Olympiad problems

2004 Unirea, 3

Hello, I've been trying to solve this for a while now, but no success! I mean, I have an expression for this but not a closed one. I derived something in terms of Tchebychev Polynomials : cos(nx) = P_n(cos(x)). Any help is appreciated. Compute the following primitive: \[ \int \frac{x\sin\left(2004 x\right)}{\tan x}\ dx\]

1967 IMO Longlists, 28

Tags: parameterization , calculus , trigonometry , algebra , trigonometric identities

Find values of the parameter $u$ for which the expression \[y = \frac{ \tan(x-u) + \tan(x) + \tan(x+u)}{ \tan(x-u)\tan(x)\tan(x+u)}\] does not depend on $x.$

1974 IMO Longlists, 8

Tags: trigonometry , trigonometric identities , trigonometric substitution , equation , algebra

Let $x, y, z$ be real numbers each of whose absolute value is different from $\frac{1}{\sqrt 3}$ such that $x + y + z = xyz$. Prove that \[\frac{3x - x^3}{1-3x^2} + \frac{3y - y^3}{1-3y^2} + \frac{3z -z^3}{1-3z^2} = \frac{3x - x^3}{1-3x^2} \cdot \frac{3y - y^3}{1-3y^2} \cdot \frac{3z - z^3}{1-3z^2}\]

2012 Online Math Open Problems, 7

Tags: probability , trigonometry , function , inequalities , number theory , relatively prime

Two distinct points $A$ and $B$ are chosen at random from 15 points equally spaced around a circle centered at $O$ such that each pair of points $A$ and $B$ has the same probability of being chosen. The probability that the perpendicular bisectors of $OA$ and $OB$ intersect strictly inside the circle can be expressed in the form $\frac{m}{n}$, where $m,n$ are relatively prime positive integers. Find $m+n$. [i]Ray Li.[/i]

2000 Harvard-MIT Mathematics Tournament, 20

Tags: geometry , perimeter , hmmt , trigonometry

What is the minimum possible perimeter of a triangle two of whose sides are along the x- and y-axes and such that the third contains the point $(1,2)$?

2007 Tournament Of Towns, 7

Tags: ratio , trigonometry , geometric transformation , reflection , geometry

$T$ is a point on the plane of triangle $ABC$ such that $\angle ATB = \angle BTC = \angle CTA = 120^\circ$. Prove that the lines symmetric to $AT, BT$ and $CT$ with respect to $BC, CA$ and $AB$, respectively, are concurrent.

2010 Today's Calculation Of Integral, 539

Tags: calculus , integration , trigonometry , calculus computations , logarithm

Evaluate $ \int_0^{\frac{\pi}{4}} \frac{\sin ^ 2 x}{\cos ^ 3 x}\ dx$.

2010 Contests, 2

Tags: inequalities , trigonometry , calculus , derivative , algebra

Prove that for any real number $ x$ the following inequality is true: $ \max\{|\sin x|, |\sin(x\plus{}2010)|\}>\dfrac1{\sqrt{17}}$

1988 IMO Shortlist, 3

Tags: geometry , circumcircle , trigonometry , inradius , geometric inequality

The triangle $ ABC$ is inscribed in a circle. The interior bisectors of the angles $ A,B$ and $ C$ meet the circle again at $ A', B'$ and $ C'$ respectively. Prove that the area of triangle $ A'B'C'$ is greater than or equal to the area of triangle $ ABC.$

1950 Moscow Mathematical Olympiad, 178

Tags: trigonometry , angle

Let $A$ be an arbitrary angle,let $B$ and $C$ be acute angles. Is there an angle $x$ such that $$\sin x =\frac{\sin B \cdot \sin C}{1 - \cos B \cdot \cos C \cdot \cos A} ?$$

1986 IMO Longlists, 41

Tags: circumcircle , trigonometry , symmetry , geometry

Let $M,N,P$ be the midpoints of the sides $BC, CA, AB$ of a triangle $ABC$. The lines $AM, BN, CP$ intersect the circumcircle of $ABC$ at points $A',B', C'$, respectively. Show that if $A'B'C'$ is an equilateral triangle, then so is $ABC.$

2005 India Regional Mathematical Olympiad, 5

Tags: trigonometry , exterior angle , eq triangle

In a triangle ABC, D is midpoint of BC . If $\angle ADB = 45 ^{\circ}$ and $\angle ACD = 30^{\circ}$, determine $\angle BAD.$

2009 Princeton University Math Competition, 4

Tags: trigonometry

Find the smallest positive $\alpha$ (in degrees) for which all the numbers \[\cos{\alpha},\cos{2\alpha},\ldots,\cos{2^n\alpha},\ldots\] are negative.

2003 India Regional Mathematical Olympiad, 1

Tags: trigonometry , geometry

Let $ABC$ be a triangle in which $AB =AC$ and $\angle CAB = 90^{\circ}$. Suppose that $M$ and $N$ are points on the hypotenuse $BC$ such that $BM^2 + CN^2 = MN^2$. Prove that $\angle MAN = 45^{\circ}$.

