Consider those functions $ f$ that satisfy $ f(x \plus{} 4) \plus{} f(x \minus{} 4) \equal{} f(x)$ for all real $ x$. Any such function is periodic, and there is a least common positive period $ p$ for all of them. Find $ p$. $ \textbf{(A)}\ 8\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 16\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 32$

Find all functions $f : \mathbb{N} \rightarrow \mathbb{R}$ such that for all triples $a,b,c$ of positive integers the following holds : $$f(ac)+f(bc)-f(c)f(ab) \ge 1$$ Proposed by [i]Mojtaba Zare[/i]

Let \[P(x)=24x^{24}+\sum_{j=1}^{23}(24-j)(x^{24-j}+x^{24+j}). \] Let $z_{1},z_{2},\ldots,z_{r}$ be the distinct zeros of $P(x),$ and let $z_{k}^{2}=a_{k}+b_{k}i$ for $k=1,2,\ldots,r,$ where $i=\sqrt{-1},$ and $a_{k}$ and $b_{k}$ are real numbers. Let \[\sum_{k=1}^{r}|b_{k}|=m+n\sqrt{p}, \] where $m,$ $n,$ and $p$ are integers and $p$ is not divisible by the square of any prime. Find $m+n+p.$

Let $1,2,3,4,5,6,7,8,9,11,12,\cdots$ be the sequence of all positive integers which do not contain the digit zero. Write $\{a_n\}$ for this sequence. By comparing with a geometric series, show that $\sum_{k=1}^n \frac{1}{a_k} < 90$.

Let $f: \ [0,\ 1] \rightarrow \mathbb{R}$ be an increasing function satisfying the following conditions: a) $f(0)=0$; b) $f\left(\frac{x}{3}\right)=\frac{f(x)}{2}$; c) $f(1-x)=1-f(x)$. Determine $f\left(\frac{18}{1991}\right)$.

$f:\mathbb{R}^2 \to \mathbb{R}^2$ is injective and surjective. Distance of $X$ and $Y$ is not less than distance of $f(X)$ and $f(Y)$. Prove for $A$ in plane: \[ S(A) \geq S(f(A))\] where $S(A)$ is area of $A$

Last year, Master Cheung is famous for multi-rotation. This year, he comes to DAMO to make noodles for sweeping monk. One day, software engineer Xiao Li talks with Master Cheung about his job. Xiao Li mainly researches and designs the algorithm to adjust the paramter of different kinds of products. These paramters can normally be obtainly by minimising loss function $f$ on $\mathbb{R}^n$. In the recent project of Xiao Li, this loss function is obtained by other topics. For safety consideration and technique reasons, this topic makes Xiao Li difficult to find the interal details of the function. They only provide a port to calculate the value of $f(\text x)$ for any $\text x\in\mathbb{R}^n$. Therefore, Xiao Li must only use the value of the function to minimise $f$. Also, every times calculating the value of $f$ will use a lot of calculating resources. It is good to know that the dimension $n$ is not very high (around $10$). Also, colleague who provides the function tells Xiao Li to assume $f$ is smooth first. This problem reminds Master Cheung of his antique radio. If you want to hear a programme from the radio, you need to turn the knob of the radio carefully. At the same time, you need to pay attention to the quality of the radio received, until the quality is the best. In this process, no one knows the relationship between the angle of turning the knob and the quality of the radio received. Master Cheung and Xiao Li realizes that minimising $f$ is same as adjusting the machine with multiple knobs: Assume every weight of $\text x$ is controlled by a knob. $f(\text x)$ is a certain performance of the machine. We only need to adjust every knobs again and again and observes the value of $f$ in the same time. Maybe there is hope to find the best $\text x$. As a result, two people suggest an iteration algorithm (named Automated Forward/Backward Tuning, $\text{AFBT}$, to minimise $f$. In $k$-th iteration, the algorithm adjusts the individual weight of $\text{x}_k$ to $2n$ points $\{\text x_k\pm t_k\text e^i:i=1,...,n\}$, where $t_k$ is the step size; then, make $y_k$ be the smallest one among the value of the function of thosse points. Then check if $\text y_k$ sufficiently makes $f$ decrease; then, take $\text x_{k+1}=\text y_k$, then make the step size doubled. Otherwise, make $\text x_{k+1}=\text x_k$ and makes the step size decrease in half. In the algorithm, $\text e^i$ is the $i$-th coordinate vector in $\mathbb{R}^n$. The weight of $i$-th is $1$. Others are $0$; $\mathbf{1}(\cdot)$ is indicator function. If $f(\text x_k)-f(\text y_k)$ is at least the square of $t_k$, then take the value of $\mathbf{1}(f(\text k)-f(y_k)\ge t^2_k)$ as $1$. Otherwise, take it as $0$. $\text{AFBT}$ algorithm Input $\text{x}_0\in \mathbb{R}^n$, $t_0>0$. For $k=0, 1, 2, ...$, perform the following loop: 1: #Calculate loss function. 2: $s_k:=\mathbb{1}[f(\text{x}_k)-f(\text{y}_k)\ge t^2_k]$ #Is it sufficiently decreasing? Yes: $s_k=1$; No: $s_k=0$. 3: $\text{x}_{k+1}:=(1-s_k)\text{x}_k+s_k\text{y}_k$ #Update the point of iteration. 4: $t_{k+1}:=2^{2S_k-1}t_k$ #Update step size. $s_k=1$: Step size doubles; $s_k=0$: Step size decreases by half. Now, we made assumption to the loss function $f:\mathbb{R}^n\to \mathbb{R}$. Assumption 1. Let $f$ be a convex function. For any $\text{x}, \text{y}\in \mathbb{R}^n$ and $\alpha \in [0, 1]$, we have $f((1-\alpha)\text{x}+\text{y})\le (1-\alpha)f(\text{x})+\alpha f(\text{y})$. Assumption 2. $f$ is differentiable on $\mathbb{R}^n$ and $\nabla f$ is L-Lipschitz continuous on $\mathbb{R}^n$. Assumption 3. The level set of $f$ is bounded. For any $\lambda\in\mathbb{R}$, set $\{\text x\in \mathbb{R}^n:f(\text x)\le \lambda\}$ is all bounded. Based on assumption 1 and 2, we can prove that $\left\langle \nabla f(\text x),\text y-\text x \right\rangle \le f(\text y)-f(\text x)\le \left\langle \nabla f(\text x),\text y-\text x\right\rangle+\frac{L}{2}||\text x-\text y||^2$ You can refer to any convex analysis textbook for more properties of convex function. Prove that under the assumption 1-3, for $AFBT$, $\lim_{k \to \infty}f(\text{x}_k)=f^*$

