Found problems: 85335
Let $ABC$ be a triangle with $\angle A = 60^\circ$, and $AA', BB', CC'$ be its internal angle bisectors. Prove that $\angle B'A'C' \le 60^\circ$.
suppose that $v(x)=\sum_{p\le x,p\in \mathbb P}log(p)$ (here $\mathbb P$ denotes the set of all positive prime numbers). prove that the two statements below are equivalent:
[b]a)[/b] $v(x) \sim x$ when $x \longrightarrow \infty$
[b]b)[/b] $\pi (x) \sim \frac{x}{ln(x)}$ when $x \longrightarrow \infty$. (here $\pi (x)$ is number of the prime numbers less than or equal to $x$).
A triangle $ABC$ is given. A circle $\gamma$ centered at $A$ meets segments $AB$ and $AC$. The common chord of $\gamma$ and the circumcircle of $ABC$ meets $AB$ and $AC$ at $X$ and $Y$, respectively. The segments $CX$ and $BY$ meet $\gamma$ at point $S$ and $T$, respectively. The circumcircles of triangles $ACT$ and $BAS$ meet at points $A$ and $P$. Prove that $CX, BY$ and $AP$ concur.
How many positive perfect squares less than $2023$ are divisible by $5$?
$\textbf{(A) } 8 \qquad\textbf{(B) }9 \qquad\textbf{(C) }10 \qquad\textbf{(D) }11 \qquad\textbf{(E) } 12$
In rhombus $ABCD$, point $P$ lies on segment $\overline{AD}$ such that $BP\perp AD$, $AP = 3$, and $PD = 2$. What is the area of $ABCD$?
[asy]
import olympiad;
size(180);
real r = 3, s = 5, t = sqrt(r*r+s*s);
defaultpen(linewidth(0.6) + fontsize(10));
pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0);
draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D));
label("$A$",A,SW);
label("$B$", B, NW);
label("$C$",C,NE);
label("$D$",D,SE);
label("$P$",P,S);
[/asy]
$\textbf{(A) }3\sqrt 5 \qquad
\textbf{(B) }10 \qquad
\textbf{(C) }6\sqrt 5 \qquad
\textbf{(D) }20\qquad
\textbf{(E) }25$
Find all pairs of positive integers $(x, y)$ such that $(xy+1)(xy+x+2)$ be a perfect square .
Circles with centers $ A$ and $ B$ have radii 3 and 8, respectively. A common internal tangent intersects the circles at $ C$ and $ D$, respectively. Lines $ AB$ and $ CD$ intersect at $ E$, and $ AE \equal{} 5$. What is $ CD$?
[asy]unitsize(2.5mm);
defaultpen(fontsize(10pt)+linewidth(.8pt));
dotfactor=3;
pair A=(0,0), Ep=(5,0), B=(5+40/3,0);
pair M=midpoint(A--Ep);
pair C=intersectionpoints(Circle(M,2.5),Circle(A,3))[1];
pair D=B+8*dir(180+degrees(C));
dot(A);
dot(C);
dot(B);
dot(D);
draw(C--D);
draw(A--B);
draw(Circle(A,3));
draw(Circle(B,8));
label("$A$",A,W);
label("$B$",B,E);
label("$C$",C,SE);
label("$E$",Ep,SSE);
label("$D$",D,NW);[/asy]$ \textbf{(A) } 13\qquad \textbf{(B) } \frac {44}{3}\qquad \textbf{(C) } \sqrt {221}\qquad \textbf{(D) } \sqrt {255}\qquad \textbf{(E) } \frac {55}{3}$
Given a positive real $t$, a set $S$ of nonnegative reals is called $t$-good if for any two distinct elements $a,b$ in $S$, $\frac{a+b}2\ge\sqrt{ab}+t$. For all positive reals $N$, find the maximum number of elements a $t$-good set can have, if all elements are at most $N$.
