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

A circle passes through the three vertices of an isosceles triangle that has two sides of length $ 3$ and a base of length $ 2$. What is the area of this circle? $ \textbf{(A)}\ 2\pi\qquad \textbf{(B)}\ \frac {5}{2}\pi\qquad \textbf{(C)}\ \frac {81}{32}\pi\qquad \textbf{(D)}\ 3\pi\qquad \textbf{(E)}\ \frac {7}{2}\pi$

Let $ABCD$ be a tetrahedron with $\angle BAD = 60^{\cdot}$, $\angle BAC = 40^{\cdot}$, $\angle ABD = 80^{\cdot}$, $\angle ABC = 70^{\cdot}$. Prove that the lines $AB$ and $CD$ are perpendicular.

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 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]

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}$

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.

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} $

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 $\tan^2 (\varphi/2) = \tan (C/2)$.

Find the maximum value of \[ \sin{2\alpha} + \sin{2\beta} + \sin{2\gamma} \] where $\alpha,\beta$ and $\gamma$ are positive and $\alpha + \beta + \gamma = 180^{\circ}$.

Consider a regular heptagon ( polygon of $7$ equal sides and angles) $ABCDEFG$ as in the figure below:- $(a).$ Prove $\frac{1}{\sin\frac{\pi}{7}}=\frac{1}{\sin\frac{2\pi}{7}}+\frac{1}{\sin\frac{3\pi}{7}}$ $(b).$ Using $(a)$ or otherwise, show that $\frac{1}{AG}=\frac{1}{AF}+\frac{1}{AE}$ [asy] draw(dir(360/7)..dir(2*360/7),blue); draw(dir(2*360/7)..dir(3*360/7),blue); draw(dir(3*360/7)..dir(4*360/7),blue); draw(dir(4*360/7)..dir(5*360/7),blue); draw(dir(5*360/7)..dir(6*360/7),blue); draw(dir(6*360/7)..dir(7*360/7),blue); draw(dir(7*360/7)..dir(360/7),blue); draw(dir(2*360/7)..dir(4*360/7),blue); draw(dir(4*360/7)..dir(1*360/7),blue); label("$A$",dir(4*360/7),W); label("$B$",dir(5*360/7),S); label("$C$",dir(6*360/7),S); label("$D$",dir(7*360/7),E); label("$E$",dir(1*360/7),E); label("$F$",dir(2*360/7),N); label("$G$",dir(3*360/7),W); [/asy]

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 $ CL$ be a bisector of triangle $ ABC$. Points $ A_1$ and $ B_1$ are the reflections of $ A$ and $ B$ in $ CL$, points $ A_2$ and $ B_2$ are the reflections of $ A$ and $ B$ in $ L$. Let $ O_1$ and $ O_2$ be the circumcenters of triangles $ AB_1B_2$ and $ BA_1A_2$ respectively. Prove that angles $ O_1CA$ and $ O_2CB$ are equal.

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 \[ xy \plus{} yz \plus{} zy \equal{} 1.\]

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$.

In triangle $ABC$, $\angle ABC$ is obtuse. Point $D$ lies on side $AC$ such that $\angle ABD$ is right, and point $E$ lies on side $AC$ between $A$ and $D$ such that $BD$ bisects $\angle EBC$. Find $CE$ given that $AC=35$, $BC=7$, and $BE=5$.

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 $A_{1}B_{1}C_{1}$.

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 $|AB|=|CE|$.

In the figure below, $\triangle ABC$ is an equilateral triangle. [asy] import graph; unitsize(60); axes("$x$", "$y$", (0, 0), (1.5, 1.5), EndArrow); real w = sqrt(3) - 1; pair A = (1, 1); pair B = (0, w); pair C = (w, 0); draw(A -- B -- C -- cycle); dot(Label("$A(1, 1)$", A, NE), A); dot(Label("$B$", B, W), B); dot(Label("$C$", C, S), C); [/asy] Point $A$ has coordinates $(1, 1)$, point $B$ is on the positive $y$-axis, and point $C$ is on the positive $x$-axis. What is the area of $\triangle ABC$?

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.

Given a triangle $ ABC$, let $ r$ be the external bisector of $ \angle ABC$. $ P$ and $ Q$ are the feet of the perpendiculars from $ A$ and $ C$ to $ r$. If $ CP \cap BA \equal{} M$ and $ AQ \cap BC\equal{}N$, show that $ MN$, $ r$ and $ AC$ concur.

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

Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ may be written in the form $a+\sqrt{b},$ where $a$ and $b$ are integers. Find $a+b.$ [asy] import graph; size(7cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(0.75,1.63),SE*lsf); dot((1,3),ds); label("$A$",(0.96,3.14),NE*lsf); dot((0,0),ds); label("$B$",(-0.15,-0.18),NE*lsf); dot((2,0),ds); label("$C$",(2.06,-0.18),NE*lsf); dot((1,0),ds); label("$M$",(0.97,-0.27),NE*lsf); dot((1,0.7),ds); label("$D$",(1.05,0.77),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]

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

Given a triangle $ABC$ with $BC=a$, $CA=b$, $AB=c$, $\angle BAC = \alpha$, $\angle CBA = \beta$, $\angle ACB = \gamma$. Prove that $$ a \sin(\beta-\gamma) + b \sin(\gamma-\alpha) +c\sin(\alpha-\beta) = 0.$$