Found problems: 4776
Find all positive integer solutions $(a, b, c)$ to the function $a^{2} + b^{2} + c^{2} = 2005$, where $a \leq b \leq c$.
For each positive integer $n\ge2$, find a polynomial $P_n(x)$ with rational coefficients such that $\displaystyle P_n(\sqrt[n]2)=\frac1{1+\sqrt[n]2}$. (Note that $\sqrt[n]2$ denotes the positive $n^\text{th}$ root of $2$.)
Let $ D \subset \mathbb {R} ^ {2} $ be a finite Lebesgue measure of a connected open set and $ u: D \rightarrow \mathbb {R} $ a harmonic function. Show that it is either a constant $ u $ or for almost every $ p \in D $
$$
f ^ {\prime} (t) = (\operatorname {grad} u) (f (t)), \quad f (0) = p
$$has no initial value problem(differentiable everywhere) solution to $ f:[0,\infty) \rightarrow D $.
Let $ a$, $ b$, $ c$ be positive real numbers such that $ a \plus{} b \plus{} c \equal{} 1$. Prove inequality:
\[ \frac{1}{bc \plus{} a \plus{} \frac{1}{a}} \plus{} \frac{1}{ac \plus{} b \plus{} \frac{1}{b}} \plus{} \frac{1}{ab \plus{} c \plus{} \frac{1}{c}} \leqslant \frac{27}{31}.\]
A positive proper divisor is a positive divisor of a number, excluding itself. For positive integers $n \ge 2$, let $f(n)$ denote the number that is one more than the largest proper divisor of $n$. Determine all positive integers $n$ such that $f(f(n)) = 2$.
Let $g$ be the multiplicative function given by $$g(p^{\alpha}) = \alpha p^{\alpha-1},$$ for all $\alpha\in\mathbb Z^+$ and primes $p$. Prove that there exist infinitely many integers $n$ such that $$g(n+1) = g(n) + g(1).$$
Find the derivative of the solution of the equation $\ddot{x} =x + A\dot{x}^2$, with initial conditions $x(0) = 1$, $\dot{x}(0) = 0$, with respect to the parameter $A$ for $A = 0$.
Show that there is no function $f : \mathbb N \to \mathbb N$ satisfying $f(f(n)) = n + 1$ for each positive integer $n.$
Let $f(x)$ be a continuous function defined on $0\leq x\leq \pi$ and satisfies $f(0)=1$ and
\[\left\{\int_0^{\pi} (\sin x+\cos x)f(x)dx\right\}^2=\pi \int_0^{\pi}\{f(x)\}^2dx.\]
Evaluate $\int_0^{\pi} \{f(x)\}^3dx.$
Examine the behavior of the expression
$ \sum_{\nu\equal{}1}^{n\minus{}1}\frac{\log(n\minus{}\nu)}{\nu}\minus{}\log^2 n$
as $ n\rightarrow \infty$
Find all $k>0$ such that there exists a function $f : [0,1]\times[0,1] \to [0,1]$ satisfying the following conditions:
$f(f(x,y),z)=f(x,f(y,z))$;
$f(x,y) = f(y,x)$;
$f(x,1)=x$;
$f(zx,zy) = z^{k}f(x,y)$, for any $x,y,z \in [0,1]$
Study the derivatives of the function
\[y=\sqrt{x^3-4x}\]
and sketch its graph on the real line.
Consider an integrable function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that $af(a)+bf(b)=0$ when $ab=1$. Find the value of the following integration:
$$ \int_{0}^{\infty} f(x) \,dx $$
Let $\mathbb R$ be the set of real numbers. Determine all functions $f:\mathbb R\to\mathbb R$ that satisfy the equation
$$\sum_{i=1}^{2015} f(x_i + x_{i+1}) + f\left( \sum_{i=1}^{2016} x_i \right) \le \sum_{i=1}^{2016} f(2x_i)$$
for all real numbers $x_1, x_2, ... , x_{2016}.$
Determine all real functions $f(x)$ that are defined and continuous on the interval $(-1, 1)$ and that satisfy the functional equation
\[f(x+y)=\frac{f(x)+f(y)}{1-f(x) f(y)} \qquad (x, y, x + y \in (-1, 1)).\]
Consider all functions $f:\mathbb{N}\to\mathbb{N}$ satisfying $f(t^2 f(s)) = s(f(t))^2$ for all $s$ and $t$ in $N$. Determine the least possible value of $f(1998)$.
A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k\geq 1$, the circles in $\textstyle\bigcup_{j=0}^{k-1} L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\textstyle\bigcup_{j=0}^6 L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is \[\sum_{C\in S}\dfrac1{\sqrt{r(C)}}?\]
[asy]
import olympiad;
size(350);
defaultpen(linewidth(0.7));
// define a bunch of arrays and starting points
pair[] coord = new pair[65];
int[] trav = {32,16,8,4,2,1};
coord[0] = (0,73^2); coord[64] = (2*73*70,70^2);
// draw the big circles and the bottom line
path arc1 = arc(coord[0],coord[0].y,260,360);
path arc2 = arc(coord[64],coord[64].y,175,280);
fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.78));
fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.78));
draw(arc1^^arc2);
draw((-930,0)--(70^2+73^2+850,0));
// We now apply the findCenter function 63 times to get
// the location of the centers of all 63 constructed circles.
