Let $ABC$ be an equilateral triangle . Let point $D$ lie on side $AB,E$ lie on side $AC, D_1$ and $E_1$ lie on side BC such that $AB=DB+BD_1$ and $AC=CE+CE_1$. Calculate the smallest angle between the lines $DE_1$ and $ED_1$.

An isosceles triangle $ABC$ is given($AB=BC$). Point $D$ lies inside of it such that $\angle ADC=150$, $E$ lies on $CD$ such that $AE=AB$. It turned out that $\angle EBC+\angle BAE=60$. Prove that $\angle BDC+\angle CAE=90$ [i]D. Vasilyev[/i]

Let $ABCDEF$ be a regular hexagon with sidelength s. The points $P$ and $Q$ are on the diagonals $BD$ and $DF$, respectively, such that $BP = DQ = s$. Prove that the three points $C$, $P$ and $Q$ are on a line. [i](Walther Janous)[/i]

Let $ABC$ be an acute triangle with $AC > AB > BC$. The perpendicular bisectors of $AC$ and $AB$ cut line $BC$ at $D$ and $E$ respectively. Let $P$ and $Q$ be points on lines $AC$ and $AB$ respectively, both different from $A$, such that $AB = BP$ and $AC = CQ$, and let $K$ be the intersection of lines $EP$ and $DQ$. Let $M$ be the midpoint of $BC$. Show that $\angle DKA = \angle EKM$.

Let $ABC$ be an acute-angled scalene triangle. Let $P$ be a point on the extension of $AB$ past $B$, and $Q$ a point on the extension of $AC$ past $C$ such that $BPQC$ is a cyclic quadrilateral. Let $N$ be the foot of the perpendicular from $A$ to $BC$. If $NP = NQ$ then prove that $N$ is also the centre of the circumcircle of $APQ$.

In the plane, three distinct non-collinear points $A,B,C$ are marked. In each step, Ege can do one of the following: [list] [*] For marked points $X,Y$, mark the reflection of $X$ across $Y$. [*]For distinct marked points $X,Y,Z,T$ which do not form a parallelogram, mark the center of spiral similarity which takes segment $XY$ to $ZT$. [*] For distinct marked points $X,Y,Z,T$, mark the intersection of lines $XY$ and $ZT$. [/list] No matter how the points $A,B,C$ are marked in the beginning, can Ege always mark, after finitely many moves, a) The circumcenter of $\triangle ABC$. b) The incenter of $\triangle ABC$. Proposed by [i]Deniz Can Karaçelebi[/i]

Point $O$ is the center of the regular octagon $ABCDEFGH$, and $X$ is the midpoint of the side $\overline{AB}.$ What fraction of the area of the octagon is shaded? $\textbf{(A) }\frac{11}{32} \qquad\textbf{(B) }\frac{3}{8} \qquad\textbf{(C) }\frac{13}{32} \qquad\textbf{(D) }\frac{7}{16}\qquad \textbf{(E) }\frac{15}{32}$ [asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("$A$",A,dir(45)); dot("$B$",B,dir(90)); dot("$C$",C,dir(135)); dot("$D$",D,dir(180)); dot("$E$",E,dir(-135)); dot("$F$",F,dir(-90)); dot("$G$",G,dir(-45)); dot("$H$",H,dir(0)); dot("$X$",X,dir(135/2)); dot("$O$",O,dir(0)); draw(E--O--X); [/asy]

Let $ABCDEFGHIJ$ be a regular $10$-gon. Let $T$ be a point inside the $10$-gon, such that the $DTE$ is isosceles: $DT = ET$ , and its angle at the apex is $72^\circ$. Prove that there exists a point $S$ such that $FTS$ and $HIS$ are both isosceles, and for both of them the angle at the apex is $72^\circ$.

Prove that the six sides of any tetrahedron can be the sides of a convex hexagon.

Given a regular $n$-gon $A_{1}A_{2}...A_{n}$ (with $n\geq 3$) in a plane. How many triangles of the kind $A_{i}A_{j}A_{k}$ are obtuse ?

Alice creates the graphs $y=|x-a|$ and $y=c-|x-b|$ , where $a,b,c\in\mathbb{R^+}$. She observes that these two graphs and $x$ axis divides the positive side of the plane ($x,y>0$) into two triangles and a quadrilateral. Find the ratio of sums of two triangles' areas to the area of quadrilateral. [hide=There might be a translation error] In the original statement,it says $XOY$ plane,instead of positive side of the plane. I think these 2 are the same,but I might be wrong [/hide]

