$D, \: E , \: F$ 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 arbitary triangle, and construct $P,Q,R$ so that each of the angles marked is $30^\circ$. Prove that $\triangle PQR$ is an equilateral triangle. [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair ext30(pair pt1, pair pt2) { pair r1 = pt1+rotate(-30)*(pt2-pt1), r2 = pt2+rotate(30)*(pt1-pt2); draw(anglemark(r1,pt1,pt2,25)); draw(anglemark(pt1,pt2,r2,25)); return intersectionpoints(pt1--r1, pt2--r2)[0]; } pair A = (0,0), B=(10,0), C=(3,7), P=ext30(B,C), Q=ext30(C,A), R=ext30(A,B); draw(A--B--C--A--R--B--P--C--Q--A); draw(P--Q--R--cycle, linetype("8 8")); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$P$", P, NE); label("$Q$", Q, NW); label("$R$", R, S);[/asy]

Consider a triangle $ABC$ with $AB=AC$, and $D$ the foot of the altitude from the vertex $A$. The point $E$ lies on the side $AB$ such that $\angle ACE= \angle ECB=18^{\circ}$. If $AD=3$, find the length of the segment $CE$.

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.

In $\triangle ABC$, $M$ is the midpoint of side $BC$, $AN$ bisects $\angle BAC$, $BN\perp AN$ and $\theta$ is the measure of $\angle BAC$. If sides $AB$ and $AC$ have lengths $14$ and $19$, respectively, then length $MN$ equals [asy] size(230); defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, A=14*dir(36), C=intersectionpoint(B--(9001,0), Circle(A,19)), M=midpoint(B--C), b=A+14*dir(A--C), N=foot(A, B, b); draw(N--B--A--N--M--C--A^^B--M); markscalefactor=0.1; draw(rightanglemark(B,N,A)); pair point=N; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$M$", M, S); label("$N$", N, dir(30)); label("$19$", (A+C)/2, dir(A--C)*dir(90)); label("$14$", (A+B)/2, dir(A--B)*dir(270)); [/asy] $\displaystyle \text{(A)} \ 2 \qquad \text{(B)} \ \frac{5}{2} \qquad \text{(C)} \ \frac{5}{2} - \sin \theta \qquad \text{(D)} \ \frac{5}{2} - \frac{1}{2} \sin \theta \qquad \text{(E)} \ \frac{5}{2} - \frac{1}{2} \sin \left(\frac{1}{2} \theta\right)$

Given that $\sum_{k=1}^{35}\sin 5k=\tan \frac mn,$ where angles are measured in degrees, and $m$ and $n$ are relatively prime positive integers that satisfy $\frac mn<90,$ find $m+n.$

If, in a triangle of sides $a, b, c$, the incircle has radius $\frac{b+c-a}{2}$, what is the magnitude of $\angle A$?

Let $ C_1$ and $ C_2$ be externally tangent circles with radius 2 and 3, respectively. Let $ C_3$ be a circle internally tangent to both $ C_1$ and $ C_2$ at points $ A$ and $ B$, respectively. The tangents to $ C_3$ at $ A$ and $ B$ meet at $ T$, and $ TA \equal{} 4$. Determine the radius of $ C_3$.

Prove that $\cos{\frac{\pi}{7}}-\cos{\frac{2\pi}{7}}+\cos{\frac{3\pi}{7}}=\frac{1}{2}$

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 ABCD be a convex quadrilateral such that $AB + CD = \sqrt{2}AC$ and $BC + DA = \sqrt{2}BD$. Prove that ABCD is a parallelogram.

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 $\alpha + \beta + \gamma = 180^{\circ}$.

A quadrilateral $ABCD$ is inscribed in a circle $\omega$. The tangent to $\omega$ at $A$ intersects the ray $CB$ at $K$, and the tangent to $\omega$ at $B$ intersects the ray $DA$ at $M$. Prove that if $AM=AD$ and $BK=BC$, then $ABCD$ is a trapezoid.

Let $ABC$ be a triangle with $AB = 42$, $AC = 39$, $BC = 45$. Let $E$, $F$ be on the sides $\overline{AC}$ and $\overline{AB}$ such that $AF = 21, AE = 13$. Let $\overline{CF}$ and $\overline{BE}$ intersect at $P$, and let ray $AP$ meet $\overline{BC}$ at $D$. Let $O$ denote the circumcenter of $\triangle DEF$, and $R$ its circumradius. Compute $CO^2-R^2$. [i]Proposed by Yang Liu[/i]

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]

A convex polygon $\mathcal{P}$ in the plane is dissected into smaller convex polygons by drawing all of its diagonals. The lengths of all sides and all diagonals of the polygon $\mathcal{P}$ are rational numbers. Prove that the lengths of all sides of all polygons in the dissection are also rational numbers.

In a triangle $ABC$, let $D$ be the point on the segment $BC$ such that $AB+BD=AC+CD$. Suppose that the points $B$, $C$ and the centroids of triangles $ABD$ and $ACD$ lie on a circle. Prove that $AB=AC$.

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.

Let $ ABCD$ be a cyclic quadrilateral. The side lengths of $ ABCD$ are distinct integers less than $ 15$ such that $ BC\cdot CD\equal{}AB\cdot DA$. What is the largest possible value of $ BD$? $ \textbf{(A)}\ \sqrt{\frac{325}{2}} \qquad \textbf{(B)}\ \sqrt{185} \qquad \textbf{(C)}\ \sqrt{\frac{389}{2}} \qquad \textbf{(D)}\ \sqrt{\frac{425}{2}} \qquad \textbf{(E)}\ \sqrt{\frac{533}{2}}$