Let $ ABC$ be an isosceles triangle with $ AB \equal{} AC$. $ M$ is the midpoint of $ BC$ and $ O$ is the point on the line $ AM$ such that $ OB$ is perpendicular to $ AB$. $ Q$ is an arbitrary point on $ BC$ different from $ B$ and $ C$. $ E$ lies on the line $ AB$ and $ F$ lies on the line $ AC$ such that $ E, Q, F$ are distinct and collinear. Prove that $ OQ$ is perpendicular to $ EF$ if and only if $ QE \equal{} QF$.

A circle of radius 2 is centered at $ O$. Square $ OABC$ has side length 1. Sides $ \overline{AB}$ and $ \overline{CB}$ are extended past $ b$ to meet the circle at $ D$ and $ E$, respectively. What is the area of the shaded region in the figure, which is bounded by $ \overline{BD}$, $ \overline{BE}$, and the minor arc connecting $ D$ and $ E$? [asy] defaultpen(linewidth(0.8)); pair O=origin, A=(1,0), C=(0,1), B=(1,1), D=(1, sqrt(3)), E=(sqrt(3), 1), point=B; fill(Arc(O, 2, 0, 90)--O--cycle, mediumgray); clip(B--Arc(O, 2, 30, 60)--cycle); draw(Circle(origin, 2)); draw((-2,0)--(2,0)^^(0,-2)--(0,2)); draw(A--D^^C--E); label("$A$", A, dir(point--A)); label("$C$", C, dir(point--C)); label("$O$", O, dir(point--O)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$B$", B, SW);[/asy] $ \textbf{(A) } \frac {\pi}3 \plus{} 1 \minus{} \sqrt {3} \qquad \textbf{(B) } \frac {\pi}2\left( 2 \minus{} \sqrt {3}\right) \qquad \textbf{(C) } \pi\left(2 \minus{} \sqrt {3}\right) \qquad \textbf{(D) } \frac {\pi}{6} \plus{} \frac {\sqrt {3} \minus{} 1}{2} \\ \qquad \indent \textbf{(E) } \frac {\pi}{3} \minus{} 1 \plus{} \sqrt {3}$

In quadrilateral $ABCD$, $BC=8$, $CD=12$, $AD=10$, and $m\angle A= m\angle B = 60^\circ$. Given that $AB=p + \sqrt{q}$, where $p$ and $q$ are positive integers, find $p+q$.

Let $ ABC$ be a triangle, let $ AB > AC$. Its incircle touches side $ BC$ at point $ E$. Point $ D$ is the second intersection of the incircle with segment $ AE$ (different from $ E$). Point $ F$ (different from $ E$) is taken on segment $ AE$ such that $ CE \equal{} CF$. The ray $ CF$ meets $ BD$ at point $ G$. Show that $ CF \equal{} FG$.

Triangle $ABC$ is a right triangle with $AC=7,$ $BC=24,$ and right angle at $C.$ Point $M$ is the midpoint of $AB,$ and $D$ is on the same side of line $AB$ as $C$ so that $AD=BD=15.$ Given that the area of triangle $CDM$ may be expressed as $\frac{m\sqrt{n}}{p},$ where $m,$ $n,$ and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime, find $m+n+p.$

Let $ y$ be the function of $ x$ satisfying the differential equation $ y'' \minus{} y \equal{} 2\sin x$. (1) Let $ y \equal{} e^xu \minus{} \sin x$, find the differential equation with which the function $ u$ with respect to $ x$ satisfies. (2) If $ y(0) \equal{} 3,\ y'(0) \equal{} 0$, then determine $ y$.

Evaluate $\lim_{n\to \infty} cos\frac{x}{2}cos\frac{x}{2^2} cos\frac{x}{2^3}...cos\frac{x}{2^n}$

Points $ K$, $ L$, $ M$, and $ N$ lie in the plane of the square $ ABCD$ so that $ AKB$, $ BLC$, $ CMD$, and $ DNA$ are equilateral triangles. If $ ABCD$ has an area of $ 16$, find the area of $ KLMN$. [asy]unitsize(2cm); defaultpen(fontsize(8)+linewidth(0.8)); pair A=(-0.5,0.5), B=(0.5,0.5), C=(0.5,-0.5), D=(-0.5,-0.5); pair K=(0,1.366), L=(1.366,0), M=(0,-1.366), N=(-1.366,0); draw(A--N--K--A--B--K--L--B--C--L--M--C--D--M--N--D--A); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$K$",K,NNW); label("$L$",L,E); label("$M$",M,S); label("$N$",N,W);[/asy] $ \textbf{(A)}\ 32 \qquad \textbf{(B)}\ 16 \plus{} 16\sqrt {3} \qquad \textbf{(C)}\ 48 \qquad \textbf{(D)}\ 32 \plus{} 16\sqrt {3} \qquad \textbf{(E)}\ 64$

In triangle $ABC$, $\angle ABC = 120^{\circ}$, $AB = 3$ and $BC = 4$. If perpendiculars constructed to $\overline{AB}$ at $A$ and to $\overline{BC}$ at $C$ meet at $D$, then $CD = $ $ \textbf{(A)}\ 3\qquad\textbf{(B)}\ \frac{8}{\sqrt{3}}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ \frac{11}{2}\qquad\textbf{(E)}\ \frac{10}{\sqrt{3}} $

Let $ABCD$ be a trapezoid (quadrilateral with one pair of parallel sides) such that $AB < CD$. Suppose that $AC$ and $BD$ meet at $E$ and $AD$ and $BC$ meet at $F$. Construct the parallelograms $AEDK$ and $BECL$. Prove that $EF$ passes through the midpoint of the segment $KL$.

