Shining tells Prajit a positive integer $n \ge 2025$. Prajit then tries to place n points such that no four points are concyclic and no $3$ points are collinear in Euclidean plane, such that Shining cannot find a group of three points such that their circumcircle contains none of the other remaining points. Is he always able to do so? [i](Prajit Adhikari, Nepal and Shining Sun, USA)[/i]

Given is a natural number $n>4$. There are $n$ points marked on the plane, no three of which lie on the same line. Vasily draws one by one all the segments connecting pairs of marked points. At each step, drawing the next segment $S$, Vasily marks it with the smallest natural number, which hasn't appeared on a drawn segment that has a common end with $S$. Find the maximal value of $k$, for which Vasily can act in such a way that he can mark some segment with the number $k$?

Find all positive integers $p,q,r,s>1$ such that $$p!+q!+r!=2^s.$$

Consider the equilateral triangular lattice in the complex plane defined by the Eisenstein integers; let the ordered pair $(x,y)$ denote the complex number $x+y\omega$ for $\omega=e^{2\pi i/3}$. We define an $\omega$-chessboard polygon to be a (non self-intersecting) polygon whose sides are situated along lines of the form $x=a$ or $y=b$, where $a$ and $b$ are integers. These lines divide the interior into unit triangles, which are shaded alternately black and white so that adjacent triangles have different colors. To tile an $\omega$-chessboard polygon by lozenges is to exactly cover the polygon by non-overlapping rhombuses consisting of two bordering triangles. Finally, a [i]tasteful tiling[/i] is one such that for every unit hexagon tiled by three lozenges, each lozenge has a black triangle on its left (defined by clockwise orientation) and a white triangle on its right (so the lozenges are BW, BW, BW in clockwise order). a) Prove that if an $\omega$-chessboard polygon can be tiled by lozenges, then it can be done so tastefully. b) Prove that such a tasteful tiling is unique. [i]Victor Wang.[/i]

Find out all the integer pairs $(m,n)$ such that there exist two monic polynomials $P(x)$ and $Q(x)$ ,with $\deg{P}=m$ and $\deg{Q}=n$,satisfy that $$P(Q(t))\not=Q(P(t))$$ holds for any real number $t$.

Two triangles $ABC, A’B’C’$ have the same orthocenter $H$ and the same circumcircle with center $O$. Letting $PQR$ be the triangle formed by $AA’, BB’, CC’$, prove that the circumcenter of $PQR$ lies on $OH$.

Determine all the integers $a$ and $b$, such that $\sqrt{2010 + 2 \sqrt{2009}}$ be a solution of the equation $x^2 + ax + b = 0$. Prove that for such $a$ and $b$ the number$\sqrt{2010 - 2 \sqrt{2009}}$ is not a solution to the given equation.

Let $\Delta_0$ be an equilateral triangle with incircle $\omega$. A point on $\omega$ is reflected in the sides of $\Delta_0$ to obtain a new triangle $\Delta_1$. The same point is then reflected over the sides of $\Delta_1$ to obtain another triangle $\Delta_2$. Prove that the circumcircle of $\Delta_2$ is tangent to $\omega$. [i]Proposed by Siddharth Choppara[/i]

Fill in each white hexagon with a positive digit from $1$ to $9$. Some digits have been given to you. Each of the seven gray hexagons touches six hexagons; these six hexagons must contain six distinct digits, and the sum of these six digits must equal the number inside the gray hexagon. [img]https://cdn.artofproblemsolving.com/attachments/2/8/a00d0605bc81430e8ed4ae108befb037c4b00b.png[/img] 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.)

If $p$, $q$, and $r$ are nonzero integers satisfying \[p^2+q^2 = r^2,\] compute the smallest possible value of $(p+q+r)^2$. [i]Proposed by David Altizio[/i]

A sequence $(a_n)_{n=1}^{\infty}$ of positive integers satisfies the condition $a_{n+1} = a_n +\tau (n)$ for all positive integers $n$ where $\tau (n)$ is the number of positive integer divisors of $n$. Determine whether two consecutive terms of this sequence can be perfect squares.

Let $x,y,z$ be real numbers from the interval $[0,1]$. Determine the maximum value of expression $$W=y\cdot \sqrt{1-x}+z\cdot\sqrt{1-y}+x\cdot\sqrt{1-z}$$

Prove that the sum of the squares of $1984$ consecutive positive integers cannot be the square of an integer.

Let real $a$, $b$, and $c$ satisfy $$abc+a+b+c=ab+bc+ca+5.$$ Find the least possible value of $a^2+b^2+c^2$.

