A fancy bathroom scale is calibrated in Newtons. This scale is put on a ramp, which is at a $40^\circ$ angle to the horizontal. A box is then put on the scale and the box-scale system is then pushed up the ramp by a horizontal force $F$. The system slides up the ramp at a constant speed. If the bathroom scale reads $R$ and the coefficient of static friction between the system and the ramp is $0.40$, what is $\frac{F}{R}$? Round to the nearest thousandth. [i](Proposed by Ahaan Rungta)[/i]

Let $ f(x)\equal{}\frac{1}{\sin x\sqrt{1\minus{}\cos x}}\ (0<x<\pi)$. (1) Find the local minimum value of $ f(x)$. (2) Evaluate $ \int_{\frac{\pi}{2}}^{\frac{2\pi}{3}} f(x)\ dx$.

A circle with center $O$ and a point $A$ in this circle are given. Let $P_{B}$ is the intersection point of $[AB]$ and the internal bisector of $\angle AOB$ where $B$ is a point on the circle such that $B$ doesn't lie on the line $OA$, Find the locus of $P_{B}$ as $B$ varies.

In a triangle $ABC$, let $I$ denote its incenter. Points $D, E, F$ are chosen on the segments $BC, CA, AB$, respectively, such that $BD + BF = AC$ and $CD + CE = AB$. The circumcircles of triangles $AEF, BFD, CDE$ intersect lines $AI, BI, CI$, respectively, at points $K, L, M$ (different from $A, B, C$), respectively. Prove that $K, L, M, I$ are concyclic.

Construct outer squares $ABMN$, $BCKL$, $ACPQ$ on sides $[AB]$, $[BC]$, $[CA]$ of triangle $ABC$, respectively. Construct squares $NQZT$ and $KPYX$ on segments $[NQ]$ and $[KP]$. If $Area(ABMN) - Area(BCKL)=1$, what is $Area(NQZT)-Area(KPYX)$? $ \textbf{(A)}\ \dfrac 34 \qquad\textbf{(B)}\ \dfrac 53 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4 $

For the vertex $ A$ of a triangle $ ABC$, let $ l_a$ be the distance between the projections on $ AB$ and $ AC$ of the intersection of the angle bisector of ∠$ A$ with side $ BC$. Define $ l_b$ and $ l_c$ analogously. If $ l$ is the perimeter of triangle $ ABC$, prove that $ \frac{l_a l_b l_c}{l^3}\le\frac{1}{64}$.

From the altitudes of an acute-angled triangle, a triangle can be composed. Prove that a triangle can be composed from the bisectors of this triangle.

Given that triangle $ABC$ has side lengths $a=7$, $b=8$ , $c=5$, find $$(\sin (A)+\sin (B)+\sin (C)) \cdot \left(\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}\right).$$ .

$u$ is a real parameter such that $0<u<1$. For $0\le x \le u$, $f(x)=0$. For $u\le x \le n$, $f(x)=1-\left(\sqrt{ux}+\sqrt{(1-u)(1-x)}\right)^2$. The sequence $\{u_n\}$ is define recursively as follows: $u_1=f(1)$ and $u_n=f(u_{n-1})$ $\forall n\in \mathbb{N}, n\neq 1$. Show that there exists a positive integer $k$ for which $u_k=0$.

Compute \[\lim_{h\to 0}\dfrac{\sin(\frac{\pi}{3}+4h)-4\sin(\frac{\pi}{3}+3h)+6\sin(\frac{\pi}{3}+2h)-4\sin(\frac{\pi}{3}+h)+\sin(\frac{\pi}{3})}{h^4}.\]

Evaluate $\int_{0}^{\infty}\frac{\tan^{-1}(\pi x)-\tan^{-1}x}{x}dx$

The measures, in degrees, of the angles , $\alpha, \beta$ and $\theta$ are greater than $0$ less than $60$. Find the value of $\theta$ knowing, also, that $\alpha + \beta = 2\theta$ and that $$\sin \alpha \sin \beta \sin \theta = \sin(60 - \alpha ) \sin(60 - \beta) \sin(60 - \theta ).$$

Given an equilateral $\triangle ABC$, find the locus of points $P$ such that $\angle APB=\angle BPC$.

$AD$ is the altitude on side $BC$ of triangle $ABC$. If $BC+AD-AB-AC = 0$, find the range of $\angle BAC$. [i]Alternative formulation.[/i] Let $AD$ be the altitude of triangle $ABC$ to the side $BC$. If $BC+AD=AB+AC$, then find the range of $\angle{A}$.

