Quadratic trinomial $f(x)$ is allowed to be replaced by one of the trinomials $x^2f(1+\frac{1}{x})$ or $(x-1)^2f(\frac{1}{x-1})$. With the use of these operations, is it possible to go from $x^2+4x+3$ to $x^2+10x+9$?

Find the last four digits of $5^{5555}$.

Let $n$ be a positive integer, $n \geq 2$, and consider the polynomial equation \[x^n - x^{n-2} - x + 2 = 0.\] For each $n,$ determine all complex numbers $x$ that satisfy the equation and have modulus $|x| = 1.$

If $ a,b,c>0 $ and $ ab+bc+ca+abc=4, $ then $ \sqrt{ab} +\sqrt{bc} +\sqrt{ca} \le 3\le a+b+c. $

Denote $S$ as the subset of $\{1,2,3,\dots,1000\}$ with the property that none of the sums of two different elements in $S$ is in $S$. Find the maximum number of elements in $S$.

Let $\mathbb{R}$ denote the set of all real numbers. Determine all functions $f:\mathbb{R}\to\mathbb{R}$ such that for all real numbers $x$ and $y$, \[f(xy+xf(x))=f(x)\left(f(x)+f(y)\right).\]

Show that the number of 5-tuples ($a$, $b$, $c$, $d$, $e$) such that $abcde = 5(bcde + acde + abde + abce + abcd)$ is odd

The two heights in the triangle are not less than the respective sides. Find the angles.

$ABC$ is a triangle with $AB = 15$, $BC = 14$, and $CA = 13$. The altitude from $A$ to $BC$ is extended to meet the circumcircle of $ABC$ at $D$. Find $AD$.

(a) Prove that the set $X = (1,2,....100)$ cannot be partitoned into THREE subsets such that two numbers differing by a square belong to different subsets. (b) Prove that $X$ can so be partitioned into $5$ subsets.

How many pairs of positive integers $(x, y)$ are there, those satisfy the identity $2^x - y^2 = 4$? (A): $1$ (B): $2$ (C): $3$ (D): $4$ (E): None of the above.

Let $E$ and $F$ on $AC$ and $AB$ respectively in $\triangle ABC$ such that $DE || BC$ then draw line $l$ through $A$ such that $l || BC$ let $D'$ and $E'$ reflection of $D$ and $E$ to $l$ respectively prove that $D'B, E'C$ and $l$ are congruence.

Three circles are constructed on the medians of a triangle as on diameters. It is known that they intersect in pairs. Let $C_1$ be the intersection point of the circles built on the medians $AM_1$ and $BM_2$, which is more distant from the vertex $C$. Points $A_1$ and $B_1$ are defined similarly. Prove that the lines $AA_1, BB_1$ and $CC_1$ intersect at one point. (D. Tereshin)

Two congruent circles centered at points $A$ and $B$ each pass through the other circle's center. The line containing both $A$ and $B$ is extended to intersect the circles at points $C$ and $D$. The circles intersect at two points, one of which is $E$. What is the degree measure of $\angle CED$? $\textbf{(A) }90\qquad\textbf{(B) }105\qquad\textbf{(C) }120\qquad\textbf{(D) }135\qquad \textbf{(E) }150$

Determine \[\left\lfloor\prod_{n=2}^{2022}\frac{2n+2}{2n+1}\right\rfloor,\] given that the answer is relatively prime to $2022$.

Let $ABCD$ be a tetragonal. [b](a)[/b] If the plane $(P)$ cuts $ABCD,$ find the necessary and sufficient condition such that the area formed from the intersection of the plane $(P)$ and the tetragonal be a parallelogram. Prove that the problem has three solutions in this case. [b](b)[/b] Consider one of the solutions of [b](a)[/b]. Find the situation of the plane $(P)$ for which the parallelogram has maximum area. [b](c)[/b] Find a plane $(P)$ for which the parallelogram be a lozenge and then find the length side of his lozenge in terms of the length of the edges of $ABCD.$

Misha has a $100$x$100$ chessboard and a bag with $199$ rooks. In one move he can either put one rook from the bag on the lower left cell of the grid, or remove two rooks which are on the same cell, put one of them on the adjacent square which is above it or right to it, and put the other in the bag. Misha wants to place exactly $100$ rooks on the board, which don't beat each other. Will he be able to achieve such arrangement?

Point $M$ on side $AB$ of quadrilateral $ABCD$ is such that quadrilaterals $AMCD$ and $BMDC$ are circumscribed around circles centered at $O_1$ and $O_2$ respectively. Line $O_1O_2$ cuts an isosceles triangle with vertex $M$ from angle $CMD$. prove that $ABCD$ is a cyclc quadrilateral.

Compute the integer $$\frac{2^{\left(2^5-2\right)/5-1}-2}{5}.$$ [i]2016 CCA Math Bonanza Individual Round #1[/i]

Let P be a set of 4n points in the plane such that none of the three points are collinear. Prove that if n is large enough, then the following two statements are equivalent. (i) P can be divided into n four-element subsets such that each subset forms the vertices of a convex quadrilateral. (ii) P can not be split into two sets A and B, each with an odd number of elements, so that each convex quadrilateral whose vertices are in P has an even number of vertices in A and B.

Find the positive integers $n$ with $n \geq 4$ such that $[\sqrt{n}]+1$ divides $n-1$ and $[\sqrt{n}]-1$ divides $n+1$. [hide="Remark"]This problem can be solved in a similar way with the one given at [url=http://www.mathlinks.ro/Forum/resources.php?c=1&cid=97&year=2006]Cono Sur Olympiad 2006[/url], problem 5.[/hide]

28. The numbers $1-10$ are written in a circle randomly. Find the expected number of numbers which are at least $2$ larger than an adjacent number. 29. We want to design a new chess piece, the American, with the property that (i) the American can never attack itself, and (ii) if an American $A_1$ attacks another American $A_2$, then $A_2$ also attacks $A_1$. Let $m$ be the number of squares that an American attacks when placed in the top left corner of an 8 by 8 chessboard. Let $n$ be the maximal number of Americans that can be placed on the $8$ by $8$ chessboard such that no Americans attack each other, if one American must be in the top left corner. Find the largest possible value of $mn$. 30. On the blackboard, Amy writes $2017$ in base-$a$ to get $133201_a$. Betsy notices she can erase a digit from Amy’s number and change the base to base-$b$ such that the value of the the number remains the same. Catherine then notices she can erase a digit from Betsy’s number and change the base to base-$c$ such that the value still remains the same. Compute, in decimal, $a + b + c$.

Find all primes $p$ and positive integers $n$ satisfying \[n \cdot 5^{n-n/p} = p! (p^2+1) + n.\]

Let $ABC$ be an acute triangle such that $\angle B > \angle C$. Let $D$ and $E$ be the points on the segments $BC$ and $CA$, respectively, such that $AD$ bisects $\angle A$ and $BE\perp AC$. Finally, let $M$ be the midpoint of the side $BC$. Suppose that the circumcircle of $\triangle CDE$ intersects $AD$ again at a point $X$ different from $D$. Prove that $\angle XME = 90^{\circ} - \angle BAC$.

In the table shown in the figure, Zosia replaced eight numbers with their negatives. It turned out that each row and each column contained exactly two negative numbers. Prove that after this change, the sum of all sixteen numbers in the table is equal to $0$. [center] [img] https://wiki-images.artofproblemsolving.com//2/2e/17-3-5.png [/img] [/center]