In a triangle $ABC$ the bisectors $BB'$ and $CC'$ are drawn. After that, the whole picture except the points $A, B'$, and $C'$ is erased. Restore the triangle using a compass and a ruler. (A.Karlyuchenko)

Let $P$ be a point inside the triangle $ABC$ such that $AB = BC$, $\angle ABC = 80^o$, $\angle PAC = 40^o$ and $\angle ACP = 30^o$. Find $\angle BPC$. (G Galperin)

Let scalene triangle $ABC$ have altitudes $AD, BE, CF$ and circumcenter $O$. The circumcircles of $\triangle ABC$ and $\triangle ADO$ meet at $P \ne A$. The circumcircle of $\triangle ABC$ meets lines $PE$ at $X \ne P$ and $PF$ at $Y \ne P$. Prove that $XY \parallel BC$. [i]Proposed by Daniel Hu[/i]

Triangle $\triangle ABC$ in the figure has area $10$. Points $D$, $E$ and $F$, all distinct from $A$, $B$ and $C$, are on sides $AB$, $BC$ and $CA$ respectively, and $AD = 2$, $DB = 3$. If triangle $\triangle ABE$ and quadrilateral $DBEF$ have equal areas, then that area is [asy] size(200); defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(10,0), C=(8,7), F=7*dir(A--C), E=(10,0)+4*dir(B--C), D=4*dir(A--B); draw(A--B--C--A--E--F--D); pair point=incenter(A,B,C); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$2$", (2,0), S); label("$3$", (7,0), S);[/asy] $ \textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ \frac{5}{3}\sqrt{10}\qquad\textbf{(E)}\ \text{not uniquely determined}$

Four points are chosen uniformly and independently at random in the interior of a given circle. Find the probability that they are the vertices of a convex quadrilateral.

An equilateral triangle with side length 9 is divided into 81 congruent triangles with segments which are parallel to the sides of the triangle. Prove that it cannot be divided into more than 18 parallelograms with sides 1 and 2. [I]Proposed by O.Vanyushina[/i]

a) We are given $n$ circles $O_1, O_2, . . . , O_n$, passing through one point $O$. Let $A_1, . . . , A_n$ denote the second intersection points of $O_1$ with $O_2, O_2$ with $O_3$, etc., $O_n$ with $O_1$, respectively. We choose an arbitrary point $B_1$ on $O_1$ and draw a line segment through $A_1$ and $B_1$ to the second intersection with $O_2$ at $B_2$, then draw a line segment through $A_2$ and $B_2$ to the second intersection with $O_3$ at $B_3$, etc., until we get a point $B_n$ on $O_n$. We draw the line segment through $B_n$ and $A_n$ to the second intersection with $O_1$ at $B_{n+1}$. If $B_k$ and $A_k$ coincide for some $k$, we draw the tangent to $O_k$ through $A_k$ until this tangent intersects $O_{k+1}$ at $B_{k+1}$. Prove that $B_{n+1}$ coincides with $B_1$. b) for $n=3$ the same problem.

Through vertices $A$ and $B$ of triangle $ABC$ are constructed two lines which divide the triangle into four regions (three triangles and one quadrilateral). It is known that three of them have equal area. Prove that one of these three regions is the quadrilateral . (G . Galperin , A . Savin, Moscow)

The circle inscribed in triangle $ABC$ touches $AC$ at point $F$. The perpendicular from point $F$ on $BC$ intersects the bisector of angle $C$ at point $N$. Prove that segment $FN$ is equal to the radius of the circle inscribed in triangle $ABC$. (Oleksii Karliuchenko)