The polygon enclosed by the solid lines in the figure consists of $ 4$ congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing? [asy]unitsize(10mm); defaultpen(fontsize(10pt)); pen finedashed=linetype("4 4"); filldraw((1,1)--(2,1)--(2,2)--(4,2)--(4,3)--(1,3)--cycle,grey,black+linewidth(.8pt)); draw((0,1)--(0,3)--(1,3)--(1,4)--(4,4)--(4,3)-- (5,3)--(5,2)--(4,2)--(4,1)--(2,1)--(2,0)--(1,0)--(1,1)--cycle,finedashed); draw((0,2)--(2,2)--(2,4),finedashed); draw((3,1)--(3,4),finedashed); label("$1$",(1.5,0.5)); draw(circle((1.5,0.5),.17)); label("$2$",(2.5,1.5)); draw(circle((2.5,1.5),.17)); label("$3$",(3.5,1.5)); draw(circle((3.5,1.5),.17)); label("$4$",(4.5,2.5)); draw(circle((4.5,2.5),.17)); label("$5$",(3.5,3.5)); draw(circle((3.5,3.5),.17)); label("$6$",(2.5,3.5)); draw(circle((2.5,3.5),.17)); label("$7$",(1.5,3.5)); draw(circle((1.5,3.5),.17)); label("$8$",(0.5,2.5)); draw(circle((0.5,2.5),.17)); label("$9$",(0.5,1.5)); draw(circle((0.5,1.5),.17));[/asy] $ \textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$

Let $C_1$ be an arbitrary point on the side $AB$ of triangle $ABC$. Points $A_1$ and $B_1$ on the rays $BC$ and $AC$ are such that $\angle AC_1B_1 = \angle BC_1A_1 = \angle ACB$. The lines $AA_1$ and $BB_1$ meet in point $C_2$. Prove that all the lines $C_1C_2$ have a common point.

In a circle $S$, a chord $AB$ is drawn and let $M$ be the midpoint of the arc $AB$. Let $P$ be a point in segment $AB$ other than its midpoint. The extension of the segment $MP$ cuts $S$ in $Q$. Let $S_1$ be the circle that is tangent to the AP segments and $MP$, and also is tangent to $S$, and let $S_2$ be the circle that is tangent to the segments $BP$ and $MP$, and also tangent to $S$. The common outer tangent lines to the circles $S_1$ and $S_2$ are cut at $C$. Prove that $\angle MQC = 90^o$.

Let $ABC$ be a triangle with $AB = 5$, $BC = 6$, and $AC = 7$. Let its orthocenter be $H$ and the feet of the altitudes from $A, B, C$ to the opposite sides be $D, E, F$ respectively. Let the line $DF$ intersect the circumcircle of $AHF$ again at $X$. Find the length of $EX$.

In triangle $ABC$, $X$ and $Y$ are the tangency points of incircle (with center $I$) with sides $AB$ and $AC$ respectively. A tangent line to the circumcircle of triangle $ABC$ (with center $O$) at point $A$, intersects the extension of $BC$ at $D$. If $D,X$ and $Y$ are collinear then prove that $D,I$ and $O$ are also collinear. [i]proposed by Amirhossein Zabeti[/i]

A hemisphere is inscribed in a cone so that its base lies on the base of the cone. The ratio of the area of the entire surface of the cone to the area of the hemisphere (without the base) is $\frac{18}5$. Compute the angle at the vertex of the cone.

Opposite sides of a convex hexagon are parallel. Let's call the "height" of such a hexagon a segment with ends on straight lines containing opposite sides and perpendicular to them. Prove that a circle can be circumscribed around this hexagon if and only if its "heights" can be parallelly moved so that they form a triangle. (A. Zaslavsky)

Let $ABCD$ be a circumscribed quadrilateral. Its incircle $\omega$ touches the sides $BC$ and $DA$ at points $E$ and $F$ respectively. It is known that lines $AB,FE$ and $CD$ concur. The circumcircles of triangles $AED$ and $BFC$ meet $\omega$ for the second time at points $E_1$ and $F_1$. Prove that $EF$ is parallel to $E_1 F_1$.

$A, B, C, D$ are points along a circle, in that order. $AC$ intersects $BD$ at $X$. If $BC=6$, $BX=4$, $XD=5$, and $AC=11$, find $AB$

In a triangle ABC we have $\angle C = \angle A + 90^o$. The point $D$ on the continuation of $BC$ is given such that $AC = AD$. A point $E$ in the side of $BC$ in which $A$ doesn’t lie is chosen such that $\angle EBC = \angle A, \angle EDC = \frac{1}{2} \angle A$ . Prove that $\angle CED = \angle ABC$. by Morteza Saghafian

A unit square is cut into $m$ quadrilaterals $Q_1,\ldots ,Q_m$. For each $i=1,\ldots ,m$ let $S_i$ be the sum of the squares of the four sides of $Q_i$. Prove that \[S_1+\ldots +S_m\ge 4\]

Cut the $9 \times 10$ grid rectangle along the grid lines into several squares so that there are exactly two of them with odd sidelengths.

Let $M$ be the midpoint of cathetus $AB$ of triangle $ABC$ with right angle $A$. Point $D$ lies on the median $AN$ of triangle $AMC$ in such a way that the angles $ACD$ and $BCM$ are equal. Prove that the angle $DBC$ is also equal to these angles.

In a scalene triangle $ABC$ with centroid $G$ and circumcircle $\omega$ centred at $O$, the extension of $AG$ meets $\omega$ at $M$; lines $AB$ and $CM$ intersect at $P$; and lines $AC$ and $BM$ intersect at $Q$. Suppose the circumcentre $S$ of the triangle $APQ$ lies on $\omega$ and $A, O, S$ are collinear. Prove that $\angle AGO = 90^{o}$.