// The complicated array setup ensures that all the circles
// will be taken in the right order
for(int i = 0;i<=5;i=i+1)
{
int skip = trav[i];
for(int k=skip;k<=64 - skip; k = k + 2*skip)
{
pair cent1 = coord[k-skip], cent2 = coord[k+skip];
real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2);
real shiftx = cent1.x + sqrt(4*r1*rn);
coord[k] = (shiftx,rn);
}
// Draw the remaining 63 circles
}
for(int i=1;i<=63;i=i+1)
{
filldraw(circle(coord[i],coord[i].y),gray(0.78));
}[/asy]
$\textbf{(A) }\dfrac{286}{35}\qquad\textbf{(B) }\dfrac{583}{70}\qquad\textbf{(C) }\dfrac{715}{73}\qquad\textbf{(D) }\dfrac{143}{14}\qquad\textbf{(E) }\dfrac{1573}{146}$
The altitude $CH$ of a right triangle $ABC$, with $\angle{C}=90$, cut the angles bisectors $AM$ and $BN$ at $P$ and $Q$, and let $R$ and $S$ be the midpoints of $PM$ and $QN$. Prove that $RS$ is parallel to the hypotenuse of $ABC$
In the plane, there are two circles $\Gamma_1, \Gamma_2$ intersecting each other at two points $A$ and $B$. Tangents of $\Gamma_1$ at $A$ and $B$ meet each other at $K$. Let us consider an arbitrary point $M$ (which is different of $A$ and $B$) on $\Gamma_1$. The line $MA$ meets $\Gamma_2$ again at $P$. The line $MK$ meets $\Gamma_1$ again at $C$. The line $CA$ meets $\Gamma_2 $ again at $Q$. Show that the midpoint of $PQ$ lies on the line $MC$ and the line $PQ$ passes through a fixed point when $M$ moves on $\Gamma_1$.
[color=red][Moderator edit: This problem was also discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=21414 .][/color]
Define a regular $n$-pointed star to be the union of $n$ line segments $P_1P_2, P_2P_3,\ldots, P_nP_1$ such that
$\bullet$ the points $P_1, P_2,\ldots, P_n$ are coplanar and no three of them are collinear,
$\bullet$ each of the $n$ line segments intersects at least one of the other line segments at a point other than an endpoint,
$\bullet$ all of the angles at $P_1, P_2,\ldots, P_n$ are congruent,
$\bullet$ all of the $n$ line segments $P_2P_3,\ldots, P_nP_1$ are congruent, and
$\bullet$ the path $P_1P_2, P_2P_3,\ldots, P_nP_1$ turns counterclockwise at an angle of less than 180 degrees at each vertex.
There are no regular 3-pointed, 4-pointed, or 6-pointed stars. All regular 5-pointed stars are similar, but there are two non-similar regular 7-pointed stars. How many non-similar regular 1000-pointed stars are there?
For a function $f(x)=\ln (1+\sqrt{1-x^2})-\sqrt{1-x^2}-\ln x\ (0<x<1)$, answer the following questions:
(1) Find $f'(x)$.
(2) Sketch the graph of $y=f(x)$.
(3) Let $P$ be a mobile point on the curve $y=f(x)$ and $Q$ be a point which is on the tangent at $P$ on the curve $y=f(x)$ and such that $PQ=1$. Note that the $x$-coordinate of $Q$ is les than that of $P$. Find the locus of $Q$.
Let $f : \mathbb{R}^+\to\mathbb{R}$ satisfying $f(x)=f(x+2)+2f(x^2+2x)$. Prove that if for all $x>1400^{2021}$, $xf(x) \le 2021$, then $xf(x) \le 2021$ for all $x \in \mathbb {R}^+$
Proposed by [i]Navid Safaei[/i]
Let $ \mathbb{Q}$ and $ \mathbb{R}$ be the set of rational numbers and the set of real numbers, respectively, and let $ f : \mathbb{Q} \rightarrow \mathbb{R}$ be a function with the following property. For every $ h \in \mathbb{Q} , \;x_0 \in \mathbb{R}$, \[ f(x\plus{}h)\minus{}f(x) \rightarrow 0\] as $ x \in \mathbb{Q}$ tends to $ x_0$. Does it follow that $ f$ is bounded on some interval?
[i]M. Laczkovich[/i]
Find all real solutions to the following system of equations. Carefully justify your answer.
\[ \left\{ \begin{array}{c} \displaystyle\frac{4x^2}{1+4x^2} = y \\ \\ \displaystyle\frac{4y^2}{1+4y^2} = z \\ \\ \displaystyle\frac{4z^2}{1+4z^2} = x \end{array} \right. \]
Let $n>2,n \in \mathbb{N}$ and $a>0,a \in \mathbb{R}$ such that $2^a + \log_2 a = n^2$. Prove that: \[ 2 \cdot \log_2 n>a>2 \cdot \log_2 n -\frac{1}{n} . \]
[i]Radu Gologan[/i]