Triangle $ABC$ is given with its centroid $G$ and cicumcentre $O$ is such that $GO$ is perpendicular to $AG$. Let $A'$ be the second intersection of $AG$ with circumcircle of triangle $ABC$. Let $D$ be the intersection of lines $CA'$ and $AB$ and $E$ the intersection of lines $BA'$ and $AC$. Prove that the circumcentre of triangle $ADE$ is on the circumcircle of triangle $ABC$.

Let $\vartriangle BCD$ be an equilateral triangle and $A$ be a point on the circumcircle of $\vartriangle BCD$ such that $A$ is on the minor arc $BD$. Then, let $P$ be the intersection of $\overline{AB}$ with $\overline{CD}$, $Q$ be the intersection of $\overline{AC}$ with $\overline{DB}$, and $R$ be the intersection of $\overline{AD}$ with $\overline{BC}$. Finally, let $X$, $Y$ , and $Z$ be the feet of the altitudes from $P$, $Q$, and $R$, respectively, in triangle $\vartriangle PQR$. Given $BQ = 3 -\sqrt5$ and $BC = 2$, compute the product of the areas $[\vartriangle XCD] \cdot [\vartriangle Y DB] \cdot [\vartriangle ZBC]$.

In an acute triangle $ABC$ let $K,L,$ and $M$ be the midpoints of sides $AB,BC,$ and $CA,$ respectively. From each of $K,L,$ and $M$ drop two perpendiculars to the other two sides of the triangle; e.g., drop perpendiculars from $K$ to sides $BC$ and $CA,$ etc. The resulting $6$ perpendiculars intersect at points $Q,S,$ and $T$ as in the figure to form a hexagon $KQLSMT$ inside triangle $ABC.$ Prove that the area of this hexagon $KQLSMT$ is half of the area of the original triangle $ABC.$ [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra; diagram by adihaya*/ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 11.888712276357234, xmax = 17.841346447833423, ymin = 10.61620970860601, ymax = 15.470685507068502; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pair A = (12.488234161849352,12.833838721895551), B = (16.50823416184936,15.093838721895553), C = (16.28823416184936,11.353838721895551), K = (14.498234161849355,13.963838721895552), L = (16.39823416184936,13.223838721895552), M = (14.388234161849356,12.093838721895551), D = (13.615830174638527,13.467760858438725), F = (15.75135711740064,11.562938202365055), G = (15.625830174638523,14.597760858438724), H = (16.435061748056253,13.849907687412797), T = (14.02296781802369,12.74356027153236), Q = (16.032967818023693,13.873560271532357), O = (16.325061748056253,11.979907687412794); draw(A--B--C--cycle, zzttqq); draw((13.426050287639166,13.361068683160477)--(13.532742462917415,13.171288796161116)--(13.722522349916774,13.277980971439364)--D--cycle, qqwuqq); draw((14.054227993863618,12.223925334689998)--(14.133240861538676,12.426796211152979)--(13.930369985075695,12.505809078828037)--(13.851357117400637,12.302938202365056)--cycle, qqwuqq); draw((16.337846386707046,12.19724654447628)--(16.12050752964356,12.210031183127075)--(16.107722890992765,11.992692326063588)--O--cycle, qqwuqq); draw((15.830369985075697,11.765809078828037)--(15.627499108612716,11.844821946503092)--(15.54848624093766,11.641951070040111)--F--cycle, qqwuqq); draw((15.436050287639164,14.491068683160476)--(15.542742462917412,14.301288796161115)--(15.73252234991677,14.407980971439365)--G--cycle, qqwuqq); draw((16.217722890992764,13.86269232606359)--(16.20493825234197,13.645353469000101)--(16.42227710940546,13.63256883034931)--H--cycle, qqwuqq); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 1., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw(A--B, zzttqq); draw(B--C, zzttqq); draw(C--A, zzttqq); draw(M--D); draw(K--(13.851357117400637,12.302938202365056)); draw(F--L); draw(L--G); draw(K--H); draw(M--O); /* dots and labels */ dot(A,dotstyle); label("$A$", (12.52502834296331,12.93568440300881), NE * labelscalefactor); dot(B,dotstyle); label("$B$", (16.548187989892043,15.193580123223922), NE * labelscalefactor); dot(C,dotstyle); label("$C$", (16.332661580235147,11.457789022504372), NE * labelscalefactor); dot(K,linewidth(3.pt) + dotstyle); label("$K$", (14.536608166427676,14.02357961365791), NE * labelscalefactor); dot(L,linewidth(3.pt) + dotstyle); label("$L$", (16.43529320388129,13.28463192340569), NE * labelscalefactor); dot(M,linewidth(3.pt) + dotstyle); label("$M$", (14.433976542781535,12.155684063298134), NE * labelscalefactor); dot(D,linewidth(3.pt) + dotstyle); dot((13.851357117400637,12.302938202365056),linewidth(3.pt) + dotstyle); dot(F,linewidth(3.pt) + dotstyle); dot(G,linewidth(3.pt) + dotstyle); dot(H,linewidth(3.pt) + dotstyle); dot((15.922967818023695,12.003560271532354),linewidth(3.pt) + dotstyle); label("$S$", (15.96318773510904,12.063315602016607), NE * labelscalefactor); dot(T,linewidth(3.pt) + dotstyle); label("$T$", (14.064502697655428,12.802263292268826), NE * labelscalefactor); dot(Q,linewidth(3.pt) + dotstyle); label("$Q$", (16.076082521119794,13.931211152376383), NE * labelscalefactor); dot(O,linewidth(3.pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy]

