Inscribed in a triangle $ABC$ with the center of the circle $I$ touch the sides $AB$ and $AC$ at points $C_{1}$ and $B_{1}$, respectively. The point $M$ divides the segment $C_{1}B_{1}$ in a 3:1 ratio, measured from $C_{1}$. $N$ - the midpoint of $AC$. Prove that the points $I, M, B_{1}, N$ lie on a circle, if you know that $AC = 3 (BC-AB)$.

Let $x=\frac{\displaystyle\sum_{n=1}^{44} \cos n^\circ}{\displaystyle \sum_{n=1}^{44} \sin n^\circ}.$ What is the greatest integer that does not exceed $100x$?

Evaluate \[\sum_{n=1}^{\infty} \int_{2n\pi}^{2(n+1)\pi} \frac{x\sin x+\cos x}{x^2}\ dx\ (n=1,2,\cdots)\]

Prove that for every triangle the following inequality holds: \[\frac{ab+bc+ca}{4S} \geq \cot \frac{\pi}{6}.\] where $a, b, c$ are lengths of the sides and $S$ is the area of the triangle.

Evaluate $\displaystyle\sum_{n=0}^\infty \dfrac{\cos n\theta}{2^n}$, where $\cos\theta = \dfrac{1}{5}$.

Prove that for all real numbers $x$ of the interval $0 < x <\pi$ the inequality $$\sin x +\frac12 \sin 2x +\frac13 \sin 3x > 0$$ holds.

Evaluate $ \int_{\frac {\pi}{4}}^{\frac {3}{4}\pi } \cos \frac {1}{\sin \left(\frac {1}{\sin x}\right)}\cdot \cos \left(\frac {1}{\sin x}\right)\cdot \frac {\cos x}{\sin ^ 2 x\cdot \sin ^ 2 \left(\frac {1}{\sin x }\right)}\ dx$ Last Edited, Sorry kunny

If \[x \equal{} \frac { \minus{} 1 \plus{} i\sqrt3}{2}\qquad\text{and}\qquad y \equal{} \frac { \minus{} 1 \minus{} i\sqrt3}{2},\] where $ i^2 \equal{} \minus{} 1$, then which of the following is [i]not[/i] correct? $ \textbf{(A)}\ x^5 \plus{} y^5 \equal{} \minus{} 1 \qquad \textbf{(B)}\ x^7 \plus{} y^7 \equal{} \minus{} 1 \qquad \textbf{(C)}\ x^9 \plus{} y^9 \equal{} \minus{} 1$ $ \textbf{(D)}\ x^{11} \plus{} y^{11} \equal{} \minus{} 1 \qquad \textbf{(E)}\ x^{13} \plus{} y^{13} \equal{} \minus{} 1$

Let $ABCD$ be a convex quadrilateral. Assume line $AB$ and $CD$ intersect at $E$, and $B$ lies between $A$ and $E$. Assume line $AD$ and $BC$ intersect at $F$, and $D$ lies between $A$ and $F$. Assume the circumcircles of $\triangle BEC$ and $\triangle CFD$ intersect at $C$ and $P$. Prove that $\angle BAP=\angle CAD$ if and only if $BD\parallel EF$.

[asy] /*Using regular asymptote, this diagram would take 30 min to make. Using cse5, this takes 5 minutes. Conclusion? CSE5 IS THE BEST PACKAGE EVER CREATED!!!!*/ size(100); import cse5; pathpen=black; anglefontpen=black; pointpen=black; anglepen=black; dotfactor=3; pair A=(0,0),B=(0.5,0.5*sqrt(3)),C=(3,0),D=(1.7,0),EE; EE=(B+C)/2; D(MP("$A$",A,W)--MP("$B$",B,N)--MP("$C$",C,E)--cycle); D(MP("$E$",EE,N)--MP("$D$",D,S)); D(D);D(EE); MA("80^\circ",8,D,EE,C,0.1); MA("20^\circ",8,EE,C,D,0.3,2,shift(1,3)*C); draw(arc(shift(-0.1,0.05)*C,0.25,100,180),arrow =ArcArrow()); MA("100^\circ",8,A,B,C,0.1,0); MA("60^\circ",8,C,A,B,0.1,0); //Credit to TheMaskedMagician for the diagram [/asy] In $\triangle ABC$, $E$ is the midpoint of side $BC$ and $D$ is on side $AC$. If the length of $AC$ is $1$ and $\measuredangle BAC = 60^\circ$, $\measuredangle ABC = 100^\circ$, $\measuredangle ACB = 20^\circ$ and $\measuredangle DEC = 80^\circ$, then the area of $\triangle ABC$ plus twice the area of $\triangle CDE$ equals $\textbf{(A) }\frac{1}{4}\cos 10^\circ\qquad\textbf{(B) }\frac{\sqrt{3}}{8}\qquad\textbf{(C) }\frac{1}{4}\cos 40^\circ\qquad\textbf{(D) }\frac{1}{4}\cos 50^\circ\qquad\textbf{(E) }\frac{1}{8}$

Find all arcs $\theta$ such that $\frac{1}{\sin ^2 \theta}, \frac{1}{\cos ^2 \theta} $ are integer numbers and roots of equation $$x^2-ax+a=0.$$

The cosines of the angles of one triangle are respectively equal to the sines of the angles of the other triangle. Find the largest of these six angles of triangles.

