Incircle of $\triangle ABC$ has centre $I$ and touches sides $AC, AB$ at $E,F$, respectively. The perpendicular bisector of segment $AI$ intersects side $AC$ at $P$. On side $AB$ a point $Q$ is chosen so that $QI \perp FP$. Prove that $EQ \perp AB$.

Point $M$ is the midpoint of the side $AB$ of an acute triangle $ABC$. Point $P$ lies on the segment $AB$, and points $S_1$ and $S_2$ are the centers of the circumcircles of $APC$ and $BPC$, respectively. Show that the midpoint of segment $S_1S_2$ lies on the perpendicular bisector of segment $CM$.

A circle $S$ bisects a circle $S'$ if it cuts $S'$ at opposite ends of a diameter. $S_A$, $S_B$,$S_C$ are circles with distinct centers $A, B, C$ (respectively). Show that $A, B, C$ are collinear iff there is no unique circle $S$ which bisects each of $S_A$, $S_B$,$S_C$ . Show that if there is more than one circle $S$ which bisects each of $S_A$, $S_B$,$S_C$ , then all such circles pass through two fixed points. Find these points. [b]Original Statement:[/b] A circle $S$ is said to cut a circle $\Sigma$ [b]diametrically[/b] if and only if their common chord is a diameter of $\Sigma.$ Let $S_A, S_B, S_C$ be three circles with distinct centres $A,B,C$ respectively. Prove that $A,B,C$ are collinear if and only if there is no unique circle $S$ which cuts each of $S_A, S_B, S_C$ diametrically. Prove further that if there exists more than one circle $S$ which cuts each $S_A, S_B, S_C$ diametrically, then all such circles $S$ pass through two fixed points. Locate these points in relation to the circles $S_A, S_B, S_C.$

Let $ABC$ be a scalene triangle with circumcenter $O$, and let $D$ and $E$ be points inside angle $\measuredangle BAC$ such that $A$ lies on line $DE$, and $\angle ADB=\angle CBA$ and $\angle AEC=\angle BCA$. Let $M$ be the midpoint of $BC$ and $K$ be a point such that $OK$ is perpendicular to $AO$ and $\angle BAK=\angle MAC$. Finally, let $P$ be the intersection of the perpendicular bisectors of $BD$ and $CE$. Show that $KO=KP$. [i]Proposed by Victor Domínguez[/i]

A rectangle intersects a circle as shown: $AB=4$, $BC=5$, and $DE=3$. Then $EF$ equals: [asy]size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F); draw(D--G--H--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, SW); label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); [/asy] $\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$

Triangle $ABC$ is inscribed in circle $\omega$. A circle with chord $BC$ intersects segments $AB$ and $AC$ again at $S$ and $R$, respectively. Segments $BR$ and $CS$ meet at $L$, and rays $LR$ and $LS$ intersect $\omega$ at $D$ and $E$, respectively. The internal angle bisector of $\angle BDE$ meets line $ER$ at $K$. Prove that if $BE = BR$, then $\angle ELK = \tfrac{1}{2} \angle BCD$. [i]Proposed by Evan Chen[/i]

Let $ABC$ be a isosceles triangle, with $AB=AC$. A line $r$ that pass through the incenter $I$ of $ABC$ touches the sides $AB$ and $AC$ at the points $D$ and $E$, respectively. Let $F$ and $G$ be points on $BC$ such that $BF=CE$ and $CG=BD$. Show that the angle $\angle FIG$ is constant when we vary the line $r$.

Let the incircle $k$ of the triangle $ABC$ touch its side $BC$ at $D$. Let the line $AD$ intersect $k$ at $L \neq D$ and denote the excentre of $ABC$ opposite to $A$ by $K$. Let $M$ and $N$ be the midpoints of $BC$ and $KM$ respectively. Prove that the points $B, C, N,$ and $L$ are concyclic.

On the plane, given an angle $ xOy$. $ M$ be a mobile point on ray $ Ox$ and $ N$ a mobile point on ray $ Oy$. Let $ d$ be the external angle bisector of angle $ xOy$ and $ I$ be the intersection of $ d$ with the perpendicular bisector of $ MN$. Let $ P$, $ Q$ be two points lie on $ d$ such that $ IP \equal{} IQ \equal{} IM \equal{} IN$, and let $ K$ the intersection of $ MQ$ and $ NP$. $ 1.$ Prove that $ K$ always lie on a fixed line. $ 2.$ Let $ d_1$ line perpendicular to $ IM$ at $ M$ and $ d_2$ line perpendicular to $ IN$ at $ N$. Assume that there exist the intersections $ E$, $ F$ of $ d_1$, $ d_2$ from $ d$. Prove that $ EN$, $ FM$ and $ OK$ are concurrent.

Consider the cyclic quadrilateral with side lengths $1$, $4$, $8$, $7$ in that order. What is its circumdiameter? Let the answer be of the form $a\sqrt b+c$, for $b$ squarefree. Find $a+b+c$.

In triangle $ABC$, $\angle B = 2\angle C$. Let $P$ and $Q$ be points on the perpendicular bisector of segment $BC$ such that rays $AP$ and $AQ$ trisect $\angle A$. Prove that $PQ < AB$ if and only if $\angle B$ is obtuse.

