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 $KL \parallel O_1O_2.$

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}$

Given circle $O$ with radius $R$, the inscribed triangle $ABC$ is an acute scalene triangle, where $AB$ is the largest side. $AH_A, BH_B,CH_C$ are heights on $BC,CA,AB$. Let $D$ be the symmetric point of $H_A$ with respect to $H_BH_C$, $E$ be the symmetric point of $H_B$ with respect to $H_AH_C$. $P$ is the intersection of $AD,BE$, $H$ is the orthocentre of $\triangle ABC$. Prove: $OP\cdot OH$ is fixed, and find this value in terms of $R$. (Edited)

The incenter of $\triangle A_1A_2A_3$ is $I$, and the center of the $A_i$-excircle is $J_i$ ($i=1,2,3$). Let $B_i$ be the intersection point of side $A_{i+1}A_{i+2}$ and the bisector of $\angle A_{i+1}IA_{i+2}$ ($A_{i+3}:=A_i$ $\forall i$). Prove that the three lines $B_iJ_i$ are concurrent.

In a non-isosceles triangle $ABC$ the bisectors of angles $A$ and $B$ are inversely proportional to the respective sidelengths. Find angle $C$.

$ABC$ 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.

Convex quadrilateral $ ABCD$ is inscribed in a circle, $ \angle{A}\equal{}60^o$, $ BC\equal{}CD\equal{}1$, rays $ AB$ and $ DC$ intersect at point $ E$, rays $ BC$ and $ AD$ intersect each other at point $ F$. It is given that the perimeters of triangle $ BCE$ and triangle $ CDF$ are both integers. Find the perimeter of quadrilateral $ ABCD$.

The base of an isosceles triangle is $ 6$ inches and one of the equal sides is $ 12$ inches. The radius of the circle through the vertices of the triangle is: $ \textbf{(A)}\ \frac{7\sqrt{15}}{5} \qquad\textbf{(B)}\ 4\sqrt{3} \qquad\textbf{(C)}\ 3\sqrt{5} \qquad\textbf{(D)}\ 6\sqrt{3} \qquad\textbf{(E)}\ \text{none of these}$

The points $D, E$ on the side $AB$ of the triangle $ABC$ are such that $\frac{AD}{DB}\frac{AE}{EB} = \left(\frac{AC}{CB}\right)^2$. Show that $\angle ACD = \angle BCE$.

Three congruent isosceles triangles are constructed with their bases on the sides of an equilateral triangle of side length $1$. The sum of the areas of the three isosceles triangles is the same as the area of the equilateral triangle. What is the length of one of the two congruent sides of one of the isosceles triangles? $\textbf{(A) }\dfrac{\sqrt3}4\qquad \textbf{(B) }\dfrac{\sqrt3}3\qquad \textbf{(C) }\dfrac23\qquad \textbf{(D) }\dfrac{\sqrt2}2\qquad \textbf{(E) }\dfrac{\sqrt3}2$

Two congruent $ 30^{\circ}$-$ 60^{\circ}$-$ 90^{\circ}$ are placed so that they overlap partly and their hypotenuses coincide. If the hypotenuse of each triangle is 12, the area common to both triangles is $ \textbf{(A)}\ 6\sqrt3 \qquad \textbf{(B)}\ 8\sqrt3 \qquad \textbf{(C)}\ 9\sqrt3 \qquad \textbf{(D)}\ 12\sqrt3 \qquad \textbf{(E)}\ 24$

Let $ABC$ be a triangle, and $M$ an interior point such that $\angle MAB=10^\circ$, $\angle MBA=20^\circ$, $\angle MAC=40^\circ$ and $\angle MCA=30^\circ$. Prove that the triangle is isosceles.

