Does there exist a nonzero algebraic number $\alpha$ with $|\alpha| \neq 1$ such that there exists infinitely many positive integers $n$ for which there's $\beta_n \in \mathbb{C}$ with $\beta_n \in \mathbb{Q}(\alpha)$ and $\beta_n^n = \alpha$?

Given is natural number $n$. Sasha claims that for any $n$ rays in space, no two of which have a common point, he will be able to mark on these rays $k$ points lying on one sphere. What is the largest $k$ for which his statement is true?

Show that if the sides $a, b, c$ of a triangle satisfy the equation \[2(ab^2 + bc^2 + ca^2) = a^2b + b^2c + c^2a + 3abc,\] then the triangle is equilateral. Show also that the equation can be satisfied by positive real numbers that are not the sides of a triangle.

Call a rational number $r$ [i]powerful[/i] if $r$ can be expressed in the form $\dfrac{p^k}{q}$ for some relatively prime positive integers $p, q$ and some integer $k >1$. Let $a, b, c$ be positive rational numbers such that $abc = 1$. Suppose there exist positive integers $x, y, z$ such that $a^x + b^y + c^z$ is an integer. Prove that $a, b, c$ are all [i]powerful[/i]. [i]Jeck Lim, Singapore[/i]

Let $P(x)$ be a nonzero polynomial with integer coefficients. Let $a_{0}=0$ and for $i \ge 0$ define $a_{i+1}=P(a_{i})$. Show that $\gcd ( a_{m}, a_{n})=a_{ \gcd (m, n)}$ for all $m, n \in \mathbb{N}$.

Inside the parallelogram $ABCD$, point $M$ is chosen, and inside the triangle $AMD$, point $N$ is chosen in such a way that $$\angle MNA + \angle MCB =\angle MND + \angle MBC = 180^o.$$ Prove that lines $MN$ and $AB$ are parallel.

Prove that in any set of $2000$ distinct real numbers there exist two pairs $a>b$ and $c>d$ with $a \neq c$ or $b \neq d $, such that \[ \left| \frac{a-b}{c-d} - 1 \right|< \frac{1}{100000}. \]

Find all pairs $(a,b,c)$ of positive integers such that $$\frac{a}{2^a}=\frac{b}{2^b}+\frac{c}{2^c}$$

Positive numbers $a_1,a_2,...,a_n, b_1, b_2,...,b_n$ satisfy the condition $a_1+a_2+...+a_n=b_1+ b_2+...+b_n=1$. Find the smallest possible value of the sum $$\frac{a_1^2}{a_1+b_1}+\frac{a_2^2}{a_2+b_2}+...+\frac{a_n^2}{a_n+b_n}$$ (V.Kolbun)

Let $A(x)=\lfloor\frac{x^2-20x+16}{4}\rfloor$, $B(x)=\sin\left(e^{\cos\sqrt{x^2+2x+2}}\right)$, $C(x)=x^3-6x^2+5x+15$, $H(x)=x^4+2x^3+3x^2+4x+5$, $M(x)=\frac{x}{2}-2\lfloor\frac{x}{2}\rfloor+\frac{x}{2^2}+\frac{x}{2^3}+\frac{x}{2^4}+\ldots$, $N(x)=\textrm{the number of integers that divide }\left\lfloor x\right\rfloor$, $O(x)=|x|\log |x|\log\log |x|$, $T(x)=\sum_{n=1}^{\infty}\frac{n^x}{\left(n!\right)^3}$, and $Z(x)=\frac{x^{21}}{2016+20x^{16}+16x^{20}}$ for any real number $x$ such that the functions are defined. Determine $$C(C(A(M(A(T(H(B(O(N(A(N(Z(A(2016)))))))))))))).$$ [i]2016 CCA Math Bonanza Lightning #5.3[/i]