Find the positive integer $N$ for which there exist reals $\alpha, \beta, \gamma, \theta$ which obey \begin{align*} 0.1 &= \sin \gamma \cos \theta \sin \alpha, \\ 0.2 &= \sin \gamma \sin \theta \cos \alpha, \\ 0.3 &= \cos \gamma \cos \theta \sin \beta, \\ 0.4 &= \cos \gamma \sin \theta \cos \beta, \\ 0.5 &\ge \left\lvert N-100 \cos2\theta \right\rvert. \end{align*}[i]Proposed by Evan Chen[/i]

Let $ ABCD$ be a rectangle with center $ O$, $ AB\neq BC$. The perpendicular from $ O$ to $ BD$ cuts the lines $ AB$ and $ BC$ in $ E$ and $ F$ respectively. Let $ M,N$ be the midpoints of the segments $ CD,AD$ respectively. Prove that $ FM \perp EN$.

Let $\alpha, \beta, \gamma$ be the angles of some triangle. Prove that there is a triangle whose sides are equal to $\sin \alpha$, $\sin \beta$, $\sin \gamma$.

Let $f(x)$ be a continuous function satisfying $f(x)=1+k\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} f(t)\sin (x-t)dt\ (k:constant\ number)$ Find the value of $k$ for which $\int_0^{\pi} f(x)dx$ is maximized.

Given a solid $R$ contained in a semi cylinder with the hight $1$ which has a semicircle with radius $1$ as the base. The cross section at the hight $x\ (0\leq x\leq 1)$ is the form combined with two right-angled triangles as attached figure as below. Answer the following questions. (1) Find the cross-sectional area $S(x)$ at the hight $x$. (2) Find the volume of $R$. If necessary, when you integrate, set $x=\sin t.$

Find the amplitude of $y = 4 \sin (x) + 3 \cos (x)$.

Prove that if $ 0 \leq \alpha < \frac{\pi}{2} $ and $ 0 \leq \beta < \frac{\pi}{2} $, then $$ tg \frac{\alpha + \beta}{2} \leq \frac{tg \alpha + tg \beta}{2}.$$

Suppose $P(x)$ is a polynomial with real coefficients, satisfying the condition $P(\cos \theta+\sin \theta)=P(\cos \theta-\sin \theta)$, for every real $\theta$. Prove that $P(x)$ can be expressed in the form$$P(x)=a_0+a_1(1-x^2)^2+a_2(1-x^2)^4+\dots+a_n(1-x^2)^{2n}$$for some real numbers $a_0, a_1, \dots, a_n$ and non-negative integer $n$. [i]Proposed by C.R. Pranesacher[/i]

In a triangle, a symmedian is a line through a vertex that is symmetric to the median with the respect to the internal bisector (all relative to the same vertex). In the triangle $ABC$, the median $m_a$ meets $BC$ at $A'$ and the circumcircle again at $A_1$. The symmedian $s_a$ meets $BC$ at $M$ and the circumcircle again at $A_2$. Given that the line $A_1A_2$ contains the circumcenter $O$ of the triangle, prove that: [i](a) [/i]$\frac{AA'}{AM} = \frac{b^2+c^2}{2bc} ;$ [i](b) [/i]$1+4b^2c^2 = a^2(b^2+c^2)$

Let $ABCD$ be a convex quadrilateral, with $\angle BAC=\angle DAC$ and $M$ a point inside such that $\angle MBA=\angle MCD$ and $\angle MBC=\angle MDC$. Show that the angle $\angle ADC$ is equal to $\angle BMC$ or $\angle AMB$.

In a cyclic quadrilateral $ ABCD, a,b,c,d$ are its side lengths, $ Q$ its area, and $ R$ its circumradius. Prove that: $ R^2\equal{}\frac{(ab\plus{}cd)(ac\plus{}bd)(ad\plus{}bc)}{16Q^2}$. Deduce that $ R \ge \frac{(abcd)^{\frac{3}{4}}}{Q\sqrt{2}}$ with equality if and only if $ ABCD$ is a square.

Let $ABCDEF$ be a convex hexagon in which diagonals $AD, BE, CF$ are concurrent at $O$. Suppose $[OAF]$ is geometric mean of $[OAB]$ and $[OEF]$ and $[OBC]$ is geometric mean of $[OAB]$ and $[OCD]$. Prove that $[OED]$ is the geometric mean of $[OCD]$ and $[OEF]$. (Here $[XYZ]$ denotes are of $\triangle XYZ$)

Let $ x,y,z$ be positive real numbers satisfying $ x\plus{}y\plus{}z\equal{}xyz$. Prove that: \[\frac1{\sqrt{1+x^2}}+\frac1{\sqrt{1+y^2}}+\frac1{\sqrt{1+z^2}}\leq\frac{3}{2}\]