The lengths of the sides of a convex hexagon $ ABCDEF$ satisfy $ AB \equal{} BC$, $ CD \equal{} DE$, $ EF \equal{} FA$. Prove that: \[ \frac {BC}{BE} \plus{} \frac {DE}{DA} \plus{} \frac {FA}{FC} \geq \frac {3}{2}. \]

A triangle $ABC$ is given, in which the segment $BC$ touches the incircle and the corresponding excircle in points $M$ and $N$. If $\angle BAC = 2 \angle MAN$, show that $BC = 2MN$. (N.Beluhov)

Let $ABC$ be an acute triangle with $AB < AC$. Let $J$ be the center of the $A$-excircle of $ABC$. Let $D$ be the projection of $J$ on line $BC$. The internal bisectors of angles $BDJ$ and $JDC$ intersectlines $BJ$ and $JC$ at $X$ and $Y$, respectively. Segments $XY$ and $JD$ intersect at $P$. Let $Q$ be the projection of $A$ on line $BC$. Prove that the internal angle bisector of $QAP$ is perpendicular to line $XY$. [i]Proposed by Dominik Burek, Poland[/i]

Points $A,B,C$, and $M$ are given in the plane. (a) Let $D$ be the point in the plane such that $DA\le CA$ and $DB\le CB$. Prove that there exists point $N$ satisfying $NA\le MA,NB\le MB$, and $ND\le MC$. (b) Let $A',B',C'$ be the points in the plane such that $A'B'\le AB,A'C'\le AC,B'C'\le BC$. Does there exist a point $M'$ which satisfies the inequalities $M'A'\le MA,M'B'\le MB,M'C'\le MC$?

In quadrilateral $ABCD$, diagonals $AC$ and $BD$ intersect at $O$. Denote by $P, Q, R, S$ the orthogonal projections of $O$ onto $AB$ , $BC$ ,$CD$ , $DA$, respectively. Prove that $$PA \cdot AB + RC \cdot CD =\frac12 (AD^2 + BC^2)$$ if and only if $$QB \cdot BC + SD \cdot DA = \frac12(AB ^2 + CD^2)$$

Let $K$ be a point on the side $[AB]$, and $L$ be a point on the side $[BC]$ of the square $ABCD$. If $|AK|=3$, $|KB|=2$, and the distance of $K$ to the line $DL$ is $3$, what is $|BL|:|LC|$? $ \textbf{(A)}\ \frac78 \qquad\textbf{(B)}\ \frac{\sqrt 3}2 \qquad\textbf{(C)}\ \frac 87 \qquad\textbf{(D)}\ \frac 38 \qquad\textbf{(E)}\ \frac{\sqrt 2}2 $

Let $ABCD$ be a rectangle with $AB>BC$. Let points $E,F$ be on side $CD$ such that $CE=ED$ and $BC=CF$. Show that if $AC$ is prependicular to $BE$ then $AB=BF$.

On the infinite chessboard several rectangular pieces are placed whose sides run along the grid lines. Each two have no squares in common, and each consists of an odd number of squares. Prove that these pieces can be painted in four colours such that two pieces painted in the same colour do not share any boundary points.

Suppose $\omega$ is a circle centered at $O$ with radius $8$. Let $AC$ and $BD$ be perpendicular chords of $\omega$. Let $P$ be a point inside quadrilateral $ABCD$ such that the circumcircles of triangles $ABP$ and $CDP$ are tangent, and the circumcircles of triangles $ADP$ and $BCP$ are tangent. If $AC = 2\sqrt{61}$ and $BD = 6\sqrt7$,then $OP$ can be expressed as $\sqrt{a}-\sqrt{b}$ for positive integers $a$ and $b$. Compute $100a + b$.

