Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is $\textbf{(A)}\ 1 \qquad \textbf{(B)}\ \dfrac{1+\sqrt{3}}{\sqrt{2}} \qquad \textbf{(C)}\ \sqrt{3} \qquad \textbf{(D)}\ 2 \qquad \textbf{(E)}\ \dfrac{4}{\sqrt{3}}$

See all the problems from 5-th Kyiv math festival [url=http://www.mathlinks.ro/Forum/viewtopic.php?p=506789#p506789]here[/url] Let $O$ be the circumcenter and $H$ be the intersection point of the altitudes of acute triangle $ABC.$ The straight lines $BH$ and $CH$ intersect the segments $CO$ and $BO$ at points $D$ and $E$ respectively. Prove that if triangles $ODH$ and $OEH$ are isosceles then triangle $ABC$ is isosceles too.

Triangle $ ABC$ has $ AB \equal{} 2 \cdot AC$. Let $ D$ and $ E$ be on $ \overline{AB}$ and $ \overline{BC}$, respectively, such that $ \angle{BAE} \equal{} \angle{ACD}.$ Let $ F$ be the intersection of segments $ AE$ and $ CD$, and suppose that $ \triangle{CFE}$ is equilateral. What is $ \angle{ACB}$? $ \textbf{(A)}\ 60^{\circ}\qquad \textbf{(B)}\ 75^{\circ}\qquad \textbf{(C)}\ 90^{\circ}\qquad \textbf{(D)}\ 105^{\circ}\qquad \textbf{(E)}\ 120^{\circ}$

Sides $AB,~ BC,$ and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5,$ and $20$, respectively. If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5} $, then side $AD$ has length $\textbf{(A) }24\qquad\textbf{(B) }24.5\qquad\textbf{(C) }24.6\qquad\textbf{(D) }24.8\qquad\textbf{(E) }25$ [size=70]*A polygon is called “simple” if it is not self intersecting.[/size]

Let $ABCD$ be a circumscribed quadrilateral and let $P$ be the orthogonal projection of its in center on $AC$. Prove that $\angle {APB}=\angle {APD}$

In the figure, $\triangle ABC$ has $\angle A =45^{\circ}$ and $\angle B =30^{\circ}$. A line $DE$, with $D$ on $AB$ and $\angle ADE =60^{\circ}$, divides $\triangle ABC$ into two pieces of equal area. (Note: the figure may not be accurate; perhaps $E$ is on $CB$ instead of $AC$.) The ratio $\frac{AD}{AB}$ is [asy] size((220)); draw((0,0)--(20,0)--(7,6)--cycle); draw((6,6)--(10,-1)); label("A", (0,0), W); label("B", (20,0), E); label("C", (7,6), NE); label("D", (9.5,-1), W); label("E", (5.9, 6.1), SW); label("$45^{\circ}$", (2.5,.5)); label("$60^{\circ}$", (7.8,.5)); label("$30^{\circ}$", (16.5,.5)); [/asy] $ \textbf{(A)}\ \frac{1}{\sqrt{2}} \qquad\textbf{(B)}\ \frac{2}{2+\sqrt{2}} \qquad\textbf{(C)}\ \frac{1}{\sqrt{3}} \qquad\textbf{(D)}\ \frac{1}{\sqrt[3]{6}} \qquad\textbf{(E)}\ \frac{1}{\sqrt[4]{12}} $

Consider a triangle $ABC$ with incenter $I$. Let $D$ be the intersection of $BI,AC$ and $CI$ intersects the circumcircle of $ABC$ at $M$. Point $K$ lies on the line $MD$ and $\angle KIA=90^\circ$. Let $F$ be the reflection of $B$ about $C$. Prove that $BIKF$ is cyclic.

In $\triangle ABC$, the length of altitude $AD$ is $12$, and the bisector $AE$ of $\angle A$ is $13$. Denote by $m$ the length of median $AF$. Find the range of $m$ when $\angle A$ is acute, orthogonal and obtuse respectively.

In convex quadrilateral $ ABCD$, $ CB,DA$ are external angle bisectors of $ \angle DCA,\angle CDB$, respectively. Points $ E,F$ lie on the rays $ AC,BD$ respectively such that $ CEFD$ is cyclic quadrilateral. Point $ P$ lie in the plane of quadrilateral $ ABCD$ such that $ DA,CB$ are external angle bisectors of $ \angle PDE,\angle PCF$ respectively. $ AD$ intersects $ BC$ at $ Q.$ Prove that $ P$ lies on $ AB$ if and only if $ Q$ lies on segment $ EF$.