[i](Proposed by Ho Janson)[/i]
Let be $x_1=\frac{1}{\sqrt[n+1]{n!}}$ and $x_2=\frac{1}{\sqrt[n+1]{(n-1)!}}$ for all $n\in \mathbb{N}^*$ and $f:\left(\left .\frac{1}{\sqrt[n+1]{(n+1)!}},1\right.\right] \to \mathbb{R}$ where $$f(x)=\frac{n+1}{x\ln (n+1)!+(n+1)\ln \left(x^x\right)}$$Prove that the sequence $(a_n)_{n\geq1}$ when $a_n=\int \limits_{x_1}^{x_2}f(x)dx$ is convergent and compute $$\lim \limits_{n \to \infty}a_n$$
An integer $n>1$ is given . Find the smallest positive number $m$ satisfying the following conditions: for any set $\{a,b\}$ $\subset \{1,2,\cdots,2n-1\}$ ,there are non-negative integers $ x, y$ ( not all zero) such that $2n|ax+by$ and $x+y\leq m.$
For nonzero real numbers $ r,\ l$ and the positive constant number $ c$, consider the curve on the $ xy$ plane : $ y \equal{} \left\{ \begin{array}{ll} x^2 & (0\leq x\leq r)\quad \\
r^2 & (r\leq x\leq l \plus{} r)\quad \\
(x \minus{} l \minus{} 2r)^2 & (l \plus{} r\leq x\leq l \plus{} 2r)\quad \end{array} \right.$
Denote $ V$ the volume of the solid by revolvering the curve about the $ x$ axis. Let $ r,\ l$ vary in such a way that $ r^2 \plus{} l \equal{} c$. Find the values of $ r,\ l$ which gives the maxmimum volume.
Suppose that a sequence $a_1,a_2,\ldots$ of positive real numbers satisfies \[a_{k+1}\geq\frac{ka_k}{a_k^2+(k-1)}\] for every positive integer $k$. Prove that $a_1+a_2+\ldots+a_n\geq n$ for every $n\geq2$.
How many moles of ions are present in $250$ mL of a $4.4$ M solution of sodium sulfate?
${ \textbf{(A)}\hspace{.05in}1.1 \qquad\textbf{(B)}\hspace{.05in}2.2 \qquad\textbf{(C)}\hspace{.05in}3.3 \qquad\textbf{(D)}\hspace{.05in}13}\qquad $
Find the number of positive integer solutions of $(x^2 + 2)(y^2 + 3)(z^2 + 4) = 60xyz$.
Let $ABC$ be a triangle with $\angle ABC$ obtuse. The [i]$A$-excircle[/i] is a circle in the exterior of $\triangle ABC$ that is tangent to side $BC$ of the triangle and tangent to the extensions of the other two sides. Let $E$, $F$ be the feet of the altitudes from $B$ and $C$ to lines $AC$ and $AB$, respectively. Can line $EF$ be tangent to the $A$-excircle?
[i]Proposed by Ankan Bhattacharya, Zack Chroman, and Anant Mudgal[/i]
A magician intends to perform the following trick. She announces a positive integer $n$, along with $2n$ real numbers $x_1 < \dots < x_{2n}$, to the audience. A member of the audience then secretly chooses a polynomial $P(x)$ of degree $n$ with real coefficients, computes the $2n$ values $P(x_1), \dots , P(x_{2n})$, and writes down these $2n$ values on the blackboard in non-decreasing order. After that the magician announces the secret polynomial to the audience. Can the magician find a strategy to perform such a trick?
Find all postitive integers n such that
$$\left\lfloor \frac{n}{2} \right\rfloor \cdot \left\lfloor \frac{n}{3} \right\rfloor \cdot \left\lfloor \frac{n}{4} \right\rfloor=n^2$$
where $\lfloor x \rfloor$ represents the largest integer less than the real number $x$.
A bug travels from $A$ to $B$ along the segments in the hexagonal lattice pictured below. The segments marked with an arrow can be traveled only in the direction of the arrow, and the bug never travels the same segment more than once. How many different paths are there?
