$I$ is the incenter of triangle $ABC$. perpendicular from $I$ to $AI$ meet $AB$ and $AC$ at ${B}'$ and ${C}'$ respectively . Suppose that ${B}''$ and ${C}''$ are points on half-line $BC$ and $CB$ such that $B{B}''=BA$ and $C{C}''=CA$. Suppose that the second intersection of circumcircles of $A{B}'{B}''$ and $A{C}'{C}''$ is $T$. Prove that the circumcenter of $AIT$ is on the $BC$.

Let $AB=AC$ in $\triangle ABC$, and let $D$ be a point on segment $AB$. The tangent at $D$ to the circumcircle $\omega$ of $BCD$ hits $AC$ at $E$. The other tangent from $E$ to $\omega$ touches it at $F$, and $G=BF \cap CD$, $H=AG \cap BC$. Prove that $BH=2HC$. [i]Proposed by David Stoner[/i]

Show that $\frac{1}{\pi} \arccos \left( \frac{1}{\sqrt{2003}} \right)$ is irrational.

Evaluate $\int_0^{\frac{\pi}{4}} \frac{1}{1-\sin x}\sqrt{\frac{\cos x}{1+\cos x+\sin x}}dx.$ Own

Consider a composition of functions $\sin, \cos, \tan, \cot, \arcsin, \arccos, \arctan, \arccos$, applied to the number $1$. Each function may be applied arbitrarily many times and in any order. (ex: $\sin \cos \arcsin \cos \sin\cdots 1$). Can one obtain the number $2010$ in this way?

Given a regular tetrahedron $OABC$. Take points $P,\ Q,\ R$ on the sides $OA,\ OB,\ OC$ respectively. Note that $P,\ Q,\ R$ are different from the vertices of the tetrahedron $OABC$. If $\triangle{PQR}$ is an equilateral triangle, then prove that three sides $PQ,\ QR,\ RP$ are pararell to three sides $AB,\ BC,\ CA$ respectively. 30 points

Let $D$ be a point on the side $[AB]$ of the isosceles triangle $ABC$ such that $|AB|=|AC|$. The parallel line to $BC$ passing through $D$ intersects $AC$ at $E$. If $m(\widehat A) = 20^\circ$, $|DE|=1$, $|BC|=a$, and $|BE|=a+1$, then which of the followings is equal to $|AB|$? $ \textbf{(A)}\ 2a \qquad\textbf{(B)}\ a^2-a \qquad\textbf{(C)}\ a^2+1 \qquad\textbf{(D)}\ (a+1)^2 \qquad\textbf{(E)}\ a^2+a $

In a triangle $ABC$, points $X$ and $Y$ are on $BC$ and $CA$ respectively such that $CX=CY$,$AX$ is not perpendicular to $BC$ and $BY$ is not perpendicular to $CA$.Let $\Gamma$ be the circle with $C$ as centre and $CX$ as its radius.Find the angles of triangle $ABC$ given that the orthocentres of triangles $AXB$ and $AYB$ lie on $\Gamma$.

In trapezoid $ABCD$, $AD \parallel BC$ and $\angle ABC + \angle CDA = 270^{\circ}$. Compute $AB^2$ given that $AB \cdot \tan(\angle BCD) = 20$ and $CD = 13$. [i]Proposed by Lewis Chen[/i]

Let $ l_0,c,\alpha,g$ be positive constants, and let $ x(t)$ be the solution of the differential equation \[ ([l_0\plus{}ct^{\alpha}] ^2x')'\plus{}g[l_0\plus{}ct^{\alpha}] \sin x\equal{}0, \;t \geq 0,\ \;\minus{}\frac{\pi}{2} <x< \frac{\pi}{2},\] satisfying the initial conditions $ x(t_0)\equal{}x_0, \;x'(t_0)\equal{}0$. (This is the equation of the mathematical pendulum whose length changes according to the law $ l\equal{}l_0\plus{}ct^{\alpha}$.) Prove that $ x(t)$ is defined on the interval $ [t_0,\infty)$; furthermore, if $ \alpha >2$ then for every $ x_0 \not\equal{} 0$ there exists a $ t_0$ such that \[ \liminf_{t \rightarrow \infty} |x(t)| >0.\] [i]L. Hatvani[/i]

A circle is inscribed in quadrilateral $ABCD,$ tangent to $\overline{AB}$ at $P$ and to $\overline{CD}$ at $Q.$ Given that $AP=19, PB=26, CQ=37,$ and $QD=23,$ find the square of the radius of the circle.

For a real number $x$, let $f(x)=\int_0^{\frac{\pi}{2}} |\cos t-x\sin 2t|\ dt$. (1) Find the minimum value of $f(x)$. (2) Evaluate $\int_0^1 f(x)\ dx$. [i]2011 Tokyo Institute of Technology entrance exam, Problem 2[/i]

[b]i)[/b]If $X,Y,Z$ be the angles of a triangle then show that \[\tan {\frac{X}{2}}\tan {\frac{Y}{2}}+\tan {\frac{Y}{2}}\tan {\frac{Z}{2}}+\tan {\frac{Z}{2}}\tan {\frac{X}{2}}=1\] [b]ii)[/b] Prove using [b](i)[/b] or otherwise that \[\tan {\frac{X}{2}}\tan {\frac{Y}{2}}\tan {\frac{Z}{2}}\leq\frac {1}{3\sqrt{3}}\]

Find the number of ordered pairs of real numbers $ (a,b)$ such that $ (a \plus{} bi)^{2002} \equal{} a \minus{} bi$. $ \textbf{(A)}\ 1001\qquad \textbf{(B)}\ 1002\qquad \textbf{(C)}\ 2001\qquad \textbf{(D)}\ 2002\qquad \textbf{(E)}\ 2004$

A circle and a point $P$ higher than the circle lie in the same vertical plane. A particle moves along a straight line under gravity from $P$ to a point $Q$ on the circle. Given that the distance travelled from $P$ in time $t$ is equal to $\dfrac{1}{2}gt^2 \sin{\alpha}$, where $\alpha$ is the angle of inclination of the line $PQ$ to the horizontal, give a geometrical characterization of the point $Q$ for which the time taken from $P$ to $Q$ is a minimum.

