text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2$

A square $ABCD$ is given. A point $P$ is chosen inside the triangle $ABC$ such that $\angle CAP = 15^\circ = \angle BCP$. A point $Q$ is chosen such that $APCQ$ is an isosceles trapezoid: $PC \parallel AQ$, and $AP=CQ, AP\nparallel CQ$. Denote by $N$ the midpoint of $PQ$. Find the angles of the triangle $CAN$.

Fill in each empty cell of the grid with a digit from 1 to 8 so that every row and every column contains each of these digits exactly once. Some diagonally adjacent cells have been joined together. For these pairs of joined cells, the same number must be written in both. [asy] filldraw((0,0)--(0,8)--(8,8)--(8,0)--cycle,white); path removex(pair p) { return ((p.x-0.5, p.y)--(p.x+0.5,p.y)); } path removey(pair p) { return ((p.x, p.y-0.5)--(p.x,p.y+0.5)); } unitsize(1cm); draw((0,0)--(8,0)--(8,8)--(0,8)--cycle, linewidth(2)); for(int i = 0; i < 8; ++i){ draw((0,i)--(8,i)); } for(int j = 0 ; j<8; ++j){ draw((j,0)--(j,8)); } pair [] pointsa = {(1,2),(3,1),(5,7),(7,6)}; pair [] pointsb= {(1,5),(4,4),(2,7),(6,1),(7,3)}; for(int q = 0; q<4; ++q){ draw(removex(pointsa[q]), white+linewidth(2)); draw(removey(pointsa[q]),white+linewidth(2)); draw(arc(pointsa[q]+(0.5,-0.5),0.5,90,180)); draw(arc(pointsa[q]-(0.5,-0.5),0.5,270,0,CCW)); draw(pointsa[q]+(-0.5,0)--pointsa[q]+(-1,0)); draw(pointsa[q]+(0.5,0)--pointsa[q]+(1,0)); draw(pointsa[q]+(0,0.5)--pointsa[q]+(0,1)); draw(pointsa[q]+(0,-0.5)--pointsa[q]+(0,-1)); } for(int q = 0; q<5; ++q){ draw(removex(pointsb[q]), white+linewidth(2)); draw(removey(pointsb[q]),white+linewidth(2)); draw(arc(pointsb[q]+(0.5,0.5),0.5,180,270,CCW)); draw(arc(pointsb[q]-(0.5,0.5),0.5,0,90,CCW)); draw(pointsb[q]+(-0.5,0)--pointsb[q]+(-1,0)); draw(pointsb[q]+(0.5,0)--pointsb[q]+(1,0)); draw(pointsb[q]+(0,0.5)--pointsb[q]+(0,1)); draw(pointsb[q]+(0,-0.5)--pointsb[q]+(0,-1)); } int [][] x = { {1,0,0,0,0,0,0,0}, {2,3,0,0,0,0,0,0}, {0,4,5,0,0,0,0,0}, {0,0,6,0,1,0,0,0}, {0,0,0,7,0,1,0,0}, {0,0,0,0,0,3,4,0}, {0,0,0,0,0,0,2,8}, {0,0,0,0,0,0,0,5} }; for(int k = 0; k<8; ++k){ for(int l = 0; l<8; ++l){ if(x[k][l]!=0){ label(string(x[k][l]), (l+0.5,-k+7.5), fontsize(24pt)); } } } [/asy] There is a unique solution, but you do not need to prove that your answer is the only one possible. You merely need to find an answer that satisfies the constraints above. (Note: In any other USAMTS problem, you need to provide a full proof. Only in this problem is an answer without justification acceptable.)

A polynomial $P (x)$ with real coefficients and of degree $n \ge 3$ has $n$ real roots $x_1 <x_2 < \cdots < x_n$ such that \[x_2 - x_1 < x_3 - x_2 < \cdots < x_n - x_{n-1} \] Prove that the maximum value of $|P (x)|$ on the interval $[x_1 , x_n ]$ is attained in the interval $[x_{n-1} , x_n ]$.