[asy]
size(10cm);
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));
draw((0.0,0.0)--(1.0,1.7320508075688772)--(3.0,1.7320508075688772)--(4.0,3.4641016151377544)--(6.0,3.4641016151377544)--(7.0,5.196152422706632)--(9.0,5.196152422706632)--(10.0,6.928203230275509)--(12.0,6.928203230275509));
draw((3.0,-1.7320508075688772)--(4.0,0.0)--(6.0,0.0)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,3.4641016151377544)--(12.0,3.464101615137755)--(13.0,5.196152422706632)--(15.0,5.196152422706632));
draw((6.0,-3.4641016151377544)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,0.0)--(12.0,0.0)--(13.0,1.7320508075688772)--(15.0,1.7320508075688776)--(16.0,3.464101615137755)--(18.0,3.4641016151377544));
draw((9.0,-5.196152422706632)--(10.0,-3.464101615137755)--(12.0,-3.464101615137755)--(13.0,-1.7320508075688776)--(15.0,-1.7320508075688776)--(16.0,0)--(18.0,0.0)--(19.0,1.7320508075688772)--(21.0,1.7320508075688767));
draw((12.0,-6.928203230275509)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544)--(19.0,-1.7320508075688772)--(21.0,-1.7320508075688767)--(22.0,0));
draw((0.0,-0.0)--(1.0,-1.7320508075688772)--(3.0,-1.7320508075688772)--(4.0,-3.4641016151377544)--(6.0,-3.4641016151377544)--(7.0,-5.196152422706632)--(9.0,-5.196152422706632)--(10.0,-6.928203230275509)--(12.0,-6.928203230275509));
draw((3.0,1.7320508075688772)--(4.0,-0.0)--(6.0,-0.0)--(7.0,-1.7320508075688772)--(9.0,-1.7320508075688772)--(10.0,-3.4641016151377544)--(12.0,-3.464101615137755)--(13.0,-5.196152422706632)--(15.0,-5.196152422706632));
draw((6.0,3.4641016151377544)--(7.0,1.7320508075688772)--(9.0,1.7320508075688772)--(10.0,-0.0)--(12.0,-0.0)--(13.0,-1.7320508075688772)--(15.0,-1.7320508075688776)--(16.0,-3.464101615137755)--(18.0,-3.4641016151377544));
draw((9.0,5.1961524)--(10.0,3.464101)--(12.0,3.46410)--(13.0,1.73205)--(15.0,1.732050)--(16.0,0)--(18.0,-0.0)--(19.0,-1.7320)--(21.0,-1.73205080));
draw((12.0,6.928203)--(13.0,5.1961524)--(15.0,5.1961524)--(16.0,3.464101615)--(18.0,3.4641016)--(19.0,1.7320508)--(21.0,1.732050)--(22.0,0));
dot((0,0));
dot((22,0));
label("$A$",(0,0),WNW);
label("$B$",(22,0),E);
filldraw((2.0,1.7320508075688772)--(1.6,1.2320508075688772)--(1.75,1.7320508075688772)--(1.6,2.232050807568877)--cycle,black);
filldraw((5.0,3.4641016151377544)--(4.6,2.9641016151377544)--(4.75,3.4641016151377544)--(4.6,3.9641016151377544)--cycle,black);
filldraw((8.0,5.196152422706632)--(7.6,4.696152422706632)--(7.75,5.196152422706632)--(7.6,5.696152422706632)--cycle,black);
filldraw((11.0,6.928203230275509)--(10.6,6.428203230275509)--(10.75,6.928203230275509)--(10.6,7.428203230275509)--cycle,black);
filldraw((4.6,0.0)--(5.0,-0.5)--(4.85,0.0)--(5.0,0.5)--cycle,white);
filldraw((8.0,1.732050)--(7.6,1.2320)--(7.75,1.73205)--(7.6,2.2320)--cycle,black);
filldraw((11.0,3.4641016)--(10.6,2.9641016)--(10.75,3.46410161)--(10.6,3.964101)--cycle,black);
filldraw((14.0,5.196152422706632)--(13.6,4.696152422706632)--(13.75,5.196152422706632)--(13.6,5.696152422706632)--cycle,black);
filldraw((8.0,-1.732050)--(7.6,-2.232050)--(7.75,-1.7320508)--(7.6,-1.2320)--cycle,black);
filldraw((10.6,0.0)--(11,-0.5)--(10.85,0.0)--(11,0.5)--cycle,white);