There is a finite number of polygons in a plane and each two of them have a point in common. Prove that there exists a line which crosses every polygon.

In the triangle $ABC$, $\angle B = 2 \angle C$, $AD$ is altitude, $M$ is the midpoint of the side $BC$. Prove that $AB = 2DM$.

In quadrilateral $ABCD$, angles $A$ and $C$ are equal. Angle bisector of $B$ intersects line $AD$ at point $P$. Perpendicular on $BP$ passing through point $A$ intersects line $BC$ at point $Q$. Prove that the lines $PQ$ and $CD$ are parallel.

In triangle $\triangle ABC$, $D$ lies between $A$ and $C$ and $AC=3AD$, $E$ lies between $B$ and $C$ and $BC=4EC$. $B,G,F,D$ in that order, are on a straight line and $BD=5GF=5FD$. Suppose the area of $\triangle ABC$ is $900$, find the area of the triangle $\triangle EFG$.

In the non-isosceles triangle $ABC$ an altitude from $A$ meets side $BC$ in $D$ . Let $M$ be the midpoint of $BC$ and let $N$ be the reflection of $M$ in $D$ . The circumcirle of triangle $AMN$ intersects the side $AB$ in $P\ne A$ and the side $AC$ in $Q\ne A$ . Prove that $AN,BQ$ and $CP$ are concurrent.

A circle of radius $1$ rolls around a circle of radius $\sqrt{2}$. Initially, the tangent point is colored red. Afterwards, the red points map from one circle to another by contact. How many red points will be on the bigger circle when the center of the smaller one has made $n$ circuits around the bigger one?

In rectangle $ABCD$, $AB=6$ and $AD=8$. Point $M$ is the midpoint of $\overline{AD}$. What is the area of $\triangle AMC$? $\textbf{(A) }12\qquad\textbf{(B) }15\qquad\textbf{(C) }18\qquad\textbf{(D) }20\qquad \textbf{(E) }24$

Consider all pairs of points $(a,b,c)$ and $(d,e, f )$ in the $3$-D coordinate system with $ad +be +c f = -2023$. What is the least positive integer that can be the distance between such a pair of points? [i]Proposed by William Hua[/i]

Let $r_1,r_2,r$, with $r_1 < r_2 < r$, be the radii of three circles $\Gamma_1,\Gamma_2,\Gamma$, respectively. The circles $\Gamma_1,\Gamma_2$ are internally tangent to $\Gamma$ at two distinct points $A,B$ and intersect in two distinct points. Prove that the segment $AB$ contains an intersection point of $\Gamma_1$ and $\Gamma_2$ if and only if $r_1 +r_2 = r$.

The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o'clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction? [asy] size(170); defaultpen(linewidth(0.9)+fontsize(13pt)); draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1) { draw(0.9*dir(90-30*i)--dir(90-30*i)); label("$"+(string) i+"$",0.78*dir(90-30*i)); } dot(origin); draw(shift((0,1.87))*arrow); draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy] $ \textbf{(A) }\text{2 o'clock} \qquad\textbf{(B) }\text{3 o'clock} \qquad\textbf{(C) }\text{4 o'clock} \qquad\textbf{(D) }\text{6 o'clock} \qquad\textbf{(E) }\text{8 o'clock} $

Let $ABC$ be a isosceles triangle with $AB=AC$ an $D$ be the foot of perpendicular of $A$. $P$ be an interior point of triangle $ADC$ such that $m(APB)>90$ and $m(PBD)+m(PAD)=m(PCB)$. $CP$ and $AD$ intersects at $Q$, $BP$ and $AD$ intersects at $R$. Let $T$ be a point on $[AB]$ and $S$ be a point on $[AP$ and not belongs to $[AP]$ satisfying $m(TRB)=m(DQC)$ and $m(PSR)=2m(PAR)$. Show that $RS=RT$

Let $ABC$ be a triangle, let $I$ be its incentre and let $D$ be the orthogonal projection of $I$ on $BC.$ The circle $\odot(ABC)$ crosses the line $AI$ again at $M,$ and the line $DM$ again at $N.$ Prove that the lines $AN$ and $IN$ are perpendicular. [i]Freddie Illingworth & Dominic Yeo[/i]

In hexagon $ABCDEF$, which is nonconvex but not self-intersecting, no pair of opposite sides are parallel. The internal angles satisfy $\angle A=3\angle D$, $\angle C=3\angle F$, and $\angle E=3\angle B$. Furthermore $AB=DE$, $BC=EF$, and $CD=FA$. Prove that diagonals $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ are concurrent.