Each BMT, every student chooses one of three focus rounds to take. Bob plans to attend BMT for the next $4$ years and wants to gure out what focus round to take each year. Given that he wants to take each focus round at least once, how many ways can he choose which round to take each year?

Prove that $1998! \left( 1+ \frac12 + \frac13 +...+\frac{1}{1998}\right)$ is an integer divisible by $1999$.

Consider function $ f: R\to R$ which satisfies the conditions for any mutually distinct real numbers $ a,b,c,d$ satisfying $ \frac {a \minus{} b}{b \minus{} c} \plus{} \frac {a \minus{} d}{d \minus{} c} \equal{} 0$, $ f(a),f(b),f(c),f(d)$ are mutully different and $ \frac {f(a) \minus{} f(b)}{f(b) \minus{} f(c)} \plus{} \frac {f(a) \minus{} f(d)}{f(d) \minus{} f(c)} \equal{} 0.$ Prove that function $ f$ is linear

Inside the $7\times8$ rectangle below, one point is chosen a distance $\sqrt2$ from the left side and a distance $\sqrt7$ from the bottom side. The line segments from that point to the four vertices of the rectangle are drawn. Find the area of the shaded region. [asy] import graph; size(4cm); pair A = (0,0); pair B = (9,0); pair C = (9,7); pair D = (0,7); pair P = (1.5,3); draw(A--B--C--D--cycle,linewidth(1.5)); filldraw(A--B--P--cycle,rgb(.76,.76,.76),linewidth(1.5)); filldraw(C--D--P--cycle,rgb(.76,.76,.76),linewidth(1.5)); [/asy]

Jesse and Tjeerd are playing a game. Jesse has access to $n\ge 2$ stones. There are two boxes: in the black box there is room for half of the stones (rounded down) and in the white box there is room for half of the stones (rounded up). Jesse and Tjeerd take turns, with Jesse starting. Jesse grabs in his turn, always one new stone, writes a positive real number on the stone and places put him in one of the boxes that isn't full yet. Tjeerd sees all these numbers on the stones in the boxes and on his turn may move any stone from one box to the other box if it is not yet full, but he may also choose to do nothing. The game stops when both boxes are full. If then the total value of the stones in the black box is greater than the total value of the stones in the white box, Jesse wins; otherwise win Tjeerd. For every $n \ge 2$, determine who can definitely win (and give a corresponding winning strategy).

In the language of wolves has two letters $F$ and $P$, any finite sequence which forms a word. А word $Y$ is called 'subpart' of word $X$ if Y is obtained from X by deleting some letters (for example, the word $FFPF$ has 8 'subpart's: F, P, FF, FP, PF, FFP, FPF, FFF). Determine $n$ such that the $n$ is the greatest number of 'subpart's can have n-letter word language of wolves. F. Petrov, V. Volkov

Prove that : For each integer $n \ge 3$, there exists the positive integers $a_1<a_2< \cdots <a_n$ , such that for $ i=1,2,\cdots,n-2 $ , With $a_{i},a_{i+1},a_{i+2}$ may be formed as a triangle side length , and the area of the triangle is a positive integer.

Let $x_1, x_2, \ldots, x_n$ be real positive numbers such that $x_1\cdot x_2\cdots x_n=1$. Prove the inequality $\frac1{x_1(x_1+1)}+\frac1{x_2(x_2+1)}+\cdots+\frac1{x_n(x_n+1)}\geq\frac n2$

Find $ \lim_{n\to\infty} \sum_{k\equal{}1}^n \ln \left(1\plus{}\frac{k^a}{n^{a\plus{}1}}\right).$

If $x$ is the number of solutions to the equation $a^2 + b^2 + c^2 = d^2$ of the form $(a,b,c,d)$ such that $\{a,b,c\}$ are three consecutive square numbers and $d$ is also a square number, find $x$.