Two circles $S_1$ and $S_2$ with centers $O_1$ and $O_2$ respectively intersect at $A$ and $B$. The tangents at $A$ to $S_1$ and $S_2$ meet segments $BO_2$ and $BO_1$ at $K$ and $L$ respectively. Show that $KL \parallel O_1O_2.$

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]

$ABC$ is a triangle. $M$ is the midpoint of $BC$. $\angle MAB = \angle C$, and $\angle MAC = 15^{\circ}$. Show that $\angle AMC$ is obtuse. If $O$ is the circumcenter of $ADC$, show that $AOD$ is equilateral.

Let $ABC$ be a triangle with $\angle{BAC} = 40^{\circ}$ and $\angle{ABC}=60^{\circ}$. Let $D$ and $E$ be the points lying on the sides $AC$ and $AB$, respectively, such that $\angle{CBD} = 40^{\circ}$ and $\angle{BCE} = 70^{\circ}$. Let $F$ be the point of intersection of the lines $BD$ and $CE$. Show that the line $AF$ is perpendicular to the line $BC$.

Let $ABC$ be a non-equilateral triangle with integer sides. Let $D$ and $E$ be respectively the mid-points of $BC$ and $CA$ ; let $G$ be the centroid of $\Delta{ABC}$. Suppose, $D$, $C$, $E$, $G$ are concyclic. Find the least possible perimeter of $\Delta{ABC}$.

Let $ ABC$ be a triangle inscribed in a circle of radius $ R$, and let $ P$ be a point in the interior of triangle $ ABC$. Prove that \[ \frac {PA}{BC^{2}} \plus{} \frac {PB}{CA^{2}} \plus{} \frac {PC}{AB^{2}}\ge \frac {1}{R}. \] [i]Alternative formulation:[/i] If $ ABC$ is a triangle with sidelengths $ BC\equal{}a$, $ CA\equal{}b$, $ AB\equal{}c$ and circumradius $ R$, and $ P$ is a point inside the triangle $ ABC$, then prove that $ \frac {PA}{a^{2}} \plus{} \frac {PB}{b^{2}} \plus{} \frac {PC}{c^{2}}\ge \frac {1}{R}$.

A circle centered at $ O$ has radius $ 1$ and contains the point $ A$. Segment $ AB$ is tangent to the circle at $ A$ and $ \angle{AOB} \equal{} \theta$. If point $ C$ lies on $ \overline{OA}$ and $ \overline{BC}$ bisects $ \angle{ABO}$, then $ OC \equal{}$ [asy]import olympiad; unitsize(2cm); defaultpen(fontsize(8pt)+linewidth(.8pt)); labelmargin=0.2; dotfactor=3; pair O=(0,0); pair A=(1,0); pair B=(1,1.5); pair D=bisectorpoint(A,B,O); pair C=extension(B,D,O,A); draw(Circle(O,1)); draw(O--A--B--cycle); draw(B--C); label("$O$",O,SW); dot(O); label("$\theta$",(0.1,0.05),ENE); dot(C); label("$C$",C,S); dot(A); label("$A$",A,E); dot(B); label("$B$",B,E);[/asy] $ \textbf{(A)}\ \sec^2\theta \minus{} \tan\theta \qquad \textbf{(B)}\ \frac {1}{2} \qquad \textbf{(C)}\ \frac {\cos^2\theta}{1 \plus{} \sin\theta} \qquad \textbf{(D)}\ \frac {1}{1 \plus{} \sin\theta} \qquad \textbf{(E)}\ \frac {\sin\theta}{\cos^2\theta}$

The lengths of the sides of a triangle are consescutive integers, and the largest angle is twice the smallest angle. The cosine of the smallest angle is $\textbf {(A) } \frac 34 \qquad \textbf {(B) } \frac{7}{10} \qquad \textbf {(C) } \frac 23 \qquad \textbf {(D) } \frac{9}{14} \qquad \textbf {(E) } \text{none of these}$

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 $A_1, A_2, \ldots, A_n$ 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 $n$.

$D, \: E , \: F$ 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 $AK=AD.$

In triangle $ ABC$, side $ AC$ and the perpendicular bisector of $ BC$ meet in point $ D$, and $ BD$ bisects $ \angle ABC$. If $ AD \equal{} 9$ and $ DC \equal{} 7$, what is the area of triangle $ ABD$? $ \textbf{(A)}\ 14 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 14\sqrt5 \qquad \textbf{(E)}\ 28\sqrt5$

$M$ is an arbitrary point inside $\triangle ABC$. $AM$ intersects the circumcircle of the triangle again at $A_1$. Find the points $M$ that minimise $\frac{MB\cdot MC}{MA_1}$.

$M$ is an arbitrary point inside $\triangle ABC$. $AM$ intersects the circumcircle of the triangle again at $A_1$. Find the points $M$ that minimise $\frac{MB\cdot MC}{MA_1}$.

Points $D$ and $E$ divides the base $BC$ of an isosceles triangle $ABC$ into three equal parts and $D$ is between $B$ and $E.$ Show that $\angle BAD<\angle DAE.$