Which of the cones listed below can be formed from a $ 252^\circ$ sector of a circle of radius $ 10$ by aligning the two straight sides? [asy]import graph;unitsize(1.5cm);defaultpen(fontsize(8pt));draw(Arc((0,0),1,-72,180),linewidth(.8pt));draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt));label("$10$",(-0.5,0),S);draw(Arc((0,0),0.1,-72,180));label("$252^{\circ}$",(0.05,0.05),NE);[/asy] [asy] import three; picture mainframe; defaultpen(fontsize(11pt)); picture conePic(picture pic, real r, real h, real sh) { size(pic, 3cm); triple eye = (11, 0, 5); currentprojection = perspective(eye); real R = 1, y = 2; triple center = (0, 0, 0); triple radPt = (0, R, 0); triple negRadPt = (0, -R, 0); triple heightPt = (0, 0, y); draw(pic, arc(center, radPt, negRadPt, heightPt, CW)); draw(pic, arc(center, radPt, negRadPt, heightPt, CCW), linetype("8 8")); draw(pic, center--radPt, linetype("8 8")); draw(pic, center--heightPt, linetype("8 8")); draw(pic, negRadPt--heightPt--radPt); label(pic, (string) r, center--radPt, dir(270)); if (h != 0) { label(pic, (string) h, heightPt--center, dir(0)); } if (sh != 0) { label(pic, (string) sh, heightPt--radPt, dir(0)); } return pic; } picture pic1; pic1 = conePic(pic1, 6, 0, 10); picture pic2; pic2 = conePic(pic2, 6, 10, 0); picture pic3; pic3 = conePic(pic3, 7, 0, 10); picture pic4; pic4 = conePic(pic4, 7, 10, 0); picture pic5; pic5 = conePic(pic5, 8, 0, 10); picture aux1; picture aux2; picture aux3; add(aux1, pic1.fit(), (0,0), W); label(aux1, "$\textbf{(A)}$", (0,0), 22W, linewidth(4)); label(aux1, "$\textbf{(B)}$", (0,0), 3E); add(aux1, pic2.fit(), (0,0), 35E); add(aux2, aux1.fit(), (0,0), W); label(aux2, "$\textbf{(C)}$", (0,0), 3E); add(aux2, pic3.fit(), (0,0), 35E); add(aux3, aux2.fit(), (0,0), W); label(aux3, "$\textbf{(D)}$", (0,0), 3E); add(aux3, pic4.fit(), (0,0), 35E); add(mainframe, aux3.fit(), (0,0), W); label(mainframe, "$\textbf{(E)}$", (0,0), 3E); add(mainframe, pic5.fit(), (0,0), 35E); add(mainframe.fit(), (0,0), N); [/asy]

In a triangle $ABC$ with a right angle at $C$, the angle bisector $AL$ (where $L$ is on segment $BC$) intersects the altitude $CH$ at point $K$. The bisector of angle $BCH$ intersects segment $AB$ at point $M$. Prove that $CK=ML$

Let $ABCD$ be a cyclic quadrilateral. Let $P$, $Q$, $R$ be the feet of the perpendiculars from $D$ to the lines $BC$, $CA$, $AB$, respectively. Show that $PQ=QR$ if and only if the bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$.

Let $O$ be the circumcenter,$R$ be the circumradius, and $k$ be the circumcircle of a triangle $ABC$ . Let $k_1$ be a circle tangent to the rays $AB$ and $AC$, and also internally tangent to $k$. Let $k_2$ be a circle tangent to the rays $AB$ and $AC$ , and also externally tangent to $k$. Let $A_1$ and $A_2$ denote the respective centers of $k_1$ and $k_2$. Prove that: $(OA_1+OA_2)^2-A_1A_2^2 = 4R^2.$

Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$ [i]Proposed by Nazar Serdyuk, Ukraine[/i]

In the triangle $ABC$, $|AB|=15,|BC|=12,|AC|=13$. Let the median $AM$ and bisector $BK$ intersect at point $O$, where $M\in BC,K\in AC$. Let $OL\perp AB,L\in AB$. Prove that $\angle OLK=\angle OLM$.

Triangle $ABC$ has $BC=1$ and $AC=2$. What is the maximum possible value of $\hat{A}$.