In a triangle $ ABC,$ let $ D$ and $ E$ be the intersections of the bisectors of $ \angle ABC$ and $ \angle ACB$ with the sides $ AC,AB,$ respectively. Determine the angles $ \angle A,\angle B, \angle C$ if $ \angle BDE \equal{} 24 ^{\circ},$ $ \angle CED \equal{} 18 ^{\circ}.$

Consider a triangle $ABC$ and $I$ its incenter. The line $(AI)$ meets the circumcircle of $ABC$ in $D$. Let $E$ and $F$ be the orthogonal projections of $I$ on $(BD)$ and $(CD)$ respectively. Assume that $IE+IF=\frac{1}{2}AD$. Calculate $\angle{BAC}$. [color=red][Moderator edited: Also discussed at http://www.mathlinks.ro/Forum/viewtopic.php?t=5088 .][/color]

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.

Let $ ABC$ be a triangle with incenter $ I$ and let $ \Gamma$ be a circle centered at $ I$, whose radius is greater than the inradius and does not pass through any vertex. Let $ X_{1}$ be the intersection point of $ \Gamma$ and line $ AB$, closer to $ B$; $ X_{2}$, $ X_{3}$ the points of intersection of $ \Gamma$ and line $ BC$, with $ X_{2}$ closer to $ B$; and let $ X_{4}$ be the point of intersection of $ \Gamma$ with line $ CA$ closer to $ C$. Let $ K$ be the intersection point of lines $ X_{1}X_{2}$ and $ X_{3}X_{4}$. Prove that $ AK$ bisects segment $ X_{2}X_{3}$.

The isosceles right triangle $ABC$ has right angle at $C$ and area $12.5$. The rays trisecting $\angle{ACB}$ intersect $AB$ at $D$ and $E$. What is the area of $\triangle{CDE}$? $\textbf{(A) }\frac{5\sqrt{2}}{3}\qquad\textbf{(B) }\frac{50\sqrt{3}-75}{4}\qquad\textbf{(C) }\frac{15\sqrt{3}}{8}\qquad\textbf{(D) }\frac{50-25\sqrt{3}}{2}\qquad\textbf{(E) }\frac{25}{6}$

Find all ordered triples $ (x, y, z)$ of real numbers such that \[ 5 \left(x \plus{} \frac{1}{x} \right) \equal{} 12 \left(y \plus{} \frac{1}{y} \right) \equal{} 13 \left(z \plus{} \frac{1}{z} \right),\] and \[ xy \plus{} yz \plus{} zy \equal{} 1.\]

Points $A$, $B$, and $C$ lie on a circle of radius $5$ such that $AB=6$ and $AC=8$. Find the smaller of the two possible values of $BC$.

From the vertex $ A$ of the equilateral triangle $ ABC$ a line is drown that intercepts the segment $ [BC]$ in the point $ E$. The point $ M \in (AE$ is such that $ M$ external to $ ABC$, $ \angle AMB \equal{} 20 ^\circ$ and $ \angle AMC \equal{} 30 ^ \circ$. What is the measure of the angle $ \angle MAB$?

Triangle $ABC$ is isosceles with $AC = BC$ and $\angle ACB = 106^\circ$. Point $M$ is in the interior of the triangle so that $\angle MAC = 7^\circ$ and $\angle MCA = 23^\circ$. Find the number of degrees in $\angle CMB$.

Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP>CP$. Let $O_1$ and $O_2$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB=12$ and $\angle O_1 P O_2 = 120^\circ$, then $AP=\sqrt{a}+\sqrt{b}$ where $a$ and $b$ are positive integers. Find $a+b$.

Let $D$ and $E$ be points on $[BC]$ and $[AC]$ of acute $\triangle ABC$, respectively. $AD$ and $BE$ meet at $F$. If $|AF|=|CD|=2|BF|=2|CE|$, and $Area(\triangle ABF) = Area(\triangle DEC)$, then $Area(\triangle AFC)/Area(\triangle BFC) = ?$ $ \textbf{(A)}\ 4 \qquad \textbf{(B)}\ 2\sqrt2 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \sqrt2 \qquad \textbf{(E)}\ 1$