Let $P(x)= x^n+a_{1}x^{n-1}+...+a_{n-1}x+(-1)^{n}$ , $a_{i} \in C$ , $n\geq 2$ with all roots having same modulo. Prove that $P(-1) \in R$

Given two convex polygons, $A_1A_2...A_n$ and $B_1B_2...B_n$ such that $A_1A_2 = B_1B_2$, $A_2A_3 = B_2B_3$,$ ...$, $A_nA_1 = B_nB_1$ and $n - 3$ angles of one polygon are equal to the respective angles of the other. Find whether these polygons are equal.

Let $a$ and $b$ be two distinct positive integers with the same parity. Prove that the fraction $\frac{a!+b!}{2^a}$ is not an integer.

In $\vartriangle ABC$, the external bisectors of $\angle A$ and $\angle B$ meet at a point $D$. Prove that the circumcentre of $\vartriangle ABD$ and the points $C, D$ lie on the same straight line.

In a $2024 \times 2024$ grid of squares, each square is colored either black or white. An ant starts at some black square in the grid and starts walking parallel to the sides of the grid. During this walk, it can choose (not required) to turn $90^\circ$ clockwise or counterclockwise if it is currently on a black square, otherwise it must continue walking in the same direction. A coloring of the grid is called [i]simple[/i] if it is [b]not[/b] possible for the ant to arrive back at its starting location after some time. How many simple colorings of the grid are maximal, in the sense that adding any black square results in a coloring that is not simple?

$33$ counters are shown in the left grid below. Choose a counter to start at and remove it from the grid. At each subsequent step, choose a direction (up, down, left, or right), move along the grid line from your current position to the nearest counter in that direction, and remove that counter. You cannot choose a direction that reverses your previous one (e.g., left then right is not allowed). Your goal is to pick up all $33$ counters in a single sequence of steps. When you find the right sequence, write the numbers $1$ to $33$ on the counters so that $N$ is written on the $N$th counter you removed. A smaller example of a solved grid is shown to the right below. (Note that the final move from $8$ to $9$ is possible because counters $3, 4,$ and $5$ have been removed in earlier steps.) 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.) [asy] unitsize(0.5cm); string[][] x = { {"-","-","0","0","0","0","0","0","0","0","0"}, {"0","0","-","0","0","0","0","0","0","0","0"}, {"0","-","0","0","-","0","0","0","0","0","0"}, {"-","0","0","-","0","-","0","0","0","0","0"}, {"-","-","-","-","0","-","0","0","0","0","0"}, {"0","0","0","-","0","-","-","-","-","0","0"}, {"0","0","0","0","-","0","0","0","0","-","0"}, {"0","0","0","0","0","0","0","-","-","0","0"}, {"0","0","0","0","0","0","-","0","0","0","-"}, {"0","0","0","0","0","0","-","-","-","-","-"}, {"0","0","0","0","0","0","0","0","0","0","-"}, {"0","0","0","0","0","0","0","0","0","0","-"}, {"0","0","0","0","0","0","0","0","0","0","-"}, {"0","0","0","0","0","0","0","0","0","0","-"}}; void drawcircle(int x, int y) { filldraw(circle((x,y),0.45),white); } for(int i = 0; i < 11; ++i) { draw((i,-0.7)--(i,13.7),dashed); } for(int i = 0; i < 14; ++i) { draw((-0.7,i)--(10.7,i),dashed); } drawcircle(10,0); drawcircle(10,1); drawcircle(10,2); drawcircle(10,3); drawcircle(10,4); drawcircle(10,5); drawcircle(9,4); drawcircle(8,4); drawcircle(7,4); drawcircle(6,4); drawcircle(6,5); drawcircle(7,6); drawcircle(8,6); drawcircle(9,7); drawcircle(8,8); drawcircle(7,8); drawcircle(6,8); drawcircle(5,8); drawcircle(4,7); drawcircle(5,9); drawcircle(5,10); drawcircle(4,11); drawcircle(3,10); drawcircle(3,9); drawcircle(3,8); drawcircle(2,9); drawcircle(1,9); drawcircle(0,9); drawcircle(0,10); drawcircle(1,11); drawcircle(2,12); drawcircle(1,13); drawcircle(0,13); for(int k = 0; k<14; ++k){ for(int l = 0; l<11; ++l){ if(x[k][l]!="0"){ label((x[k][l]),(l,-k+13),fontsize(10pt)); } } } [/asy]

Suppose that a point in the plane is called [i]good[/i] if it has rational coordinates. Prove that all good points can be divided into three sets satisfying: 1) If the centre of the circle is good, then there are three points in the circle from each of the three sets. 2) There are no three collinear points that are from each of the three sets.

Let $S$ be the number of bijective functions $f:\{0,1,\dots,288\}\rightarrow\{0,1,\dots,288\}$ such that $f((m+n)\pmod{17})$ is divisible by $17$ if and only if $f(m)+f(n)$ is divisible by $17$. Compute the largest positive integer $n$ such that $2^n$ divides $S$.

Fifteen freshmen are sitting in a circle around a table, but the course assistant (who remains standing) has made only six copies of today’s handout. No freshman should get more than one handout, and any freshman who does not get one should be able to read a neighbor’s. If the freshmen are distinguishable but the handouts are not, how many ways are there to distribute the six handouts subject to the above conditions?

