The real numbers $a_1,a_2,a_3$ and $b{}$ are given. The equation \[(x-a_1)(x-a_2)(x-a_3)=b\]has three distinct real roots, $c_1,c_2,c_3.$ Determine the roots of the equation \[(x+c_1)(x+c_2)(x+c_3)=b.\][i]Proposed by A. Antropov and K. Sukhov[/i]

If $f(x, y) = xy + 2x + y + 1$, find $f(f(2, f(3, 4)), 5)$.

For a non-isosceles $ABC$ we have that $2AC = AB + BC$. Point $I$ is the center of the circle inscribed in $\triangle ABC$, point $K$ is the middle of the arc $\widehat{AC}$ that includes point $B$, and point $T$ is from the line $AC$, such that $\angle TIB = 90^\circ$. Prove that the line $TB$ is tangent to the circumscribed circle of $\triangle KBI$.

The smallest product one could obtain by multiplying two numbers in the set $\{ -7, -5, -1, 1, 3 \}$ is $\text{(A)}\ -35 \qquad \text{(B)}\ -21 \qquad \text{(C)}\ -15 \qquad \text{(D)}\ -1 \qquad \text{(E)}\ 3$

Let $ABC$ be a triangle with $AB < AC$, let $L$ be midpoint of arc $BC$(the point $A$ is not in this arc) of the circumcircle $w$($ABC$). Let $E$ be a point in $AC$ where $AE = \frac{AB + AC}{2}$, the line $EL$ intersects $w$ in $P$. If $M$ and $N$ are the midpoints of $AB$ and $BC$, respectively, prove that $AL, BP$ and $MN$ are concurrents

Consider a triple $(a, b, c)$ of pairwise distinct positive integers satisfying $a + b + c = 2013$. A step consists of replacing the triple $(x, y, z)$ by the triple $(y + z - x,z + x - y,x + y - z)$. Prove that, starting from the given triple $(a, b,c)$, after $10$ steps we obtain a triple containing at least one negative number.

Partition the grid into 1 by 1 squares and 1 by 2 dominoes in either orientation, marking dominoes with a line connecting the two adjacent squares, and 1 by 1 squares with an asterisk ($*$). No two 1 by 1 squares can share a side. A $border$ is a grid segment between two adjacent squares that contain dominoes of opposite orientations. All borders have been marked with thick lines in the grid. 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(1cm); // makes asterisks larger (you can remove if you want) defaultpen(fontsize(30pt)); for(int i = 0; i < 10; ++i) { for(int j = 0; j < 10; ++j) { draw((i - 0.5, j - 0.5)--(i + 0.5, j - 0.5)--(i + 0.5, j + 0.5)--(i - 0.5, j + 0.5)--(i - 0.5, j - 0.5), gray(0.5)); } } draw((0 - 0.5, 2 - 0.5)--(1 - 0.5, 2 - 0.5), gray(0) + 3); draw((0 - 0.5, 6 - 0.5)--(0.5, 5.5), black+3); draw((-0.5, 6.5)--(0.5, 6.5)--(0.5, 7.5), black+3); draw((-0.5, 8.5)--(0.5, 8.5), black+3); draw((1.5, -0.5)--(1.5, 1.5), black+3); draw((1.5, 5.5)--(1.5, 6.5)--(2.5, 6.5)--(2.5, 7.5)--(1.5, 7.5), black+3); draw((1.5, 9.5)--(1.5, 8.5), black+3); draw((2.5, -0.5)--(2.5, 0.5)--(3.5, 0.5)--(3.5, 1.5), black+3); draw((4.5, 1.5)--(5.5, 1.5)--(5.5, 2.5), black+3); draw((4.5, 6.5)--(5.5, 6.5), black+3); draw((5.5, 8.5)--(6.5, 8.5)--(6.5, 7.5), black+3); draw((6.5, -0.5)--(6.5, 0.5), black+3); draw((6.5, 1.5)--(7.5, 1.5), black+3); draw((8.5, 5.5)--(8.5, 4.5), black+3); string[] grid = { "----------", "----------", "----------", "----------", "----------", "----------", "----------", "----------", "----------", "----------" }; /* L is the left side of a domino R is the right T is the top B is the bottom */ for(int j = 9; j >= 0; --j) { for(int i = 0; i < 10; ++i) { string identifier = substr(grid[9 - j], i, 1); if (identifier == "*") label("$*$", (i, j)); else if (identifier == "L") draw((i, j)--(i + 0.5, j)); else if (identifier == "R") draw((i, j)--(i - 0.5, j)); else if (identifier == "T") draw((i, j)--(i, j - 0.5)); else if (identifier == "B") draw((i, j)--(i, j + 0.5)); } } [/asy]

By dividing the integer $m$ by the integer $n, 22$ is the quotient and $5$ the remainder. As the division of the remainder with $n$ continues, the new quotient is $0.4$ and the new remainder is $0.2$. Find $m$ and $n$.

Points $ A \equal{} (3,9), B \equal{} (1,1), C \equal{} (5,3),$ and $ D \equal{} (a,b)$ lie in the first quadrant and are the vertices of quadrilateral $ ABCD$. The quadrilateral formed by joining the midpoints of $ \overline{AB}, \overline{BC}, \overline{CD},$ and $ \overline{DA}$ is a square. What is the sum of the coordinates of point $ D$? $ \textbf{(A)} \ 7 \qquad \textbf{(B)} \ 9 \qquad \textbf{(C)} \ 10 \qquad \textbf{(D)} \ 12 \qquad \textbf{(E)} \ 16$

Let \(a,b,c\) be positive real numbers such that \(a+b+c=4\sqrt[3]{abc}\). Prove that \[2(ab+bc+ca)+4\min(a^2,b^2,c^2)\ge a^2+b^2+c^2.\]

How many figures can be obtained by intersecting the infinite-dimensional cube $|x_k| \le 1$, $k = 1,2,\ldots$ with a two-dimensional plane?

Given two distinct circles touching each other internally, show how to construct a triangle with the inner circle as its incircle and the outer circle as its nine point circle.

Prove that $n^2 + 8n + 15$ is not divisible by $n + 4$ for any positive integer $n$.

In triangle $ABC$, angle $C$ is a right angle. Found on the side $AC$ point $D$, and on the segment $BD$, point $K$ such that $\angle ABC = \angle KAD =\angle AKD$. Prove that $BK = 2DC$.

Which triplet of numbers has a sum NOT equal to 1? $ \text{(A)}\ (1/2,1/3,1/6)\qquad\text{(B)}\ (2,-2,1)\qquad\text{(C)}\ (0.1,0.3,0.6)\qquad\text{(D)}\ (1.1,-2.1,1.0)\qquad\text{(E)}\ (-3/2,-5/2,5) $

Let $\mathbb R_{>0}$ be the set of positive real numbers. Determine all functions $f \colon \mathbb R_{>0} \to \mathbb R_{>0}$ such that \[x \big(f(x) + f(y)\big) \geqslant \big(f(f(x)) + y\big) f(y)\] for every $x, y \in \mathbb R_{>0}$.

You are given a convex quadrilateral $ABCD$. It is known that $\angle CAD = \angle DBA = 40^o$, $\angle CAB = 60^o$, $\angle CBD = 20^o$. Find the angle $\angle CDB $.