Find $\lim_{n\to\infty}\sum_{k=0}^{\infty} \int_{2k\pi}^{(2k+1)\pi} xe^{-x}\sin x\ dx.$

Triangle $ ABC$ obeys $ AB = 2AC$ and $ \angle{BAC} = 120^{\circ}.$ Points $ P$ and $ Q$ lie on segment $ BC$ such that \begin{eqnarray*} AB^2 + BC \cdot CP = BC^2 \\ 3AC^2 + 2BC \cdot CQ = BC^2 \end{eqnarray*} Find $ \angle{PAQ}$ in degrees.

Squares $CAKL$ and $CBMN$ are constructed on the sides of acute-angled triangle $ABC$, outside of the triangle. Line $CN$ intersects line segment $AK$ at $X$, while line $CL$ intersects line segment $BM$ at $Y$. Point $P$, lying inside triangle $ABC$, is an intersection of the circumcircles of triangles $KXN$ and $LYM$. Point $S$ is the midpoint of $AB$. Prove that angle $\angle ACS=\angle BCP$.

Find all real solutions of the equation $x^2 + 2x \sin (xy) + 1 = 0$.

If $\tan x=\dfrac{2ab}{a^2-b^2}$ where $a>b>0$ and $0^\circ <x<90^\circ$, then $\sin x$ is equal to $\textbf{(A) }\frac{a}{b}\qquad\textbf{(B) }\frac{b}{a}\qquad\textbf{(C) }\frac{\sqrt{a^2-b^2}}{2a}\qquad\textbf{(D) }\frac{\sqrt{a^2-b^2}}{2ab}\qquad \textbf{(E) }\dfrac{2ab}{a^2+b^2}$

The sequence $ \{x_{n}\}_{n \ge 1}$ is defined by \[ x_{1} \equal{} 2, x_{n \plus{} 1} \equal{} \frac {2 \plus{} x_{n}}{1 \minus{} 2x_{n}}\;\; (n \in \mathbb{N}). \] Prove that a) $ x_{n}\not \equal{} 0$ for all $ n \in \mathbb{N}$, b) $ \{x_{n}\}_{n \ge 1}$ is not periodic.

The vertex $B$ of the triangle $ABC$ lies in the plane $P$. The plane of the triangle meets the plane in a line $L$. The angle between $L$ and $AB$ is a, and the angle between $L$ and $BC$ is $b$. The angle between the two planes is $c$. Angle $ABC$ is $90^o$. Show that $\sin^2c = \sin^2a + \sin^2b$. [img]https://cdn.artofproblemsolving.com/attachments/9/e/c0608e5408fd27a5f907a3488cce7dc2af6953.png[/img]

In the isosceles triangle $ ABC$ the angle $ BAC$ is a right angle. Point $ D$ lies on the side $ BC$ and satisfies $ BD \equal{} 2 \cdot CD$. Point $ E$ is the foot of the perpendicular of the point $ B$ on the line $ AD$. Find the angle $ CED$.

For each pair of even natural numbers $ k $, $ m $determine all real numbers $ x $that satisfy the equation $$ (\sin x)^k + (\cos x)^{-m} = (\cos x)^k + (\sin x)^{-m}$$

Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP>CP$. Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB=12$ and $\angle O_1 P O_2 = 120^\circ$, then $AP=\sqrt{a}+\sqrt{b}$ where $a$ and $b$ are positive integers. Find $a+b$.

Let $a, b, c$ be sides of a triangle whose perimeter does not exceed $2 \cdot \pi.$, Prove that $\sin a, \sin b, \sin c$ are sides of a triangle.

[b]Evaluate[/b] $\int^{\pi/2}_0 \frac{\cos^4x + \sin x \cos^3 x + \sin^2x\cos^2x + \sin^3x\cos x}{\sin^4x + \cos^4x + 2\ sinx\cos^3x + 2\sin^2x\cos^2x + 2\sin^3x\cos x} dx$

Let $ABC$ be an acute-angled triangle with $AB\not= AC$. Let $\Gamma$ be the circumcircle, $H$ the orthocentre and $O$ the centre of $\Gamma$. $M$ is the midpoint of $BC$. The line $AM$ meets $\Gamma$ again at $N$ and the circle with diameter $AM$ crosses $\Gamma$ again at $P$. Prove that the lines $AP,BC,OH$ are concurrent if and only if $AH=HN$.

In $\triangle A B C$, let $D, E$, and $F$ be the midpoints of the sides of the triangle, and let $P, Q,$ and $R$ be the midpoints of the corresponding medians, $AD ,B E,$ and $C F$, respectively, as shown in the figure at the right. Prove that the value of \[\frac{AQ^2 + A R^2 + B P^2 + B R^2 + C P^2+ C Q^2 }{A B^2 + B C^2 + C A^2}\] does not depend on the shape of $\triangle A B C$ and find that value. [asy] defaultpen(linewidth(0.7)+fontsize(10));size(200); pair A=origin, B=(14,0), C=(9,12), D=midpoint(C--B), E=midpoint(C--A), F=midpoint(A--B), R=midpoint(C--F), P=midpoint(D--A), Q=midpoint(E--B); draw(A--B--C--A, linewidth(1)); draw(A--D^^B--E^^C--F); draw(B--R--A--Q--C--P--cycle, dashed); pair point=centroid(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$P$", P, dir(40)*dir(point--P)); label("$Q$", Q, dir(40)*dir(point--Q)); label("$R$", R, dir(40)*dir(point--R)); dot(P^^Q^^R);[/asy]

The function $f$ has the property that, for each real number $x,$ \[ f(x)+f(x-1) = x^2. \] If $f(19)=94,$ what is the remainder when $f(94)$ is divided by 1000?