Find all odd primes $ p$, if any, such that $ p$ divides $ \sum_{n\equal{}1}^{103}n^{p\minus{}1}$

Define $p(x) = 4x^{3}-2x^{2}-15x+9, q(x) = 12x^{3}+6x^{2}-7x+1$. Show that each polynomial has just three distinct real roots. Let $A$ be the largest root of $p(x)$ and $B$ the largest root of $q(x)$. Show that $A^{2}+3 B^{2}= 4$.

Let $\Lambda$ denote the set of points $(x,y)$ in 2D space with integer coordinates such that $0\leq x\leq 4$ and $0\leq y\leq 2$. That is, \[ \Lambda=\{ (x,y) \in \mathbb{Z}^2: 0\leq x\leq 4, \ 0\leq y\leq 2 \}. \] Find the number of ways to connect points of $\Lambda$ with segments of length $\sqrt{2}$ or $\sqrt{5}$ such that the interior of any unit square with vertices in $\Lambda$ contains part of exactly one segment; an example is shown below (connections that differ by reflections are distinct). [asy] unitsize(1cm); dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((0,1)); dot((1,1)); dot((2,1)); dot((3,1)); dot((4,1)); dot((0,2)); dot((1,2)); dot((2,2)); dot((3,2)); dot((4,2)); draw((0,0)--(1,1)); draw((0,2)--(2,1)); draw((1,1)--(2,0)); draw((2,0)--(3,2)); draw((3,1)--(4,2)); draw((3,0)--(4,1)); [/asy]

In the plane, six circles are given so that none of the circles contain one the center of the other. Show that there is no point that lies in all the circles.

Let $a, b$ and $c$ be integers such that $a + b + c = 0$. (a) Show that $a^4 + b^4 + c^4$ is divisible by $a^2 + b^2 + c^2$. (b) Show that $a^{100} + b^{100} + c^{100}$ is divisible by $a^2 + b^2 + c^2$. .

If $a+b=b-c=c-a=3$, find $a+b+c$. $\text{(A) }3\qquad\text{(B) }4\frac{1}{2}\qquad\text{(C) }6\qquad\text{(D) }7\frac{1}{2}\qquad\text{(E) }9$

Let $n\ge 3$ be an integer and $a_1,a_2,\dots ,a_n$ be a finite sequence of positive integers, such that, for $k=2,3,\dots ,n$ $$n(a_k+1)-(n-1)a_{k-1}=1.$$ Prove that $a_n$ is not divisible by $(n-1)^2$.

Let $ A$ be a family of proper closed subspaces of the Hilbert space $ H\equal{}l^2$ totally ordered with respect to inclusion (that is , if $ L_1,L_2 \in A$, then either $ L_1\subset L_2$ or $ L_2\subset L_1$). Prove that there exists a vector $ x \in H$ not contaied in any of the subspaces $ L$ belonging to $ A$. [i]B. Szokefalvi Nagy[/i]

An equilateral triangle is dissected into white and black triangles. It is known that all white triangles are right-angled and mutually congruent, and all black triangles are isosceles and also mutually congruent. Is it necessarily true that a) all angles of white triangles are multiples of $30^{\circ}$; (4 marks) b) all angles of black triangles are multiples of $30^{\circ}$ ? (5 marks)

Let $p$ be a prime number and $n$ a positive integer. Prove that the product \[{N=\frac{1}{p^{n^2}}} \prod_{i=1;2 \nmid i}^{2n-1} \biggl[ \left( (p-1)! \right) \binom{p^2 i}{pi}\biggr]\] Is a positive integer that is not divisible by $p.$

Find all pairs $(x,y)$ of positive integers such that $x^{x+y} =y^{y-x}$.

Fins all ordered triples $ \left(a,b,c\right)$ of positive integers such that $ abc \plus{} ab \plus{} c \equal{} a^3$.

A sequence of positive integers $a_1, a_2,\ldots, a_n$ is called [i]ladder[/i] of length $n$ if it consists of $n$ consecutive integers in ascending order. (a) Prove that for every positive integer $n$ there exist two ladders of length $n$, with no elements in common, $a_1, a_2,\ldots, a_n$ and $b_1, b_2,\ldots, b_n$, such that for all $i$ between $1$ and $n$, the greatest common divisor of $a_i$ and $b_i$ is equal to $1$. (b) Prove that for every positive integer $n$ there exist two ladders of length $n$, with no elements in common, $a_1, a_2,\ldots, a_n$ and $b_1, b_2,\ldots, b_n$, such that for all $i$ between $1$ and $n$, the greatest common divisor of $a_i$ and $b_i$ is greater than $1$.

Let $ p\ge 2 $ be a natural number that divides $ \binom{p}{k} , $ for any natural number $ k $ smaller than $ p. $ Prove that: [b]a)[/b] $ p $ is prime. [b]b)[/b] $ p^2 $ divides $ -2+\binom{2p}{p} . $

Five persons of different heights stand next to the another on numbered booths to take a picture. From how many ways can be arranged so that people in positions $ 1$ and $3$ are both taller than the person in the position $2$?

Rob is helping to build the set for a school play. For one scene, he needs to build a multi-colored tetrahedron out of cloth and bamboo. He begins by fitting three lengths of bamboo together, such that they meet at the same point, and each pair of bamboo rods meet at a right angle. Three more lengths of bamboo are then cut to connect the other ends of the first three rods. Rob then cuts out four triangular pieces of fabric: a blue piece, a red piece, a green piece, and a yellow piece. These triangular pieces of fabric just fill in the triangular spaces between the bamboo, making up the four faces of the tetrahedron. The areas in square feet of the red, yellow, and green pieces are $60$, $20$, and $15$ respectively. If the blue piece is the largest of the four sides, find the number of square feet in its area.