The altitudes $AN$ and $BM$ are drawn in triangle $ABC$. Prove that the perpendicular bisector to the segment $NM$ divides the segment $AB$ in half.

Let $ABC$ be an acute triangle with orthocenter $H$. The circumcircle of the triangle $BHC$ intersects $AC$ a second time in point $P$ and $AB$ a second time in point $Q$. Prove that $H$ is the circumcenter of the triangle $APQ$. [i](Karl Czakler)[/i]

Nine lines, parallel to the base of a triangle, divide the other sides into 10 equal segments and the area into 10 distinct parts. Find the area of the original triangle, if the area of the largest of these parts is 76.

In isosceles triangle $ABC$ with $AB=AC$, consider point $D$ on the base $BC$ and point $E$ on side $AC$ such that $ \angle BAD = 2 \angle CDE$. Prove that $AD=AE$.

Find all positive integers n with the property that $n^3 - 1$ is a perfect square.

Let $ S$ be a finite family of squares on a plane such that every point on that plane is contained in at most $ k$ squares in $ S$. Prove that $ P$ can be divided into $ 4(k\minus{}1)\plus{}1$ sub-family such that in each sub-family, each pair of squares are disjoint.

Prove that for every positive integer $n,$ the set $\{2,3,4,\ldots,3n+1\}$ can be partitioned into $n$ triples in such a way that the numbers from each triple are the lengths of the sides of some obtuse triangle. [i]Proposed by Canada[/i]

[u]Round 1[/u] [b]p1.[/b] Let $ABCDEF$ be a regular hexagon. How many acute triangles have all their vertices among the vertices of $ABCDEF$? [b]p2.[/b] A rectangle has a diagonal of length $20$. If the width of the rectangle is doubled, the length of the diagonal becomes $22$. Given that the width of the original rectangle is $w$, compute $w^2$. [b]p3.[/b] The number $\overline{2022A20B22}$ is divisible by 99. What is $A + B$? [u]Round 2[/u] [b]p4.[/b] How many two-digit positive integers have digits that sum to at least $16$? [b]p5.[/b] For how many integers $k$ less than $10$ do there exist positive integers x and y such that $k =x^2 - xy + y^2$? [b]p6.[/b] Isosceles trapezoid $ABCD$ is inscribed in a circle of radius $2$ with $AB \parallel CD$, $AB = 2$, and one of the interior angles of the trapezoid equal to $110^o$. What is the degree measure of minor arc $CD$? [u]Round 3[/u] [b]p7.[/b] In rectangle $ALEX$, point $U$ lies on side $EX$ so that $\angle AUL = 90^o$. Suppose that $UE = 2$ and $UX = 12$. Compute the square of the area of $ALEX$. [b]p8.[/b] How many digits does $20^{22}$ have? [b]p9.[/b] Compute the units digit of $3 + 3^3 + 3^{3^3} + ... + 3^{3^{...{^3}}}$ , where the last term of the series has $2022$ $3$s. [u]Round 4[/u] [b]p10.[/b] Given that $\sqrt{x - 1} + \sqrt{x} = \sqrt{x + 1}$ for some real number $x$, the number $x^2$ can be written as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [b]p11.[/b] Eric the Chicken Farmer arranges his $9$ chickens in a $3$-by-$3$ grid, with each chicken being exactly one meter away from its closest neighbors. At the sound of a whistle, each chicken simultaneously chooses one of its closest neighbors at random and moves $\frac12$ of a unit towards it. Given that the expected number of pairs of chickens that meet can be written as $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers, compute $p + q$. [b]p12.[/b] For a positive integer $n$, let $s(n)$ denote the sum of the digits of $n$ in base $10$. Find the greatest positive integer $n$ less than $2022$ such that $s(n) = s(n^2)$. PS. You should use hide for answers. Rounds 5-8 have been posted [url=https://artofproblemsolving.com/community/c3h2949432p26408285]here[/url]. Collected [url=https://artofproblemsolving.com/community/c5h2760506p24143309]here[/url].