In the triangle $ABC$, $\angle A$ is biggest. On the circumcircle of $\triangle ABC$, let $D$ be the midpoint of $\widehat{ABC}$ and $E$ be the midpoint of $\widehat{ACB}$. The circle $c_1$ passes through $A,B$ and is tangent to $AC$ at $A$, the circle $c_2$ passes through $A,E$ and is tangent $AD$ at $A$. $c_1$ and $c_2$ intersect at $A$ and $P$. Prove that $AP$ bisects $\angle BAC$. [hide="Diagram"][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(14.4cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -5.23, xmax = 9.18, ymin = -2.97, ymax = 4.82; /* image dimensions */ /* draw figures */ draw(circle((-1.32,1.36), 2.98)); draw(circle((3.56,1.53), 3.18)); draw((0.92,3.31)--(-2.72,-1.27)); draw(circle((0.08,0.25), 3.18)); draw((-2.72,-1.27)--(3.13,-0.65)); draw((3.13,-0.65)--(0.92,3.31)); draw((0.92,3.31)--(2.71,-1.54)); draw((-2.41,-1.74)--(0.92,3.31)); draw((0.92,3.31)--(1.05,-0.43)); /* dots and labels */ dot((-1.32,1.36),dotstyle); dot((0.92,3.31),dotstyle); label("$A$", (0.81,3.72), NE * labelscalefactor); label("$c_1$", (-2.81,3.53), NE * labelscalefactor); dot((3.56,1.53),dotstyle); label("$c_2$", (3.43,3.98), NE * labelscalefactor); dot((1.05,-0.43),dotstyle); label("$P$", (0.5,-0.43), NE * labelscalefactor); dot((-2.72,-1.27),dotstyle); label("$B$", (-3.02,-1.57), NE * labelscalefactor); dot((2.71,-1.54),dotstyle); label("$E$", (2.71,-1.86), NE * labelscalefactor); dot((3.13,-0.65),dotstyle); label("$C$", (3.39,-0.9), NE * labelscalefactor); dot((-2.41,-1.74),dotstyle); label("$D$", (-2.78,-2.07), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

In the triangle $ABC$ we have $\angle ABC=100^\circ$, $\angle ACB=65^\circ$, $M\in AB$, $N\in AC$, and $\angle MCB=55^\circ$, $\angle NBC=80^\circ$. Find $\angle NMC$. [i]St.Petersburg folklore[/i]

Construct a right triangle with given hypotenuse $c$ such that the median drawn to the hypotenuse is the geometric mean of the two legs of the triangle.

In convex quadrilateral $ABCD$, $\angle A \cong \angle C$, $AB = CD = 180$, and $AD \neq BC$. The perimeter of $ABCD$ is 640. Find $\lfloor 1000 \cos A \rfloor$. (The notation $\lfloor x \rfloor$ means the greatest integer that is less than or equal to $x$.)

If $ \theta$ is an acute angle and $ \sin \frac12 \theta \equal{} \sqrt{\frac{x\minus{}1}{2x}}$, then $ \tan \theta$ equals $ \textbf{(A)}\ x \qquad \textbf{(B)}\ \frac1{x} \qquad \textbf{(C)}\ \frac{\sqrt{x\minus{}1}}{x\plus{}1} \qquad \textbf{(D)}\ \frac{\sqrt{x^2\minus{}1}}{x} \qquad \textbf{(E)}\ \sqrt{x^2\minus{}1}$

Let $ Q$ be the centre of the inscribed circle of a triangle $ ABC.$ Prove that for any point $ P,$ \[ a(PA)^2 \plus{} b(PB)^2 \plus{} c(PC)^2 \equal{} a(QA)^2 \plus{} b(QB)^2 \plus{} c(QC)^2 \plus{} (a \plus{} b \plus{} c)(QP)^2, \] where $ a \equal{} BC, b \equal{} CA$ and $ c \equal{} AB.$

Let $O$ be the circumcenter and $H$ the orthocenter of an acute-angled triangle $ABC$ such that $BC>CA$. Let $F$ be the foot of the altitude $CH$ of triangle $ABC$. The perpendicular to the line $OF$ at the point $F$ intersects the line $AC$ at $P$. Prove that $\measuredangle FHP=\measuredangle BAC$.

In a triangle $ABC$, let $D$ and $E$ be the feet of the angle bisectors of angles $A$ and $B$, respectively. A rhombus is inscribed into the quadrilateral $AEDB$ (all vertices of the rhombus lie on different sides of $AEDB$). Let $\varphi$ be the non-obtuse angle of the rhombus. Prove that $\varphi \le \max \{ \angle BAC, \angle ABC \}$.

Consider a triangle $ABC$ and $I$ its incenter. The line $(AI)$ meets the circumcircle of $ABC$ in $D$. Let $E$ and $F$ be the orthogonal projections of $I$ on $(BD)$ and $(CD)$ respectively. Assume that $IE+IF=\frac{1}{2}AD$. Calculate $\angle{BAC}$. [color=red][Moderator edited: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=5088 .][/color]

In quadrilateral $ABCD$, $\angle DAC = 98^{\circ}$, $\angle DBC = 82^\circ$, $\angle BCD = 70^\circ$, and $BC = AD$. Find $\angle ACD.$