filldraw((14.0,1.7320508075688772)--(13.6,1.2320508075688772)--(13.75,1.7320508075688772)--(13.6,2.232050807568877)--cycle,black);
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filldraw((11.0,-3.464101615137755)--(10.6,-3.964101615137755)--(10.75,-3.464101615137755)--(10.6,-2.964101615137755)--cycle,black);
filldraw((14.0,-1.7320508075688776)--(13.6,-2.2320508075688776)--(13.75,-1.7320508075688776)--(13.6,-1.2320508075688776)--cycle,black);
filldraw((16.6,0)--(17,-0.5)--(16.85,0)--(17,0.5)--cycle,white);
filldraw((20.0,1.7320508075688772)--(19.6,1.2320508075688772)--(19.75,1.7320508075688772)--(19.6,2.232050807568877)--cycle,black);
filldraw((14.0,-5.196152422706632)--(13.6,-5.696152422706632)--(13.75,-5.196152422706632)--(13.6,-4.696152422706632)--cycle,black);
filldraw((17.0,-3.464101615137755)--(16.6,-3.964101615137755)--(16.75,-3.464101615137755)--(16.6,-2.964101615137755)--cycle,black);
filldraw((20.0,-1.7320508075688772)--(19.6,-2.232050807568877)--(19.75,-1.7320508075688772)--(19.6,-1.2320508075688772)--cycle,black);
filldraw((2.0,-1.7320508075688772)--(1.6,-1.2320508075688772)--(1.75,-1.7320508075688772)--(1.6,-2.232050807568877)--cycle,black);
filldraw((5.0,-3.4641016)--(4.6,-2.964101)--(4.75,-3.4641)--(4.6,-3.9641016)--cycle,black);
filldraw((8.0,-5.1961524)--(7.6,-4.6961524)--(7.75,-5.19615242)--(7.6,-5.696152422)--cycle,black);
filldraw((11.0,-6.9282032)--(10.6,-6.4282032)--(10.75,-6.928203)--(10.6,-7.428203)--cycle,black);[/asy]
$ \textbf{(A)}\ 2112\qquad\textbf{(B)}\ 2304\qquad\textbf{(C)}\ 2368\qquad\textbf{(D)}\ 2384\qquad\textbf{(E)}\ 2400 $
The least positive angle $\alpha$ for which $$\left(\frac34-\sin^2(\alpha)\right)\left(\frac34-\sin^2(3\alpha)\right)\left(\frac34-\sin^2(3^2\alpha)\right)\left(\frac34-\sin^2(3^3\alpha)\right)=\frac1{256}$$ has a degree measure of $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Two points $A$ and $B$ are given on the plane. An arbitrary circle passes through $B$ and intersects the straight line $AB$ for second time at a point $K$, different from $A$. A circle passing through $A$, $K$ and the center of the first circle intersects the first one for second time at point $M$. Find the locus of points $M$.
The sequence $(p_n)_{n\in\mathbb N}$ is defined by $p_1=2$ and, for $n\ge2$, $p_n$ is the largest prime factor of $p_1p_2\cdots p_{n-1}+1$. Show that $p_n\ne5$ for all $n$.
Real numbers $x_1, x_2, x_3, \cdots , x_n$ satisfy $x_1^2 + x_2^2 + x_3^2 + \cdots + x_n^2 = 1$. Show that \[ \frac{x_1}{1+x_1^2}+\frac{x_2}{1+x_1^2+x_2^2}+\cdots+\frac{x_n}{1+ x_1^2 + x_2^2 + x_3^2 + \cdots + x_n^2} < \sqrt{\frac n2} . \]
A square of side length $1$ centimeter is cut into three convex polygons. Is it possible that the diameter of each of them does not exceed
[list][b]a)[/b] $1$ centimeter;
[b]b)[/b] $1.01$ centimeters;
[b]c)[/b] $1.001$ centimeters?[/list]
Triangle $\vartriangle ABC$ has side lengths $AB = 2$, $CA = 3$ and $BC = 4$. Compute the radius of the circle centered on $BC$ that is tangent to both $AB$ and $AC$.
Evaluate the following expression: $$0 - 1 -2 + 3 - 4 + 5 + 6 + 7 - 8 + ... + 2000$$ The terms with minus signs are exactly the powers of two.