Suppose that a function $f(x)$ defined in $-1<x<1$ satisfies the following properties (i) , (ii), (iii). (i) $f'(x)$ is continuous. (ii) When $-1<x<0,\ f'(x)<0,\ f'(0)=0$, when $0<x<1,\ f'(x)>0$. (iii) $f(0)=-1$ Let $F(x)=\int_0^x \sqrt{1+\{f'(t)\}^2}dt\ (-1<x<1)$. If $F(\sin \theta)=c\theta\ (c :\text{constant})$ holds for $-\frac{\pi}{2}<\theta <\frac{\pi}{2}$, then find $f(x)$. [i]1975 Waseda University entrance exam/Science and Technology[/i]

Consider three real numbers $ x,y,z $ satisfying $ \cos x+\cos y+\cos z =\cos 3x +\cos 3y +\cos 3z=0. $ Show that $ \cos 2x\cdot \cos 2y\cdot\cos 2z\le 0. $ [i]Bogdan Enescu[/i]

Let $ABC$ be a triangle, let $H$ be the orthocentre and $L,M,N$ the midpoints of the sides $AB, BC, CA$ respectively. Prove that \[HL^{2} + HM^{2} + HN^{2} < AL^{2} + BM^{2} + CN^{2}\] if and only if $ABC$ is acute-angled.

Let $x,y,z$ be real numbers such that $0\le x,y,z\le\frac{\pi}{2}$. Prove the inequality \[\frac{\pi}{2}+2\sin x\cos y+2\sin y\cos z\ge\sin 2x+\sin 2y+\sin 2z.\]

An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $\mathcal{C}$ to the area of shaded region $\mathcal{B}$ is $11/5$. Find the ratio of shaded region $\mathcal{D}$ to the area of shaded region $\mathcal{A}$. [asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]

Let $ABCD$ be a parallelogram. Let $M$ be a point on the side $AB$ and $N$ be a point on the side $BC$ such that the segments $AM$ and $CN$ have equal lengths and are non-zero. The lines $AN$ and $CM$ meet at $Q$. Prove that the line $DQ$ is the bisector of the angle $\measuredangle ADC$. [i]Alternative formulation.[/i] Let $ABCD$ be a parallelogram. Let $M$ and $N$ be points on the sides $AB$ and $BC$, respectively, such that $AM=CN\neq 0$. The lines $AN$ and $CM$ intersect at a point $Q$. Prove that the point $Q$ lies on the bisector of the angle $\measuredangle ADC$.

Let $ AB$ be the diameter of semicircle $O$ , $C, D $ be points on the arc $AB$, $P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ . Prove that:$CP\cdot CQ=DP \cdot DQ$.[asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy]

The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ACD? [asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0); fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); } label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy] $\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4.5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 9$

Define the sequence $\{x_{n}\}$ by $x_{0}=a\in (0,1)$ and $x_{n+1}=\frac{4}{\pi^{2}}(\cos^{-1}x_{n}+\frac{\pi}{2})\sin^{-1}x_{n}(n=0,1,2,...)$. Show that the sequence converges and find its limit.

Let $ABC$ be a triangle such that $|AC|=1$ and $|AB|=\sqrt 2$. Let $M$ be a point such that $|MA|=|AB|$, $m(\widehat{MAB}) = 90^\circ$, and $C$ and $M$ are on the opposite sides of $AB$. Let $N$ be a point such that $|NA|=|AX|$, $m(\widehat{NAC}) = 90^\circ$, and $B$ and $N$ are on the opposite sides of $AC$. If the line passing throung $A$ and the circumcenter of triangle $MAN$ meets $[BC]$ at $F$, what is $\dfrac {|BF|}{|FC|}$? $ \textbf{(A)}\ 2\sqrt 2 \qquad\textbf{(B)}\ 2\sqrt 3 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 3\sqrt 2 $

In an acute triangle $ABC$ , $AB>AC$ , $ \cos B+ \cos C=1$ , $E,F$ are on the extend line of $AB,AC$ such that $\angle ABF = \angle ACE = 90$ . (1) Prove :$BE+CF=EF$ ; (2) Assume the bisector of $\angle EBC$ meet $EF$ at $P$ , prove that $CP$ is the bisector of $\angle BCF$. [img]https://cdn.artofproblemsolving.com/attachments/a/2/c554c2bc0b4e044c45f88138568f5234d544a8.png[/img]