Let $\ell$ be the perimeter of an acute-angled triangle $ABC$ which is not an equilateral triangle. Let $P$ be a variable points inside the triangle $ABC$, and let $D,E,F$ be the projections of $P$ on the sides $BC,CA,AB$ respectively. Prove that \[ 2(AF+BD+CE ) = \ell \] if and only if $P$ is collinear with the incenter and the circumcenter of the triangle $ABC$.

Given line $AB$ and point $M$. Find all lines in space passing through $M$ at distance $d$.

Let $ABC$ be a right-angled triangle with $\angle BAC = 90^{\circ}$. Let the height from $A$ cut its side $BC$ at $D$. Let $I, I_B, I_C$ be the incenters of triangles $ABC, ABD, ACD$ respectively. Let also $EB, EC$ be the excenters of $ABC$ with respect to vertices $B$ and $C$ respectively. If $K$ is the point of intersection of the circumcircles of $E_CIB_I$ and $E_BIC_I$, show that $KI$ passes through the midpoint $M$ of side $BC$.

Let $ABC$ be a triangle of perimeter $100$ and $I$ be the point of intersection of its bisectors. Let $M$ be the midpoint of side $BC$. The line parallel to $AB$ drawn by$ I$ cuts the median $AM$ at point $P$ so that $\frac{AP}{PM} =\frac73$. Find the length of side $AB$.

Convex quadrilateral $ABCD$ is given. Lines $BC$ and $AD$ intersect at point $O$, with $B$ lying on the segment $OC$, and $A$ on the segment $OD$. $I$ is the center of the circle inscribed in the $OAB$ triangle, $J$ is the center of the circle exscribed in the triangle $OCD$ touching the side of $CD$ and the extensions of the other two sides. The perpendicular from the midpoint of the segment $IJ$ on the lines $BC$ and $AD$ intersect the corresponding sides of the quadrilateral (not the extension) at points $X$ and $Y$. Prove that the segment $XY$ divides the perimeter of the quadrilateral$ABCD$ in half, and from all segments with this property and ends on $BC$ and $AD$, segment $XY$ has the smallest length.

Given triangle $ABC$, the three altitudes intersect at point $H$. Determine all points $X$ on the side $BC$ so that the symmetric of $H$ wrt point $X$ lies on the circumcircle of triangle $ABC$.

Let $ABC$ be a triangle. Points $D$ and $E$ are placed on $\overline{AC}$ in the order $A$, $D$, $E$, and $C$, and point $F$ lies on $\overline{AB}$ with $EF\parallel BC$. Line segments $\overline{BD}$ and $\overline{EF}$ meet at $X$. If $AD = 1$, $DE = 3$, $EC = 5$, and $EF = 4$, compute $FX$.

Let $ABCD$ be a cyclic quadrilateral with circumcenter $O$, such that $CD$ is not a diameter of its circumcircle. The lines $AD$ and $BC$ intersect at point $P$, so that $A$ lies between $D$ and $P$, and $B$ lies between $C$ and $P$. Suppose triangle $PCD$ is acute and let $H$ be its orthocenter. The points $E$ and $F$ on the lines $BC$ and $AD$, respectively, are such that $BD \parallel HE$ and $AC\parallel HF$. The line through $E$, perpendicular to $BC$, intersects $AD$ at $L$, and the line through $F$, perpendicular to $AD$, intersects $BC$ at $K$. Prove that the points $K$, $L$, $O$ are collinear. [i]Amir Parsa Hosseini Nayeri, Iran[/i]

Given triangle $ABC$, $AL$ bisects angle $\angle BAC$ with $L$ on side $BC$. Lines $LR$ and $LS$ are parallel to $BA$ and $CA$ respectively, $R$ on side $AC$ and$ S$ on side $AB$, respectively. Through point $B$ draw a perpendicular on $AL$, intersecting $LR$ at $M$. If point $D$ is the midpoint of $BC$, prove that that the three points $A, M, D$ lie on a straight line.