In equilateral triangle $ABC$, $AB=2$ and $M$ is the midpoint of $AB$. A laser is shot from $M$ in a certain direction, and whenever it collides with a side of $ABC$ it will reflect off the side such that the acute angle formed by the incident ray and the side is equal to the acute angle formed by the reflected ray and the side. Once the laser coincides with a vertex, it stops. Find the sum of the smallest three possible integer distances that the laser could have traveled. [i]Proposed by Jerry Xu[/i] [hide=Solution] [i]Solution.[/i] $\boxed{21}$ Whenever the laser hits a side of the triangle, reflect the laser's path over that side so that the path of the laser forms a straight line. We want the path of the laser to coincide with a vertex of one of the reflected triangles. Thus, we can restate the problem as follows: Tessellate the plane with equilateral triangles of side length $3$. Consider one of these equilateral triangles $ABC$ with $M$ being the midpoint of $AB=2$. Find the sum of the three minimum integer distances from $M$ to any vertex in the plane. [asy] import geometry; size(8cm); pair A = (0,sqrt(3)); pair B = (-1,0); pair C = (1,0); pair M = (0,0); for (int i = -1; i <= 2; ++i) { draw((i-3,i*sqrt(3))--(-i+3,i*sqrt(3))); draw(((i-1)*2,-sqrt(3))--(i+1,(2-i)*sqrt(3))); draw((-i-1,(2-i)*sqrt(3))--((1-i)*2,-sqrt(3))); } draw(A--B--C--A, red); dot(M); label("$A$",A+(0,0.25),N); label("$B$",B-(0.25,0),SW); label("$C$",C+(0.25,0),SE); label("$M$",M,S); [/asy] It is trivial to see that the vertical distance between $M$ and a given vertex is $n\sqrt{3}$ for $n \in \mathbb{N}^{0}$. If $n$ is even, the horizontal distance between $O$ and a given vertex is $1+2m$ for $m \in \mathbb{N}^{0}$. If $n$ is odd, the horizontal distance is $2m$ for $m \in \mathbb{N}^{0}$. We consider two separate cases: $1.$ $n$ is even. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(1+2m)^2=l^2.$$Make the substitution $1+2m=k$ to get that $$3n^2+k^2=l^2.$$Notice that these equations form a family of generalized Pell equations $y^2-3x^2=N$ with $N=k^2$. We can find some set of roots to these equations using the multiplicative principle: we will use this idea to find three small $l$ values, and that gives us an upper bound on what the three $l$ values can be. From there, a simple bash of lower $l$ values to see if solutions to each generalized Pell equation not given by the multiplicative principle exist finishes this case. By the multiplicative principle some set of solutions $(x_n,y_n)$ to the above equation with sufficiently small $x_n$ follow the formula$$x_n\sqrt{3}+y_n=\left(x_0\sqrt{3}+y_0\right)\left(u_n\sqrt{3}+v_n\right),$$where $\left(x_0,y_0\right)$ is a solution to the generalized Pell equation and $\left(u_n,v_n\right)$ are solutions to the Pell equation $y^2-3x^2=1$. Remember that the solutions to this last Pell equation satisfy$$u_n\sqrt{3}+v_n=\left(u_0\sqrt{3}+v_0\right)^k$$where the trivial positive integer solution $$\left(u_0, v_0\right)=(1,2)$$(this can easily be found by inspection or by taking the convergents of the continued fraction expansion of $\sqrt{3}$). We thus get that$$\left(u_1,v_1\right)=(4,7),\left(u_2,v_2\right)=(15,26),\left(u_2,v_2\right)=(56,97)\dots$$(also don't forget that $(u,v)=(0,1)$ is another solution). From here, note that $k$ must be odd since $k=1+2m$ for $m \in \mathbb{N}^{0}$. For $k=1$, the smallest three solutions to the Pell equation with $n$ even are \begin{align*} (x,y)&=(0,1),(4,7),(56,97) \\ \longrightarrow (n,m,l)&=(0,0,1),(4,0,7),(56,0,97) \end{align*}Our current smallest three values of $l$ are thus $1,7,97$. A quick check confirms that all of these solutions are not extraneous (extraneous solutions appear when the path taken by the laser prematurely hits a vertex). For $k=3$, using the multiplicative principle we get two new smaller solutions \begin{align*} (x,y)&=(0,3),(12,21) \\ \longrightarrow (n,m,l)&=(0,1,3),(12,1,21) \end{align*}However, note that $(n,m,l)=(0,1,3)$ is extraneous since is equivalent to the path that is traced out by the solution $(n,m,l)=(0,0,1)$ found previously and will thus hit a vertex prematurely. Thus, our new three smallest values of $l$ are $1,7,21$. For $k \ge 5$, it is evident that there are no more smaller integral values of $l$ that can be found using the multiplicative principle: the solution set $(n,m,l)=\left(0,\dfrac{k-1}{2},k\right)$ is always extraneous for $k > 1$ since it is equivalent to the path traced out by $(0,0,1)$ as described above, and any other solutions will give larger values of $l$. Thus, we now only need to consider solutions to each generalized Pell equation not found by the multiplicative principle. A quick bash shows that $l=3,5,9,11$ gives no solutions for any odd $k$ and even $n$, however $n=13$ gives $k=11$ and $n=4$, a non-extraneous solution smaller than one of the three we currently have. Thus, our new three smallest $l$ values are $1,7,13$. $2$. $n$ is odd. We thus want to find $l \in \mathbb{N}$ such that $$\left(n\sqrt{3}\right)^2+(2m)^2=l^2.$$Make the substitution $2m=k$ to get that $$3n^2+k^2=l^2.$$This is once again a family of generalized Pell equations with $N=k^2$, however this time we must have $k$ even instead of $k$ odd. However, note that there are no solutions to this family of Pell equation with $n$ odd: $k^2 \equiv 0 \text{ (mod }4)$ since $k$ is even, and $3n^2 \equiv 3 \text{ (mod }4)$ since $n$ is odd, however $0+3 \equiv 3 \text{ (mod }4)$ is not a possible quadratic residue mod $4$. Thus, this case gives no solutions. Our final answer is thus $1+7+13=\boxed{21}$. [/hide]

