$AB$ is a diameter of a circle with center $O$. Let $C$ and $D$ be two different points on the circle on the same side of $AB$, and the lines tangent to the circle at points $C$ and $D$ meet at $E$. Segments $AD$ and $BC$ meet at $F$. Lines $EF$ and $AB$ meet at $M$. Prove that $E,C,M$ and $D$ are concyclic.

In $\vartriangle ABC$, $AB = 13$, $BC = 14,$ and $C A = 15$. Let $E$ and $F$ be the feet of the altitudes from $B$ onto $C A$, and $C$ onto $AB$, respectively. A line $\ell$ is parallel to $EF$ and tangent to the circumcircle of $ABC$ on minor arc $BC$. Let $X$ and $Y$ be the intersections of $\ell$ with $AB$ and $AC$ respectively. Find $X Y$ .

Point $H$ is the orthocenter of the scalene triangle $ABC.$ A line, which passes through point $H$, intersect the sides $AB$ and $AC$ at points $D$ and $E$, respectively, such that $AD=AE.$ Let $M$ be the midpoint of side $BC.$ Line $MH$ intersects the circumscribed circle of triangle $ABC$ at point $K$, which is on the smaller arc $AB$. Prove that Nibab can draw a circle through $A, D, E$ and $K.$

For each finite set $ U$ of nonzero vectors in the plane we define $ l(U)$ to be the length of the vector that is the sum of all vectors in $ U.$ Given a finite set $ V$ of nonzero vectors in the plane, a subset $ B$ of $ V$ is said to be maximal if $ l(B)$ is greater than or equal to $ l(A)$ for each nonempty subset $ A$ of $ V.$ (a) Construct sets of 4 and 5 vectors that have 8 and 10 maximal subsets respectively. (b) Show that, for any set $ V$ consisting of $ n \geq 1$ vectors the number of maximal subsets is less than or equal to $ 2n.$

A non-isosceles triangle $\triangle ABC$ has its incircle tangent to $BC, CA, AB$ at points $D, E, F$. Let the incenter be $I$. Say $AD$ hits the incircle again at $G$, at let the tangent to the incircle at $G$ hit $AC$ at $H$. Let $IH \cap AD = K$, and let the foot of the perpendicular from $I$ to $AD$ be $L$. Prove that $IE \cdot IK= IC \cdot IL$.

Given an acute triangle $ABC$ with $AB>AC$ that has circumcenter $O$. Line $BO$ and $CO$ meet the bisector of $\angle BAC$ at $P$ and $Q$, respectively. Moreover, line $BQ$ and $CP$ meet at $R$. Show that $AR$ is perpendicular to $BC$. [i]Proposer: Soewono and Fajar Yuliawan[/i]

Let $A$ and $B$ be rectangles in the plane and $f : A \rightarrow B$ be a mapping which is uniform on the interior of $A$, maps the boundary of $A$ homeomorphically to the boundary of $B$ by mapping the sides of $A$ to corresponding sides in $B$. Prove that $f$ is an affine transformation.

Let $ABC$ be a triangle with $AB=10$, $AC=11$, and circumradius $6$. Points $D$ and $E$ are located on the circumcircle of $\triangle ABC$ such that $\triangle ADE$ is equilateral. Line segments $\overline{DE}$ and $\overline{BC}$ intersect at $X$. Find $\tfrac{BX}{XC}$.

Let $ABC$ be a triangle, $O$ is the center of the circle circumscribed around it, $AD$ the diameter of this circle. It is known that the lines $CO$ and $DB$ are parallel. Prove that the triangle $ABC$ is isosceles. (Andrey Mostovy)

Andrea inscribed a circle inside a regular pentagon, circumscribed a circle around the pentagon, and calculated the area of the region between the two circles. Bethany did the same with a regular heptagon (7 sides). The areas of the two regions were $ A$ and $ B$, respectively. Each polygon had a side length of $ 2$. Which of the following is true? $ \textbf{(A)}\ A\equal{}\frac{25}{49}B\qquad \textbf{(B)}\ A\equal{}\frac{5}{7}B\qquad \textbf{(C)}\ A\equal{}B\qquad \textbf{(D)}\ A\equal{}\frac{7}{5}B\qquad \textbf{(E)}\ A\equal{}\frac{49}{25}B$

Triangle $ABC$ is inscribed into circle $k$. Points $A_1,B_1, C_1$ on its sides were marked, after this the triangle was erased. Prove that it can be restored uniquely if and only if $AA_1, BB_1$ and $CC_1$ concur.

The quadrilateral $ ABCD$ inscribed in a circle wich has diameter $ BD$. Let $ A',B'$ are symmetric to $ A,B$ with respect to the line $ BD$ and $ AC$ respectively. If $ A'C \cap BD \equal{} P$ and $ AC\cap B'D \equal{} Q$ then prove that $ PQ \perp AC$

Let $ ABCD$ be a convex quadrilateral and let $ P$ and $ Q$ be points in $ ABCD$ such that $ PQDA$ and $ QPBC$ are cyclic quadrilaterals. Suppose that there exists a point $ E$ on the line segment $ PQ$ such that $ \angle PAE \equal{} \angle QDE$ and $ \angle PBE \equal{} \angle QCE$. Show that the quadrilateral $ ABCD$ is cyclic. [i]Proposed by John Cuya, Peru[/i]

A triangle $ABC$ is given. Find all the pairs of points $X,Y$ so that $X$ is on the sides of the triangle, $Y$ is inside the triangle, and four non-intersecting segments from the set $\{XY, AX, AY, BX,BY, CX, CY\}$ divide the triangle $ABC$ into four triangles with equal areas.