Let $M$ be the intersection of two diagonal, $AC$ and $BD$, of a rhombus $ABCD$, where angle $A<90^\circ$. Construct $O$ on segment $MC$ so that $OB<OC$ and let $t=\frac{MA}{MO}$, provided that $O \neq M$. Construct a circle that has $O$ as centre and goes through $B$ and $D$. Let the intersections between the circle and $AB$ be $B$ and $X$. Let the intersections between the circle and $BC$ be $B$ and $Y$. Let the intersections of $AC$ with $DX$ and $DY$ be $P$ and $Q$, respectively. Express $\frac{OQ}{OP}$ in terms of $t$.

(B.Frenkin, 8) Does a regular polygon exist such that just half of its diagonals are parallel to its sides?

In acute triangle $\triangle ABC$, point $R$ lies on the perpendicular bisector of $AC$ such that $\overline{CA}$ bisects $\angle BAR$. Let $Q$ be the intersection of lines $AC$ and $BR$. The circumcircle of $\triangle ARC$ intersects segment $\overline{AB}$ at $P\neq A$, with $AP=1$, $PB=5$, and $AQ=2$. Compute $AR$.

Let $H$ and $O$ be the orthocenter and circumcenter of an acute-angled triangle $ABC$, respectively. The perpendicular bisector of $BH$ meets $AB$ and $BC$ at points $A_1$ and $C_1$, respectively. Prove that $OB$ bisects the angle $A_1OC_1$.

Let $ABC$ be an isosceles triangle with $AC=BC$, let $M$ be the midpoint of its side $AC$, and let $Z$ be the line through $C$ perpendicular to $AB$. The circle through the points $B$, $C$, and $M$ intersects the line $Z$ at the points $C$ and $Q$. Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$.

Consider an arbitrary equilateral triangle $KLM$, whose vertices $K, L$ and $M$ lie on the sides $AB, BC$ and $CD$, respectively, of a given square $ABCD$. Find the locus of the midpoints of the sides $KL$ of all such triangles $KLM$.

A triangle $ABC$ is given with barycentre $G$ and circumcentre $O$. The perpendicular bisectors of $GA, GB$ meet at $C_1$,of $GB,GC$ meet at $A _1$, and $GC,GA$ meet at $B_1$. Prove that $O$ is the barycenter of the triangle $A_1B_1C_1$.

You are given a ruler with two parallel straight edges a distance $d$ apart. It may be used (1) to draw the line through two points, (2) given two points a distance $\ge d$ apart, to draw two parallel lines, one through each point, (3) to draw a line parallel to a given line, a distance d away. One can also (4) choose an arbitrary point in the plane, and (5) choose an arbitrary point on a line. Show how to construct : (A) the bisector of a given angle, and (B) the perpendicular to the midpoint of a given line segment.

In triangle $ABC$, $\angle B=2\angle C$ and $\angle A>90^\circ$. Let $D$ be the point on the line $AB$ such that $CD$ is perpendicular to $AC$, and let $M$ be the midpoint of $BC$. Prove that $\angle AMB=\angle DMC$.

Let $ABCD$ be a parallelogram. Let $\omega_1$ be the circle passing through $D$ tangent to $AB$ at $A$. Let $\omega_2$ be the circle passing through $A$ tangent to $CD$ at $D$. The tangents from $B$ to $\omega_1$ touch it at $A$ and $P$. The tangents from $C$ to $\omega_2$ touch it at $D$ and $Q$. Lines $AP$ and $DQ$ intersect at $X$. The perpendicular bisector of $BC$ intersects $AD$ at $R$. Show that the circumcircles of triangles $\triangle PQX$, $\triangle BCR$ are concentric.

The vertices $ A$ and $ B$ of an equilateral triangle $ ABC$ lie on a circle $k$ of radius $1$, and the vertex $ C$ is in the interior of the circle $ k$. A point $ D$, different from $ B$, lies on $ k$ so that $ AD\equal{}AB$. The line $ DC$ intersects $ k$ for the second time at point $ E$. Find the length of the line segment $ CE$.

The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle? $ \textbf{(A)}\ \frac{3\sqrt2}{\pi} \qquad \textbf{(B)}\ \frac{3\sqrt3}{\pi} \qquad \textbf{(C)}\ \sqrt3 \qquad \textbf{(D)}\ \frac{6}{\pi} \qquad \textbf{(E)}\ \sqrt3\pi$

It is known that $ \angle BAC$ is the smallest angle in the triangle $ ABC$. The points $ B$ and $ C$ divide the circumcircle of the triangle into two arcs. Let $ U$ be an interior point of the arc between $ B$ and $ C$ which does not contain $ A$. The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AU$ at $ V$ and $ W$, respectively. The lines $ BV$ and $ CW$ meet at $ T$. Show that $ AU \equal{} TB \plus{} TC$. [i]Alternative formulation:[/i] Four different points $ A,B,C,D$ are chosen on a circle $ \Gamma$ such that the triangle $ BCD$ is not right-angled. Prove that: (a) The perpendicular bisectors of $ AB$ and $ AC$ meet the line $ AD$ at certain points $ W$ and $ V,$ respectively, and that the lines $ CV$ and $ BW$ meet at a certain point $ T.$ (b) The length of one of the line segments $ AD, BT,$ and $ CT$ is the sum of the lengths of the other two.

Find all triangles $ABC$ for which $\angle ACB$ is acute and the interior angle bisector of $BC$ intersects the trisectors $(AX, (AY$ of the angle $\angle BAC$ in the points $N,P$ respectively, such that $AB=NP=2DM$, where $D$ is the foot of the altitude from $A$ on $BC$ and $M$ is the midpoint of the side $BC$.