Let $ABC$ be an equilateral triangle whose angle bisectors of $B$ and $C$ intersect at $D$. Perpendicular bisectors of $BD$ and $CD$ intersect $BC$ at points $E$ and $Z$ respectively. a) Prove that $BE=EZ=ZC$. b) Find the ratio of the areas of the triangles $BDE$ to $ABC$

Let $ABCD$ be a convex quadrilateral. The rays $AB$ and $DC$ intersect in point $E$. Rays $AD$ and $BC$ intersect in point $F$. The angle bisector of $\angle DCF$ intersects $EF$ in point $K$. Let $I_1$ and $I_2$ be the centers of the inscribed circles in $\Delta ECB$ and $\Delta FCD$. $M$ is the projection of $I_2$ on line $CF$ and $N$ is the projection of $I_1$ on line $BC$. Let $P$ be the reflection of $N$ in $I_1$. If $P,M,K$ are colinear, prove that $ABCD$ is tangential.

In equilateral triangle $ABC$, a circle $\omega$ is drawn such that it is tangent to all three sides of the triangle. A line is drawn from $A$ to point $D$ on segment $BC$ such that $AD$ intersects $\omega$ at points $E$ and $F$. If $EF = 4$ and $AB = 8$, determine $|AE - FD|$.

Let $ABC$ be a triangle, $I$ its incenter, and $\Gamma$ its circumcircle. Let $D$ be the second point of intersection of $AI$ with $\Gamma$. The line parallel to $BC$ through $I$ intersects $AB$ and $AC$ at $P$ and $Q$, respectively. The lines $PD$ and $QD$ intersect $BC$ at $E$ and $F$, respectively. Prove that triangles $IEF$ and $ABC$ are similar.

The teacher drew a coordinate plane on the board and marked some points on this plane. Unfortunately, Vasya's second-grader, who was on duty, erased almost the entire drawing, except for two points $A (1, 2)$ and $B (3,1)$. Will the excellent Andriyko be able to follow these two points to construct the beginning of the coordinate system point $O (0, 0)$? Point A on the board located above and to the left of point $B$.

The midpoints of the sides of a regular hexagon $ABCDEF$ are joined to form a smaller hexagon. What fraction of the area of $ABCDEF$ is enclosed by the smaller hexagon? [asy] size(130); pair A, B, C, D, E, F, G, H, I, J, K, L; A = dir(120); B = dir(60); C = dir(0); D = dir(-60); E = dir(-120); F = dir(180); draw(A--B--C--D--E--F--cycle); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); G = midpoint(A--B); H = midpoint(B--C); I = midpoint(C--D); J = midpoint(D--E); K = midpoint(E--F); L = midpoint(F--A); draw(G--H--I--J--K--L--cycle); label("$A$", A, dir(120)); label("$B$", B, dir(60)); label("$C$", C, dir(0)); label("$D$", D, dir(-60)); label("$E$", E, dir(-120)); label("$F$", F, dir(180)); [/asy] $\textbf{(A)}\ \displaystyle \frac{1}{2} \qquad \textbf{(B)}\ \displaystyle \frac{\sqrt 3}{3} \qquad \textbf{(C)}\ \displaystyle \frac{2}{3} \qquad \textbf{(D)}\ \displaystyle \frac{3}{4} \qquad \textbf{(E)}\ \displaystyle \frac{\sqrt 3}{2}$

Let $\Delta ABC$ be an acute triangle with incenter $I$ and orthocenter $H$. $AI$ meets the circumcircle of $\Delta ABC$ again at $M$. Suppose the length $IM$ is exactly the circumradius of $\Delta ABC$. Show that $AH\geq AI$.

Triangle $ABC$ has side lengths $AC = 3$, $BC = 4$, and $AB = 5$. Let $I$ be the incenter of $\triangle{ABC}$, and let $\mathcal{P}$ be the parabola with focus $I$ and directrix $\overleftrightarrow{AC}$. Suppose that $\mathcal{P}$ intersects $\overline{AB}$ and $\overline{BC}$ at points $D$ and $E$, respectively. Find $DI+EI$.