Let $X$ be a bounded, nonempty set of points in the Cartesian plane. Let $f(X)$ be the set of all points that are at a distance of at most $1$ from some point in $X$. Let $f_n(X) = f(f(\cdots(f(X))\cdots))$ ($n$ times). Show that $f_n(X)$ becomes “more circular” as $n$ gets larger. In other words, if $r_n = \sup\{\text{radii of circles contained in } f_n(X) \}$ and $R_n = \inf \{\text{radii of circles containing } f_n(X)\}$, then show that $R_n/r_n$ gets arbitrarily close to $1$ as $n$ becomes arbitrarily large. [hide]I'm not sure that I'm posting this in a right forum. If it's in a wrong forum, please mods move it.[/hide]

Let $f:[-1,1]\to\mathbb{R}$ be a continuous function having finite derivative at $0$, and \[I(h)=\int^h_{-h}f(x)\text{ d}x,\ h\in [0,1].\] Prove that a) there exists $M>0$ such that $|I(h)-2f(0)h|\le Mh^2$, for any $h\in [0,1]$. b) the sequence $(a_n)_{n\ge 1}$, defined by $a_n=\sum_{k=1}^n\sqrt{k}|I(1/k)|$, is convergent if and only if $f(0)=0$. [i]Calin Popescu[/i]

Let $a$ and $b$ be positive integers such that $ab+1$ divides $a^{2}+b^{2}$. Show that \[\frac{a^{2}+b^{2}}{ab+1}\] is the square of an integer.

Let $n$ be a positive integer. Let $S$ be a subset of points on the plane with these conditions: $i)$ There does not exist $n$ lines in the plane such that every element of $S$ be on at least one of them. $ii)$ for all $X \in S$ there exists $n$ lines in the plane such that every element of $S - {X} $ be on at least one of them. Find maximum of $\mid S\mid$. [i]Proposed by Erfan Salavati[/i]

Let $\mathbb Q$ be the set of all rational numbers and $\mathbb R$ be the set of real numbers. Function $f: \mathbb Q \to \mathbb R$ satisfies the following conditions: (i) $f(0) = 0$, and for any nonzero $a \in Q, f(a) > 0.$ (ii) $f(x + y) = f(x)f(y) \qquad \forall x,y \in \mathbb Q.$ (iii) $f(x + y) \leq \max\{f(x), f(y)\} \qquad \forall x,y \in \mathbb Q , x,y \neq 0.$ Let $x$ be an integer and $f(x) \neq 1$. Prove that $f(1 + x + x^2+ \cdots + x^n) = 1$ for any positive integer $n.$

The lines $L$ and $K$ are symmetric to each other with respect to the line $y=x$. If the equation of the line $L$ is $y=ax+b$ with $a\neq 0$ and $b \neq 0$, then the equation of $K$ is $y=$ $\text{(A)}\ \frac 1ax+b \qquad \text{(B)}\ -\frac 1ax+b \qquad \text{(C)}\ \frac 1ax - \frac ba \qquad \text{(D)}\ \frac 1ax+\frac ba \qquad \text{(E)}\ \frac 1ax -\frac ba$

Find all functions $ f: \mathbb{R} \to \mathbb{R} $ such that $$ f\left(xf\left(y\right)-f\left(x\right)-y\right) = yf\left(x\right)-f\left(y\right)-x $$ holds for all $ x,y \in \mathbb{R} $