Joaquim, José and João participate of the worship of triangle $ABC$. It is well known that $ABC$ is a random triangle, nothing special. According to the dogmas of the worship, when they form a triangle which is similar to $ABC$, they will get immortal. Nevertheless, there is a condition: each person must represent a vertice of the triangle. In this case, Joaquim will represent vertice $A$, José vertice $B$ and João will represent vertice $C$. Thus, they must form a triangle which is similar to $ABC$, in this order. Suppose all three points are in the Euclidean Plane. Once they are very excited to become immortal, they act in the following way: in each instant $t$, Joaquim, for example, will move with constant velocity $v$ to the point in the same semi-plan determined by the line which connects the other two points, and which would create a triangle similar to $ABC$ in the desired order. The other participants act in the same way. If the velocity of all of them is same, and if they initially have a finite, but sufficiently large life, determine if they can get immortal. [i]Observation: Initially, Joaquim, José and João do not represent three collinear points in the plane[/i]

Given that circle $ \omega$ is tangent internally to circle $ \Gamma$ at $ S.$ $ \omega$ touches the chord $ AB$ of $ \Gamma$ at $ T$. Let $ O$ be the center of $ \omega.$ Point $ P$ lies on the line $ AO.$ Show that $ PB\perp AB$ if and only if $ PS\perp TS.$

Let the intersections of $\odot O_1$ and $\odot O_2$ be $A$ and $B$. Point $R$ is on arc $AB$ of $\odot O_1$ and $T$ is on arc $AB$ on $\odot O_2$. $AR$ and $BR$ meet $\odot O_2$ at $C$ and $D$; $AT$ and $BT$ meet $\odot O_1$ at $Q$ and $P$. If $PR$ and $TD$ meet at $E$ and $QR$ and $TC$ meet at $F$, then prove: $AE \cdot BT \cdot BR = BF \cdot AT \cdot AR$.

$S_{1}$ and $S_{2}$ are two circles that intersect at distinct points $P$ and $Q$. $\ell_{1}$ and $\ell_{2}$ are two parallel lines through $P$ and $Q$. $\ell_{1}$ intersects $S_{1}$ and $S_{2}$ at points $A_{1}$ and $A_{2}$, different from $P$, respectively. $\ell_{2}$ intersects $S_{1}$ and $S_{2}$ at points $B_{1}$ and $B_{2}$, different from $Q$, respectively. Show that the perimeters of the triangles $A_{1}QA_{2}$ and $B_{1}PB_{2}$ are equal.

The goal of this problem is to show that the maximum area of a polygon with a fixed number of sides and a fixed perimeter is achieved by a regular polygon. (a) Prove that the polygon with maximum area must be convex. (Hint: If any angle is concave, show that the polygon’s area can be increased.) (b) Prove that if two adjacent sides have different lengths, the area of the polygon can be increased without changing the perimeter. (c) Prove that the polygon with maximum area is equilateral, that is, has all the same side lengths. It is true that when given all four side lengths in order of a quadrilateral, the maximum area is achieved in the unique configuration in which the quadrilateral is cyclic, that is, it can be inscribed in a circle. (d) Prove that in an equilateral polygon, if any two adjacent angles are different then the area of the polygon can be increased without changing the perimeter. (e) Prove that the polygon of maximum area must be equiangular, or have all angles equal. (f ) Prove that the polygon of maximum area is a regular polygon. PS. You had better use hide for answers.

In $\vartriangle ABC, \angle ACB = 90^o, D$ is the foot of the altitude from $C$ to $AB$ and $E$ is the point on the side $BC$ such that $CE = BD/2$. Prove that $AD + CE = AE$.

$P$ is a point in the $\Delta ABC$, $\omega $ is the circumcircle of $\Delta ABC $. $BP \cap \omega = \left\{ {B,{B_1}} \right\}$,$CP \cap \omega = \left\{ {C,{C_1}} \right\}$, $PE \bot AC$，$PF \bot AB$. The radius of the inscribed circle and circumcircle of $\Delta ABC $ is $r,R$. Prove $\frac{{EF}}{{{B_1}{C_1}}} \geqslant \frac{r}{R}$.

$AD$ Is the angle bisector Of $\angle BAC$ Where $D$ lies on the The circumcircle of $\triangle ABC$. Show that $2AD>AB+AC$

Let $\omega=-\tfrac{1}{2}+\tfrac{1}{2}i\sqrt3.$ Let $S$ denote all points in the complex plane of the form $a+b\omega+c\omega^2,$ where $0\leq a \leq 1,0\leq b\leq 1,$ and $0\leq c\leq 1.$ What is the area of $S$? $\textbf{(A) } \frac{1}{2}\sqrt3 \qquad\textbf{(B) } \frac{3}{4}\sqrt3 \qquad\textbf{(C) } \frac{3}{2}\sqrt3\qquad\textbf{(D) } \frac{1}{2}\pi\sqrt3 \qquad\textbf{(E) } \pi$

Given a cyclic quadrilateral $ABCD$, let the diagonals $AC$ and $BD$ meet at $E$ and the lines $AD$ and $BC$ meet at $F$. The midpoints of $AB$ and $CD$ are $G$ and $H$, respectively. Show that $EF$ is tangent at $E$ to the circle through the points $E$, $G$ and $H$. [i]Proposed by David Monk, United Kingdom[/i]

An acute triangle has unit area. Show that there is a point inside the triangle whose distance from each of the vertices is at least $\frac{2}{\sqrt[4]{27}}$.