In $\triangle PQR$, $PR=15$, $QR=20$, and $PQ=25$. Points $A$ and $B$ lie on $\overline{PQ}$, points $C$ and $D$ lie on $\overline{QR}$, and points $E$ and $F$ lie on $\overline{PR}$, with $PA=QB=QC=RD=RE=PF=5$. Find the area of hexagon $ABCDEF$.

Refer to the diagram below. Let $ABCD$ be a cyclic quadrilateral with center $O$. Let the internal angle bisectors of $\angle A$ and $\angle C$ intersect at $I$ and let those of $\angle B$ and $\angle D$ intersect at $J$. Now extend $AB$ and $CD$ to intersect $IJ$ and $P$ and $R$ respectively and let $IJ$ intersect $BC$ and $DA$ at $Q$ and $S$ respectively. Let the midpoints of $PR$ and $QS$ be $M$ and $N$ respectively. Given that $O$ does not lie on the line $IJ$, show that $OM$ and $ON$ are perpendicular.

In the quadrilateral $ABCD$, $AB=AD$, $CB=CD$, $\angle ABC =90^\circ$. $E$, $F$ are on $AB$, $AD$ and $P$, $Q$ are on $EF$($P$ is between $E, Q$), satisfy $\frac{AE}{EP}=\frac{AF}{FQ}$. $X, Y$ are on $CP, CQ$ that satisfy $BX \perp CP, DY \perp CQ$. Prove that $X, P, Q, Y$ are concyclic.

What is the area of the region determined by the points outside a triangle with perimeter length $\pi$ where none of these points has a distance greater than $1$ to any corner of the triangle? $ \textbf{(A)}\ 4\pi \qquad\textbf{(B)}\ 3\pi \qquad\textbf{(C)}\ \dfrac{5\pi}2 \qquad\textbf{(D)}\ 2\pi \qquad\textbf{(E)}\ \dfrac{3\pi}2 $

Given the intersecting lines $ r$ and $s$, consider the lines $u$ and $v$ as such what: a) $u$ is symmetric to $r$ with respect to $s$, b) $v$ is symmetric to $s$ with respect to $r$ . Determine the angle that the given lines must form such that $u$ and $v$ to be coplanar.

Assume that all angles of a triangle $ABC$ are acute. Let $D$ and $E$ be points on the sides $AC$ and $BC$ of the triangle such that $A, B, D,$ and $E$ lie on the same circle. Further suppose the circle through $D,E,$ and $C$ intersects the side $AB$ in two points $X$ and $Y$. Show that the midpoint of $XY$ is the foot of the altitude from $C$ to $AB$.

Circle with center $I$, inscribed in a triangle $ABC$ , touches the sides $BC$ and $AC$ at points $A_1$ and $B_1$ respectively. On rays $A_1I$ and $B_1I$, respectively, let be the points $A_2$ and $B_2$ such that $IA_2=IB_2=R$, where $R$is the radius of the circumscribed circle of the triangle $ABC$. Prove that: a) $AA_2 = BB_2 = OI$ where $O$ is the center of the circumscribed circle of the triangle $ABC$, b) lines $AA_2$ and $BB_2$ intersect on the circumcircle of the triangle $ABC$.

About pentagon $ABCDE$ is known that angle $A$ and angle $C$ are right and that the sides $| AB | = 4$, $| BC | = 5$, $| CD | = 10$, $| DE | = 6$. Furthermore, the point $C'$ that appears by mirroring $C$ in the line $BD$, lies on the line segment $AE$. Find angle $E$.

We are given a triangle $ABC$ and two circles ($k_1$ and $k_2$) so the diameter of $k_1$ is $AB$ and the diameter of $k_2$ is $AC$. Let the intersection of $BC$ line segment and $k_1$ (that isn’t $B$) be $P,$ and the intersection of $BC$ line segment and $k_2$ (that isn’t $B$) be $Q$. We know, that $AB = 3003$ and $AC = 4004$ and $BC = 5005$. What is the distance between $P$ and $Q$?