Solve for $x \in [0,2\pi[$: \[\sin x < \cos x < \tan x < \cot x\]

Prove that the inequality $$ \frac{1}{3} \leq \frac{\tan 3\alpha}{\tan \alpha} \leq 3 $$ is not true for any value of $ \alpha $.

Let $f(x)$ be a continuous function, whose first and second derivatives are continuous on $[0,2\pi]$ and $f''(x) \geq 0$ for all $x$ in $[0,2\pi]$. Show that \[\int_{0}^{2\pi} f(x)\cos x dx \geq 0\]

Let $N_{14}$ be the answer to problem 14. Rectangle $ABCD$ has area $\sqrt{2N_{14}}$. Points $E$, $F$, $G$, and $H$ lie on the rays $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CD}$, and $\overrightarrow{DA}$, respectively, such that $EFGH$ is a rectangle with area $2\sqrt{2N_{14}}$ that contains all of $ABCD$ in its interior. If \[ \tan\angle AEH = \tan\angle BFE = \tan\angle CGF = \tan\angle DHG = \sqrt{\frac{1}{48}}, \] then $EG=\tfrac{m\sqrt{n}}{p}$, where $m$, $n$, and $p$ are positive integers, $m$ and $p$ are relatively prime, and $n$ is not divisible by the square of any prime. Compute $m + n + p$.

Find the minimum value of the following function $f(x) $ defined at $0<x<\frac{\pi}{2}$. \[f(x)=\int_0^x \frac{d\theta}{\cos \theta}+\int_x^{\frac{\pi}{2}} \frac{d\theta}{\sin \theta}\]

Let $s$ be any arc of the unit circle lying entirely in the first quadrant. Let $A$ be the area of the region lying below $s$ and above the $x$-axis and let $B$ be the area of the region lying to the right of the $y$-axis and to the left of $s$. Prove that $A+B$ depends only on the arc length, and not on the position, of $s$.

Find the minimum value of $a\ (0<a<1)$ for which the following definite integral is minimized. \[ \int_0^{\pi} |\sin x-ax|\ dx \]

Prove that if for angles $A,B,C$ of a triangle holds $$\sin^2 A+\sin^2 B +\sin^2 C=2$$ iff the triangle $ABC$ is right.

Let $A_1, A_2, \ldots, A_n$ be an $n$ -sided regular polygon. If $\frac{1}{A_1 A_2} = \frac{1}{A_1 A_3} + \frac{1}{A_1A_4}$, find $n$.

Let $ABC$ be a triangle, let $O$ be its circumcenter, and let $H$ be its orthocenter. Let $P$ be a point on the segment $OH$. Prove that $6r\leq PA+PB+PC\leq 3R$, where $r$ is the inradius and $R$ the circumradius of triangle $ABC$. [b]Moderator edit:[/b] This is true only if the point $P$ lies inside the triangle $ABC$. (Of course, this is always fulfilled if triangle $ABC$ is acute-angled, since in this case the segment $OH$ completely lies inside the triangle $ABC$; but if triangle $ABC$ is obtuse-angled, then the condition about $P$ lying inside the triangle $ABC$ is really necessary.)

$ABCD$ is a cyclic quadrilateral with circumcenter $O$ and circumradius $7$. $AB$ intersects $CD$ at $E$, $DA$ intersects $CB$ at $F$. $OE=13$, $OF=14$. Let $\cos\angle FOE=\dfrac pq$, with $p$, $q$ coprime. Find $p+q$.

Let $180^o < \theta_1 < \theta_2 <...< \theta_n = 360^o$. For $i = 1,2,..., n$, $P_i = (\cos \theta_i^o, \sin \theta_i^o)$ is a point on the circle $C$ with centre $(0,0)$ and radius $1$. Let $P$ be any point on the upper half of $C$. Find the coordinates of $P$ such that the sum of areas $[PP_1P_2] + [PP_2P_3] + ...+ [PP_{n-1}P_n]$ attains its maximum.

$D, \: E , \: F$ are points on the sides $AB, \: BC, \: CA,$ respectively, of a triangle $ABC$ such that $AD=AF, \: BD=BE,$ and $DE=DF.$ Let $I$ be the incenter of the triangle $ABC,$ and let $K$ be the point of intersection of the line $BI$ and the tangent line through $A$ to the circumcircle of the triangle $ABI.$ Show that $AK=EK$ if $AK=AD.$

Prove that for all $x_1, x_2,\ldots , x_n \in \mathbb R$ the following inequality holds: \[\sum_{n \geq i >j \geq 1} \cos^2(x_i - x_j ) \geq \frac{n(n-2)}{4}\]

Evaluate $\int_0^1 \frac{x^2+x+1}{x^4+x^3+x^2+x+1}\ dx.$