Let $ABCDE$ be a pentagon with right angles at vertices $B$ and $E$ and such that $AB = AE$ and $BC = CD = DE$. The diagonals $BD$ and $CE$ meet at point $F$. Prove that $FA = AB$.

What is the value of the expression $\frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}{1+2+3+4+5+6+7+8}$? $\textbf{(A) }1020\qquad\textbf{(B) }1120\qquad\textbf{(C) }1220\qquad\textbf{(D) }2240\qquad\textbf{(E) }3360$

Find the sum of $$\frac{\sigma(n) \cdot d(n)}{ \phi (n)}$$ over all positive $n$ that divide $ 60$. Note: The function $d(i)$ outputs the number of divisors of $i$, $\sigma (i)$ outputs the sum of the factors of $i$, and $\phi (i)$ outputs the number of positive integers less than or equal to $i$ that are relatively prime to $i$.

In a quadrilateral $ABCD$ the intersection of the diagonals is called $P$. Point $X$ is the orthocentre of triangle $PAB$. (The orthocentre of a triangle is the point where the three altitudes of the triangle intersect.) Point $Y$ is the orthocentre of triangle $PCD$. Suppose that $X$ lies inside triangle $PAB$ and $Y$ lies inside triangle $PCD$. Moreover, suppose that $P$ is the midpoint of line segment $XY$ . Prove that $ABCD$ is a parallelogram. [asy] import geometry; unitsize (1.5 cm); pair A, B, C, D, P, X, Y; A = (0,0); B = (2,-0.5); C = (3.5,2.2); D = A + C - B; P = (A + C)/2; X = orthocentercenter(A,B,P); Y = orthocentercenter(C,D,P); draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(A--extension(A,X,B,P), dotted); draw(B--extension(B,X,A,P), dotted); draw(P--extension(P,X,A,B), dotted); draw(C--extension(C,Y,D,P), dotted); draw(D--extension(D,Y,C,P), dotted); draw(P--extension(P,Y,C,D), dotted); dot("$A$", A, W); dot("$B$", B, S); dot("$C$", C, E); dot("$D$", D, N); dot("$P$", P, E); dot("$X$", X, NW); dot("$Y$", Y, SE); [/asy]

In an $n\times n$ table, it is allowed to rearrange rows, as well as rearrange columns. Asterisks are placed in some $k{}$ cells of the table. What maximum $k{}$ for which it is always possible to ensure that all the asterisks are on the same side of the main diagonal (and that there are no asterisks on the main diagonal itself)? [i]Proposed by P. Kozhevnikov[/i]

Let $ABCD$ be a trapezoid with $AB \parallel CD$ and $BC \perp CD$. There exists a point $P$ on $BC$ such that $\triangle{PAD}$ is equilateral. If $PB = 20$ and $PC = 23$, the area of $ABCD$ can be expressed in the form $\frac{a\sqrt{b}}{c}$ where $b$ is square-free and $a$ and $c$ are relatively prime. Find $a+b+c$. [i]Proposed by Andy Xu[/i]

Let $a_1$, $a_2$, $a_3$, $\ldots$ be a sequence of positive integers such that $a_1=2021$ and $$\sqrt{a_{n+1}-a_n}=\lfloor \sqrt{a_n} \rfloor. $$ Show that there are infinitely many odd numbers and infinitely many even numbers in this sequence. [i] Proposed by Li4, Tsung-Chen Chen, and Ming Hsiao.[/i]

To prepare for an Olympiad $20$ students went to a coach. The coach gave them $20$ problems and it turned out that (a) each of the students solved two problems and (b) each problem was solved by twostudents. Prove that it is possible to organize the coaching so that each student would discuss one of the problems that (s)he had solved, and so that all problems would be discussed.

A sequence of positive integers $a_1,a_2,\ldots $ is such that for each $m$ and $n$ the following holds: if $m$ is a divisor of $n$ and $m<n$, then $a_m$ is a divisor of $a_n$ and $a_m<a_n$. Find the least possible value of $a_{2000}$.

Let $A, B \in M_6 (\mathbb{Z} )$ such that $A \equiv I \equiv B \text{ mod }3$ and $A^3 B^3 A^3 = B^3$. Prove that $A = I$. Here $M_6 (\mathbb{Z} )$ indicates the $6$ by $6$ matrices with integer entries, $I$ is the identity matrix, and $X \equiv Y \text{ mod }3$ means all entries of $X-Y$ are divisible by $3$.

Exists a triangle whose three altitudes have lengths $1, 2$ and $3$ respectively?