Let $\mathcal{F}$ be the set of functions $f: [0,1]\to\mathbb{R}$ that are differentiable, with continuous derivative, and $f(0)=0$, $f(1)=1$. Find the minimum of $\int_{0}^{1}\sqrt{1+x^{2}}\cdot \big(f'(x)\big)^{2}\ dx$ (where $f\in\mathcal{F}$) and find all functions $f\in\mathcal{F}$ for which this minimum is attained. [hide="Comment"] In the contest, this was the b) point of the problem. The a) point was simply ``Prove the Cauchy inequality in integral form''. [/hide]

In the adjoining figure, two circles of radii 6 and 8 are drawn with their centers 12 units apart. At $P$, one of the points of intersection, a line is drawn in such a way that the chords $QP$ and $PR$ have equal length. Find the square of the length of $QP$. [asy]unitsize(2.5mm); defaultpen(linewidth(.8pt)+fontsize(12pt)); dotfactor=3; pair O1=(0,0), O2=(12,0); path C1=Circle(O1,8), C2=Circle(O2,6); pair P=intersectionpoints(C1,C2)[0]; path C3=Circle(P,sqrt(130)); pair Q=intersectionpoints(C3,C1)[0]; pair R=intersectionpoints(C3,C2)[1]; draw(C1); draw(C2); //draw(O2--O1); //dot(O1); //dot(O2); draw(Q--R); label("$Q$",Q,N); label("$P$",P,dir(80)); label("$R$",R,E); //label("12",waypoint(O1--O2,0.4),S);[/asy]

Let $x$ be in the interval $\left(0, \frac{\pi}{2}\right)$ such that $\sin x - \cos x = \frac12$ . Then $\sin^3 x + \cos^3 x = \frac{m\sqrt{p}}{n}$ , where $m, n$, and $p$ are relatively prime positive integers, and $p$ is not divisible by the square of any prime. Find $m + n + p$.

Prove that for all $n \in \mathbb N$ the following is true: \[2^n \prod_{k=1}^n \sin \frac{k \pi}{2n+1} = \sqrt{2n+1}\]

There are $2n$ complex numbers that satisfy both $z^{28}-z^{8}-1=0$ and $|z|=1$. These numbers have the form $z_{m}=\cos\theta_{m}+i\sin\theta_{m}$, where $0\leq\theta_{1}<\theta_{2}< \dots <\theta_{2n}<360$ and angles are measured in degrees. Find the value of $\theta_{2}+\theta_{4}+\dots+\theta_{2n}$.

Three rays are going out from point $O$ in space, forming pairwise angles $\alpha, \beta$ and $\gamma$ with $0^o<\alpha \le \beta \le \gamma <180^o$. Prove that $\sin \frac{\alpha}{2}+ \sin \frac{\beta}{2} > \sin \frac{\gamma}{2}$.

Find all triples $(x,y,z)$ of real numbers that satisfy \[ \begin{aligned} & x\left(1 - y^2\right)\left(1 - z^2\right) + y\left(1 - z^2\right)\left(1 - x^2\right) + z\left(1 - x^2\right)\left(1 - y^2\right) \\ & = 4xyz \\ & = 4(x + y + z). \end{aligned} \]

Find all $x$ for which the inequality holds $$9 \sin x +40 \cos x \ge 41.$$

Evaluate $ \int_0^{n\pi} \sqrt{1\minus{}\sin t}\ dt\ (n\equal{}1,\ 2,\ \cdots).$

Let $ M$ and $ N$ be two points inside triangle $ ABC$ such that \[ \angle MAB \equal{} \angle NAC\quad \mbox{and}\quad \angle MBA \equal{} \angle NBC. \] Prove that \[ \frac {AM \cdot AN}{AB \cdot AC} \plus{} \frac {BM \cdot BN}{BA \cdot BC} \plus{} \frac {CM \cdot CN}{CA \cdot CB} \equal{} 1. \]

$AB$ is the diameter of a circle centered at $O$. $C$ is a point on the circle such that angle $BOC$ is $60^\circ$. If the diameter of the circle is $5$ inches, the length of chord $AC$, expressed in inches, is: $\text{(A)} \ 3 \qquad \text{(B)} \ \frac{5\sqrt{2}}{2} \qquad \text{(C)} \frac{5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$