Let $f_{n}(x)=\sum_{k=1}^{n}\frac{\sin kx}{\sqrt{k(k+1)}}.$ Find $\lim_{n\to\infty}\int_{0}^{2\pi}\{f_{n}(x)\}^{2}dx.$

Let $\Gamma$ and $\Gamma'$ be two concentric circles. Let $ABC$ and $A'B'C'$ be any two equilateral triangles inscribed in $\Gamma$ and $\Gamma'$ respectively. If $P$ and $P'$ are any two points on $\Gamma$ and $\Gamma'$ respectively, show that \[ P'A^2 + P'B^2 + P'C^2 = A'P^2 + B'P^2 + C'P^2. \]

In an acute-angled $ABC$, $\angle A>60^{\circ}$, $H$ is its orthocenter. $M,N$ are two points on $AB,AC$ respectively, such that $\angle HMB=\angle HNC=60^{\circ}$. Let $O$ be the circumcenter of triangle $HMN$. $D$ is a point on the same side with $A$ of $BC$ such that $\triangle DBC$ is an equilateral triangle. Prove that $H,O,D$ are collinear.

Prove that if $ n $ is a natural number greater than $ 1 $, then $$ \cos \frac{2\pi}{n} + \cos \frac{4\pi}{n} + \cos \frac{6\pi}{n} + \ldots + \cos \frac{2n \pi}{n} = 0.$$

In triangle $ABC,$ the medians to the sides $\overline{AB}$ and $\overline{AC}$ are perpendicular. Prove that $\cot B+\cot C\ge \frac23.$

A plane is drawn through the height of a regular tetrahedron, which intersects the planes of the lateral faces along $ 3 $ lines that form angles $ \alpha $, $ \beta $, $ \gamma $ with the plane of the tetrahedron's base. Prove that $$ tg^2 \alpha + tg^2 \beta + tg^2 \gamma =12.$$

Compare the values of the following definite integrals. \[\int_{0}^{\infty}\ln \left(x+\frac{1}{x}\right)\frac{dx}{1+x^{2}},\ \ \int_{0}^{\frac{\pi}{2}}\left(\frac{\theta}{\sin \theta}\right)^{2}d\theta\]

For a non negative integer $n$, set t $I_n=\int_0^{\frac{\pi}{4}} \tan ^ n x\ dx$ to answer the following questions: (1) Calculate $I_{n+2}+I_n.$ (2) Evaluate the values of $I_1,\ I_2$ and $I_3.$ 1978 Niigata university entrance exam

Find all real values $x$ that satisfy the following equation: $$\frac{\sin 3x cos \left(\frac{\pi}{3}-4x \right)+ 1}{\sin \left(\frac{\pi}{3}-7x \right) - cos\left(\frac{\pi}{6}+x \right)+m}= 0$$ where $m$ is a given real number.

Rhombus $ABCD$ has side length $2$ and $\angle B = 120 ^\circ$. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$? $ \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad \textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad \textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad \textbf{(D)}\ 1+\frac{\sqrt{3}}{3} \qquad \textbf{(E)}\ 2 $

Let $ \alpha,\beta,\gamma$ be the angles of a triangle. Prove that $csc^2\frac{\alpha}{2}+csc^2\frac{\beta}{2}+csc^2\frac{\gamma}{2} \ge 12$ and find the conditions for equality.

Given that $\log_{10} \sin x + \log_{10} \cos x = -1$ and that $\log_{10} (\sin x + \cos x) = \textstyle \frac{1}{2} (\log_{10} n - 1)$, find $n$.

Find all pairs of real numbers $(x, y)$ that satisfy the equation $$\frac{(x+y)(2-\sin(x+y))}{4\sin^2(x+y)}=\frac{xy}{x+y}$$

Let $a, b, \alpha, \beta$ be real numbers such that $0 \leq a, b \leq 1$, and $0 \leq \alpha, \beta \leq \frac{\pi}{2}$. Show that if \[ ab\cos(\alpha - \beta) \leq \sqrt{(1-a^2)(1-b^2)}, \] then \[ a\cos\alpha + b\sin\beta \leq 1 + ab\sin(\beta - \alpha). \]