1967 IMO Longlists, 45

Tags: trigonometry , algebra , trigonometric equation , geometry

[b](i)[/b] Solve the equation: \[ \sin^3(x) + \sin^3\left( \frac{2 \pi}{3} + x\right) + \sin^3\left( \frac{4 \pi}{3} + x\right) + \frac{3}{4} \cos {2x} = 0.\] [b](ii)[/b] Supposing the solutions are in the form of arcs $AB$ with one end at the point $A$, the beginning of the arcs of the trigonometric circle, and $P$ a regular polygon inscribed in the circle with one vertex in $A$, find: 1) The subsets of arcs having the other end in $B$ in one of the vertices of the regular dodecagon. 2) Prove that no solution can have the end $B$ in one of the vertices of polygon $P$ whose number of sides is prime or having factors other than 2 or 3.

2009 Hong Kong TST, 1

Tags: trigonometry , inequalities

Let $ \theta_1, \theta_2,\ldots , \theta_{2008}$ be real numbers. Find the maximum value of $ \sin\theta_1\cos\theta_2 \plus{} \sin\theta_2\cos\theta_3 \plus{} \ldots \plus{} \sin\theta_{2007}\cos\theta_{2008} \plus{} \sin\theta_{2008}\cos\theta_1$

1994 IberoAmerican, 2

Tags: geometry , rectangle , inradius , incenter , ratio , trigonometry , perimeter

Let $ ABCD$ a cuadrilateral inscribed in a circumference. Suppose that there is a semicircle with its center on $ AB$, that is tangent to the other three sides of the cuadrilateral. (i) Show that $ AB \equal{} AD \plus{} BC$. (ii) Calculate, in term of $ x \equal{} AB$ and $ y \equal{} CD$, the maximal area that can be reached for such quadrilateral.

2002 Austrian-Polish Competition, 8

Tags: trigonometry , limit , inequalities , algebra

Determine the number of real solutions of the system \[\left\{ \begin{aligned}\cos x_{1}&= x_{2}\\ &\cdots \\ \cos x_{n-1}&= x_{n}\\ \cos x_{n}&= x_{1}\\ \end{aligned}\right.\]

2000 National Olympiad First Round, 33

Tags: geometry , trigonometry , quadratic , pythagorean theorem , angle bisector

Let $K$ be a point on the side $[AB]$, and $L$ be a point on the side $[BC]$ of the square $ABCD$. If $|AK|=3$, $|KB|=2$, and the distance of $K$ to the line $DL$ is $3$, what is $|BL|:|LC|$? $ \textbf{(A)}\ \frac78 \qquad\textbf{(B)}\ \frac{\sqrt 3}2 \qquad\textbf{(C)}\ \frac 87 \qquad\textbf{(D)}\ \frac 38 \qquad\textbf{(E)}\ \frac{\sqrt 2}2 $

IV Soros Olympiad 1997 - 98 (Russia), 11.2

Tags: inequalities , algebra , trigonometry , analytic geometry

Find the area of a figure consisting of points whose coordinates satisfy the inequality $$(y^3 - arcsin x)(x^3 + arcsin y) \ge 0.$$

2012 Turkey MO (2nd round), 2

Tags: trigonometry , circumcircle , projective geometry , geometry

Let $ABC$ be a isosceles triangle with $AB=AC$ an $D$ be the foot of perpendicular of $A$. $P$ be an interior point of triangle $ADC$ such that $m(APB)>90$ and $m(PBD)+m(PAD)=m(PCB)$. $CP$ and $AD$ intersects at $Q$, $BP$ and $AD$ intersects at $R$. Let $T$ be a point on $[AB]$ and $S$ be a point on $[AP$ and not belongs to $[AP]$ satisfying $m(TRB)=m(DQC)$ and $m(PSR)=2m(PAR)$. Show that $RS=RT$

2014 Bosnia And Herzegovina - Regional Olympiad, 3

Tags: trigonometry , angle bisector , geometry

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 $\measuredangle ADC$.

1969 Canada National Olympiad, 4

Tags: ratio , trigonometry , area of a triangle , heron's formula , geometry

Let $ABC$ be an equilateral triangle, and $P$ be an arbitrary point within the triangle. Perpendiculars $PD,PE,PF$ are drawn to the three sides of the triangle. Show that, no matter where $P$ is chosen, \[ \frac{PD+PE+PF}{AB+BC+CA}=\frac{1}{2\sqrt{3}}. \]

2014 National Olympiad First Round, 21

Tags: geometry , trapezoid , trigonometry , rectangle , geometric transformation , reflection , circumcircle

Let $ABCD$ be a trapezoid such that side $[AB]$ and side $[CD]$ are perpendicular to side $[BC]$. Let $E$ be a point on side $[BC]$ such that $\triangle AED$ is equilateral. If $|AB|=7$ and $|CD|=5$, what is the area of trapezoid $ABCD$? $ \textbf{(A)}\ 27\sqrt{3} \qquad\textbf{(B)}\ 42 \qquad\textbf{(C)}\ 24\sqrt{3} \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 36 $

2012 APMO, 4

Tags: circumcircle , trigonometry , parallelogram , geometric transformation , reflection , asymptote , geometry

Let $ ABC $ be an acute triangle. Denote by $ D $ the foot of the perpendicular line drawn from the point $ A $ to the side $ BC $, by $M$ the midpoint of $ BC $, and by $ H $ the orthocenter of $ ABC $. Let $ E $ be the point of intersection of the circumcircle $ \Gamma $ of the triangle $ ABC $ and the half line $ MH $, and $ F $ be the point of intersection (other than $E$) of the line $ ED $ and the circle $ \Gamma $. Prove that $ \tfrac{BF}{CF} = \tfrac{AB}{AC} $ must hold. (Here we denote $XY$ the length of the line segment $XY$.)