Let $O$ be an interior point of an acute triangle $ABC$. The circles with centers the midpoints of its sides and passing through $O$ mutually intersect the second time at the points $K$, $L$ and $M$ different from $O$. Prove that $O$ is the incenter of the triangle $KLM$ if and only if $O$ is the circumcenter of the triangle $ABC$.

The sum of the real numbers $x_1, x_2, ...,x_n$ belonging to the segment $[a, b]$ is equal to zero. Prove that $$x_1^2+ x_2^2+ ...+x_n^2 \le - nab.$$

If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, then $p^3+q^3+r^3$ equals $ \textbf{(A)}\ -1 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ \text{none of these} $

Tom has a scientific calculator. Unfortunately, all keys are broken except for one row: 1, 2, 3, + and -. Tom presses a sequence of $5$ random keystrokes; at each stroke, each key is equally likely to be pressed. The calculator then evaluates the entire expression, yielding a result of $E$. Find the expected value of $E$. (Note: Negative numbers are permitted, so 13-22 gives $E = -9$. Any excess operators are parsed as signs, so -2-+3 gives $E=-5$ and -+-31 gives $E = 31$. Trailing operators are discarded, so 2++-+ gives $E=2$. A string consisting only of operators, such as -++-+, gives $E=0$.) [i]Proposed by Lewis Chen[/i]

Determine all real $ x$ satisfying the equation \[ \sqrt[5]{x^3 \plus{} 2x} \equal{} \sqrt[3]{x^5\minus{}2x}.\] Odd roots for negative radicands shall be included in the discussion.

What is the area of the largest circle contained in an equilateral triangle of area $8\sqrt3$?

You are given a set of $n$ blocks, each weighing at least $1$; their total weight is $2n$. Prove that for every real number $r$ with $0 \leq r \leq 2n-2$ you can choose a subset of the blocks whose total weight is at least $r$ but at most $r + 2$.

Prove that for positive $a,b,c,d$ \[\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\ge 4\]

$AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC+AD-AB-AC = 0$, find the range of $\angle BAC$. [i]Alternative formulation.[/i] Let $AD$ be the altitude of triangle $ABC$ to the side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.

A positive integer $n$ is [i]Mozart[/i] if the decimal representation of the sequence $1, 2, \ldots, n$ contains each digit an even number of times. Prove that: 1. All Mozart numbers are even. 2. There are infinitely many Mozart numbers.

Let $n$ and $k$ be two integers with $n>k\geqslant 1$. There are $2n+1$ students standing in a circle. Each student $S$ has $2k$ [i]neighbors[/i] - namely, the $k$ students closest to $S$ on the left, and the $k$ students closest to $S$ on the right. Suppose that $n+1$ of the students are girls, and the other $n$ are boys. Prove that there is a girl with at least $k$ girls among her neighbors. [i]Proposed by Gurgen Asatryan, Armenia[/i]

Let $a_1, a_2,...,a_{2n}$ be positive real numbers such that $a_i + a_{n+i} = 1$, for all $i = 1,...,n$. Prove that there exist two different integers $1 \le j, k \le 2n$ for which $$\sqrt{a^2_j-a^2_k} < \frac{1}{\sqrt{n} +\sqrt{n - 1}}$$

Suppose $\triangle ABC$ is scalene. $O$ is the circumcenter and $A'$ is a point on the extension of segment $AO$ such that $\angle BA'A = \angle CA'A$. Let point $A_1$ and $A_2$ be foot of perpendicular from $A'$ onto $AB$ and $AC$. $H_{A}$ is the foot of perpendicular from $A$ onto $BC$. Denote $R_{A}$ to be the radius of circumcircle of $\triangle H_{A}A_1A_2$. Similiarly we can define $R_{B}$ and $R_{C}$. Show that: \[\frac{1}{R_{A}} + \frac{1}{R_{B}} + \frac{1}{R_{C}} = \frac{2}{R}\] where R is the radius of circumcircle of $\triangle ABC$.

Find the maximum value of $$\int^1_0|f'(x)|^2|f(x)|\frac1{\sqrt x}dx$$over all continuously differentiable functions $f:[0,1]\to\mathbb R$ with $f(0)=0$ and $$\int^1_0|f'(x)|^2dx\le1.$$