Consider the triangle $ABC$, where $\angle B= 30^o, \angle C = 15^o$, and $M$ is the midpoint of the side $[BC]$. Let point $N \in (BC)$ be such that $[NC] = [AB]$. Show that $[AN$ is the angle bisector of $MAC$

A honeybug crawls along the honeycombs with the unite length of their hexagons. He has moved from the node $A$ to the node $B$ along the shortest possible trajectory. Prove that the half of his way he moved in one direction.

Let $Oxy$ be a fixed rectangular coordinate system in the plane. Each ordered pair of points $A_1, A_2$ from the same plane which are different from O and have coordinates $x_1, y_1$ and $x_2, y_2$ respectively is associated with real number $f(A_1,A_2)$ in such a way that the following conditions are satisfied: (a) If $OA_1 = OB_1$, $OA_2 = OB_2$ and $A_1A_2 = B_1B_2$ then $f(A_1,A_2) = f(B_1,B_2)$. (b) There exists a polynomial of second degree $F(u,v,w,z)$ such that $f(A_1,A_2)=F(x_1,y_1,x_2,y_2)$. (c) There exists such a number $\phi \in (0,\pi)$ that for every two points $A_1, A_2$ for which $\angle A_1OA_2 = \phi$ is satisfied $f(A_1,A_2) = 0$. (d) If the points $A_1, A_2$ are such that the triangle $OA_1A_2$ is equilateral with side $1$ then$ f(A_1,A_2) = \frac12$. Prove that $f(A_1,A_2) = \overrightarrow{OA_1} \cdot \overrightarrow{OA_2}$ for each ordered pair of points $A_1, A_2$.

Let points $M$ and $N$ lie on sides $AB$ and $BC$ of triangle $ABC$ in such a way that $MN||AC$. Points $M'$ and $N'$ are the reflections of $M$ and $N$ about $BC$ and $AB$ respectively. Let $M'A$ meet $BC$ at $X$, and let $N'C$ meet $AB$ at $Y$. Prove that $A,C,X,Y$ are concyclic.

On a plane there is a finite set of $M$ points, no three of which are collinear . Some points are joined to others by line segments, with each point connected to no more than one line segment . If we have a pair of intersecting line segments $AB$ and $CD$ we decide to replace them with $AC$ and $BD$, which are opposite sides of quadrilateral $ABCD$. In the resulting system of segments we decide to perform a similar substitution, if possible, and so on . Is it possible that such substitutions can be carried out indefinitely? (V.E. Kolosov)

Let $ABC$ be a triangle, let $M$ be the midpoint of $AB$, and let $N$ be the midpoint of $CM$. Let $X$ be a point satisfying both $\angle XMC = \angle MBC$ and $\angle XCM = \angle MCB$ such that $X$ and $B$ lie on opposite sides of $CM$. Let $\omega$ be the circumcircle of triangle $AMX$. $(a)$ Show that $CM$ is tangent to $\omega$. $(b)$ Show that the lines $NX$ and $AC$ intersect on $\omega$

In the figure below the triangles $BCD, CAE$ and $ABF$ are equilateral, and the triangle $ABC$ is right-angled with $\angle A = 90^o$. Prove that $|AD| = |EF|$. [img]https://1.bp.blogspot.com/-QMMhRdej1x8/XzP18QbsXOI/AAAAAAAAMUI/n53OsE8rwZcjB_zpKUXWXq6bg3o8GUfSwCLcBGAsYHQ/s0/2019%2Bmohr%2Bp5.png[/img]

Several knights are arranged on an infinite chessboard. No square is attacked by more than one knight (in particular, a square occupied by a knight can be attacked by one knight but not by two). Sasha outlined a $ 14\times 16$ rectangle. What maximum number of knights can this rectangle contain?