Let $ABCD$ be a cyclic quadrilateral who interior angle at $B$ is $60$ degrees. Show that if $BC=CD$, then $CD+DA=AB$. Does the converse hold?

Find the maximum value of the following integral. \[\int_0^{\infty} e^{-x}\sin tx\ dx\]

Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.

Let $a_n = \frac{\pi}{2n}$, where $n$ is a natural number. Prove that for any $k = 1$,$2$,$...$, $n$ holds the equality $$\frac{\sin ka_n}{1-\cos a_n}+\frac{\sin 5ka_n}{1-\cos 5a_n}+\frac{\sin 9ka_n}{1-\cos 9a_n}+...+\frac{\sin (4n-3)a_n}{1-\cos (4n-3)a_n}=kn$$

[asy] size(5cm); pen p=linewidth(3), dark_grey=gray(0.25), ll_grey=gray(0.90), light_grey=gray(0.75); transform dishift(real x) { return shift(x,x); } // Draw the table of latch of table path ell = ((0,0)--(0,-1)--(-0.1,-1)--(-0.1,-0.1)--(-1,-0.1)--(-1,0)--cycle); // the ell shape path corner = dishift(-0.85)*ell; // define the path path table = dishift(-1)*scale(5)*ell; // define the table by scaling the pulley filldraw(corner, ll_grey, light_grey+p); // base of pulley filldraw(table, ll_grey, grey+p); // table real block_size = 1.6; // template for block path block = unitsquare; pair block_center = (0.5,0.5); /* Resting block */ transform rest = shift(-5, -0.9) * scale(block_size); // transformation for resting block filldraw(rest * block, ll_grey, light_grey+p); // draw block draw(rest*(1,0.5)--dir(110), light_grey+p); // rope fr0m midpoint of right block to wheel label("$m$", rest * block_center, fontsize(16)); // label block /* Hanging block */ transform hang = shift(0.2,-4.1) * scale(block_size); // transformation for hanging block draw((1,0)--(1,-2.5), light_grey+p); // string of pulley filldraw(hang * block, ll_grey, light_grey+p); // fill it label("$M$",hang * block_center,fontsize(16)); // label the small m // Draws the actual pulley filldraw(unitcircle, grey, p); // outer boundary of pulley wheel filldraw(scale(0.4)*unitcircle, light_grey, p); // inner boundary of pulley wheel path pulley_body=arc((0,0),0.3,-40,130)--arc((-1,-1),0.5,130,320)--cycle; // defines "arm" of pulley filldraw(pulley_body, ll_grey, dark_grey+p); // draws the arm filldraw(scale(0.18)*unitcircle, ll_grey, dark_grey+p); // inner circle of pulley [/asy] A pulley system of two blocks, shown above, is released from rest. The block on the table, which has mass $m=1.0 \, \text{kg}$ slides after the time of release and hits the pulley to come to a dead stop. There was originally a distance of $ 1.0 \, \text{m} $ between the block and the pulley, which the block fully covers during the slide. From the time of release to the time of hitting the pulley, the angle that the rope above the table makes with the horizontal axis is a, nearly constant, $10.0^\circ$. The hanging block has mass $ M = 2.0 \, \text{kg} $. The table has a coefficient of friction of $0.50$ with the block that sits on it. The pulley is frictionless. Also, assume that, during the entire slide, the block never leaves the ground. Let $t$ be the number of seconds in takes for the $1.0\text{-m}$ slide. Find $100t$, rounded to two significant figures. [i](Ahaan Rungta, 4 points)[/i]

Evaluate $50(\cos 39^{\circ}\cos21^{\circ}+\cos129^{\circ}\cos69^{\circ})$

The diagonals of convex quadrilateral $BSCT$ meet at the midpoint $M$ of $\overline{ST}$. Lines $BT$ and $SC$ meet at $A$, and $AB = 91$, $BC = 98$, $CA = 105$. Given that $\overline{AM} \perp \overline{BC}$, find the positive difference between the areas of $\triangle SMC$ and $\triangle BMT$. [i]Proposed by Evan Chen[/i]

Let $ABC$ be a triangle and $M,N,P$ be the midpoints of sides $BC, CA, AB$. The lines $AM, BN, CP$ meet the circumcircle of $ABC$ in the points $A_{1}, B_{1}, C_{1}$. Show that the area of triangle $ABC$ is at most the sum of areas of triangles $BCA_{1}, CAB_{1}, ABC_{1}$.

In an acute-angled triangle $ABC$, we consider the feet $H_a$ and $H_b$ of the altitudes from $A$ and $B$, and the intersections $W_a$ and $W_b$ of the angle bisectors from $A$ and $B$ with the opposite sides $BC$ and $CA$ respectively. Show that the centre of the incircle $I$ of triangle $ABC$ lies on the segment $H_aH_b$ if and only if the centre of the circumcircle $O$ of triangle $ABC$ lies on the segment $W_aW_b$.

For any quadratic functions $ f(x)$ such that $ f'(2)\equal{}1$, evaluate $ \int_{2\minus{}\pi}^{2\plus{}\pi}f(x)\sin\left(\frac{x}{2}\minus{}1\right) dx$.