Let $ABC$ be an isosceles triangle with $\angle{ABC} = \angle{ACB} = 80^\circ$. Let $D$ be a point on $AB$ such that $\angle{DCB} = 60^\circ$ and $E$ be a point on $AC$ such that $\angle{ABE} = 30^\circ$. Find $\angle{CDE}$ in degrees.

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

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)

Let $\triangle ABC$ be a triangle. Let $M$ be the midpoint of $BC$ and let $D$ be a point on the interior of side $AB$. The intersection of $AM$ and $CD$ is called $E$. Suppose that $|AD|=|DE|$. Prove that $|AB|=|CE|$.

Points $A, B, C$ and $D$ are on a circle of diameter $1$, and $X$ is on diameter $\overline{AD}$. If $BX=CX$ and $3 \angle BAC=\angle BXC=36^{\circ}$, then $AX=$ [asy] draw(Circle((0,0),10)); draw((-10,0)--(8,6)--(2,0)--(8,-6)--cycle); draw((-10,0)--(10,0)); dot((-10,0)); dot((2,0)); dot((10,0)); dot((8,6)); dot((8,-6)); label("A", (-10,0), W); label("B", (8,6), NE); label("C", (8,-6), SE); label("D", (10,0), E); label("X", (2,0), NW); [/asy] $ \textbf{(A)}\ \cos 6^{\circ}\cos 12^{\circ} \sec 18^{\circ} \qquad\textbf{(B)}\ \cos 6^{\circ}\sin 12^{\circ} \csc 18^{\circ} \qquad\textbf{(C)}\ \cos 6^{\circ}\sin 12^{\circ} \sec 18^{\circ} \\ \qquad\textbf{(D)}\ \sin 6^{\circ}\sin 12^{\circ} \csc 18^{\circ} \qquad\textbf{(E)}\ \sin 6^{\circ} \sin 12^{\circ} \sec 18^{\circ} $

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 $n$.

In a triangle $ABC$ with $B = 90^\circ$, $D$ is a point on the segment $BC$ such that the inradii of triangles $ABD$ and $ADC$ are equal. If $\widehat{ADB} = \varphi$ then prove that $\tan^2 (\varphi/2) = \tan (C/2)$.

If $I$ is the incenter of a triangle $ABC$ and $R$ is the radius of its circumcircle then \[AI+BI+CI\leq 3R\]

In a triangle $ABC,$ incircle touches the sides $BC, CA, AB$ at $D, E, F,$ respectively. A circle $\omega$ passing through $A$ and tangent to line $BC$ at $D$ intersects the line segments $BF$ and $CE$ at $K$ and $L,$ respectively. The line passing through $E$ and parallel to $DL$ intersects the line passing through $F$ and parallel to $DK$ at $P.$ If $R_1, R_2, R_3, R_4$ denotes the circumradius of the triangles $AFD, AED, FPD, EPD,$ respectively, prove that $R_1R_4=R_2R_3.$

Let $a$, $b$, $c$ be the three sides of a triangle, and let $\alpha$, $\beta$, $\gamma$, be the angles opposite them. If $a^2+b^2=1989c^2$, find \[ \frac{\cot \gamma}{\cot \alpha+\cot \beta}. \]

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.

Let $ABC$ and $A_{1}B_{1}C_{1}$ be two triangles. Prove that $\frac{a}{a_{1}}+\frac{b}{b_{1}}+\frac{c}{c_{1}}\leq\frac{3R}{2r_{1}}$, where $a = BC$, $b = CA$, $c = AB$ are the sidelengths of triangle $ABC$, where $a_{1}=B_{1}C_{1}$, $b_{1}=C_{1}A_{1}$, $c_{1}=A_{1}B_{1}$ are the sidelengths of triangle $A_{1}B_{1}C_{1}$, where $R$ is the circumradius of triangle $ABC$ and $r_{1}$ is the inradius of triangle $A_{1}B_{1}C_{1}$.

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$.

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

Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area? $ \textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}$

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.

In triangle $ABC$, angles $A$ and $B$ measure 60 degrees and 45 degrees, respectively. The bisector of angle $A$ intersects $\overline{BC}$ at $T$, and $AT=24.$ The area of triangle $ABC$ can be written in the form $a+b\sqrt{c},$ where $a$, $b$, and $c$ are positive integers, and $c$ is not divisible by the square of any prime. Find $a+b+c.$

Let $\triangle{ABC}$ be an acute-angled triangle and let $D$, $E$, $F$ be points on $BC$, $CA$, $AB$, respectively, such that \[\angle{AFE}=\angle{BFD}\mbox{,}\quad\angle{BDF}=\angle{CDE}\quad\mbox{and}\quad\angle{CED}=\angle{AEF}\mbox{.}\] Prove that $D$, $E$ and $F$ are the feet of the perpendiculars through $A$, $B$ and $C$ on $BC$, $CA$ and $AB$, respectively. [i](Swiss Mathematical Olympiad 2011, Final round, problem 2)[/i]

Let $\triangle ABC$ be a triangle. Let $M$ be the midpoint of $BC$ and let $D$ be a point on the interior of side $AB$. The intersection of $AM$ and $CD$ is called $E$. Suppose that $|AD|=|DE|$. Prove that $|AB|=|CE|$.

$ABCD$ is a convex quadrilateral with \[\angle BAC = 30^\circ \]\[\angle CAD = 20^\circ\]\[\angle ABD = 50^\circ\]\[\angle DBC = 30^\circ\] If the diagonals intersect at $P$, show that $PC = PD$.

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]

Given any convex polygon, show that there are three consecutive vertices such that the polygon lies inside the circle through them.

Let $ABC$ be an acute triangle and let $\Gamma$ be its circumcircle. The bisector of $\angle{A}$ intersects $BC$ at $D$, $\Gamma$ at $K$ (different from $A$), and the line through $B$ tangent to $\Gamma$ at $X$. Show that $K$ is the midpoint of $AX$ if and only if $\frac{AD}{DC}=\sqrt{2}$.

Given a triangle $ABC$, let $D$ and $E$ be points on the side $BC$ such that $\angle BAD = \angle CAE$. If $M$ and $N$ are, respectively, the points of tangency of the incircles of the triangles $ABD$ and $ACE$ with the line $BC$, then show that \[\frac{1}{MB}+\frac{1}{MD}= \frac{1}{NC}+\frac{1}{NE}. \]

In quadrilateral $ABCD$, $\angle DAC = 98^{\circ}$, $\angle DBC = 82^\circ$, $\angle BCD = 70^\circ$, and $BC = AD$. Find $\angle ACD.$