For a positive number such as 3.27, 3 is referred to as the integral part of the number and .27 as the decimal part. Find a positive number such that its decimal part, its integral part, and the number itself form a geometric progression.

Let be given a convex polygon $ M_0M_1\ldots M_{2n}$ ($ n\ge 1$), where $ 2n \plus{} 1$ points $ M_0$, $ M_1$, $ \ldots$, $ M_{2n}$ lie on a circle $ (C)$ with diameter $ R$ in an anticlockwise direction. Suppose that there is a point $ A$ inside this convex polygon such that $ \angle M_0AM_1$, $ \angle M_1AM_2$, $ \ldots$, $ \angle M_{2n \minus{} 1}AM_{2n}$, $ \angle M_{2n}AM_0$ are equal. Assume that $ A$ is not coincide with the center of the circle $ (C)$ and $ B$ be a point lies on $ (C)$ such that $ AB$ is perpendicular to the diameter of $ (C)$ passes through $ A$. Prove that \[ \frac {2n \plus{} 1}{\frac {1}{AM_0} \plus{} \frac {1}{AM_1} \plus{} \cdots \plus{} \frac {1}{AM_{2n}}} < AB < \frac {AM_0 \plus{} AM_1 \plus{} \cdots \plus{} AM_{2n}}{2n \plus{} 1} < R \]

Let $x,y,z$ be distinct positive integers such that $(y+z)(z+x)=(x+y)^2$ . Show that \[x^2+y^2>8(x+y)+2(xy+1).\] (Paolo Leonetti)

Both roots of the quadratic equation $ x^2 \minus{} 63x \plus{} k \equal{} 0$ are prime numbers. The number of possible values of $ k$ is $ \textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ \textbf{more than four}$

Let $k$ be any fixed positive integer. Let's look at integer pairs $(a, b)$, for which the quadratic equations $x^2 - 2ax + b = 0$ and $y^2 + 2ay + b = 0$ are real solutions (not necessarily different), which can be denoted by $x_1, x_2$ and $y_1, y_2$, respectively, in such an order that the equation $x_1 y_1 - x_2 y_2 = 4k$. a) Find the largest possible value of the second component $b$ of such a pair of numbers ($a, b)$. b) Find the sum of the other components of all such pairs of numbers.

The real numbers $a,b,c$ with $bc\neq0$ satisfy $\frac{1-c^2}{bc}\geq0.$ Prove that $10(a^2+b^2+c^2-bc^3)\geq2ab+5ac.$

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 $x^2+10x+9$?

Let $ n$ and $ k$ be positive integers such as either $ n$ is odd or both $ n$ and $ k$ are even. Prove that exists integers $ a$ and $ b$ such as $ GCD(a,n) \equal{} GCD(b,n) \equal{} 1$ and $ k \equal{} a \plus{} b$

Prove that, among $100000$ consecutive $100$-digit positive integers, there is an integer $n$ such that the length of the period of the decimal expansion of $\frac1n$ is greater than $2011$.

A parabola with equation $ y \equal{} x^2 \plus{} bx \plus{} c$ passes through the points $ (2,3)$ and $ (4,3)$. What is $ c$? $ \textbf{(A) } 2 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 7 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 11$

Determine all pairs of positive real numbers $(a, b)$ for which there exists a function $ f:\mathbb{R^{+}}\rightarrow\mathbb{R^{+}} $ satisfying for all positive real numbers $x$ the equation $ f(f(x))=af(x)- bx $

Find all pairs of real number $x$ and $y$ which simultaneously satisfy the following 2 relations: $x+y+4=\frac{12x+11y}{x^2+y^2}$ $y-x+3=\frac{11x-12y}{x^2+y^2}$

Let $ a,\ b$ be real numbers satisfying $ \int_0^1 (ax\plus{}b)^2dx\equal{}1$. Determine the values of $ a,\ b$ for which $ \int_0^1 3x(ax\plus{}b)\ dx$ is maximized.

While there do not exist pairwise distinct real numbers $a,b,c$ satisfying $a^2+b^2+c^2 = ab+bc+ca$, there do exist complex numbers with that property. Let $a,b,c$ be complex numbers such that $a^2+b^2+c^2 = ab+bc+ca$ and $|a+b+c| = 21$. Given that $|a-b| = 2\sqrt{3}$, $|a| = 3\sqrt{3}$, compute $|b|^2+|c|^2$. [hide="Clarifications"] [list] [*] The problem should read $|a+b+c| = 21$. An earlier version of the test read $|a+b+c| = 7$; that value is incorrect. [*] $|b|^2+|c|^2$ should be a positive integer, not a fraction; an earlier version of the test read ``... for relatively prime positive integers $m$ and $n$. Find $m+n$.''[/list][/hide] [i]Ray Li[/i]