Points $A$, $B$, $C$, $D$ lie on a circle (in that order) where $AB$ and $CD$ are not parallel. The length of arc $AB$ (which contains the points $D$ and $C$) is twice the length of arc $CD$ (which does not contain the points $A$ and $B$). Let $E$ be a point where $AC=AE$ and $BD=BE$. Prove that if the perpendicular line from point $E$ to the line $AB$ passes through the center of the arc $CD$ (which does not contain the points $A$ and $B$), then $\angle ACB = 108^\circ$.

Square $ABCD$ is shown in the diagram below. Points $E$, $F$, and $G$ are on sides $\overline{AB}$, $\overline{BC}$ and $\overline{DA}$, respectively, so that lengths $\overline{BE}$, $\overline{BF}$, and $\overline{DG}$ are equal. Points $H$ and $I$ are the midpoints of segments $\overline{EF}$ and $\overline{CG}$, respectively. Segment $\overline{GJ}$ is the perpendicular bisector of segment $\overline{HI}$. The ratio of the areas of pentagon $AEHJG$ and quadrilateral $CIHF$ can be written as $\dfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] draw((0,0)--(50,0)--(50,50)--(0,50)--cycle); label("$A$",(0,50),NW); label("$B$",(50,50),NE); label("$C$",(50,0),SE); label("$D$",(0,0),SW); label("$E$",(0,100/3-1),W); label("$F$",(100/3-1,0),S); label("$G$",(20,50),N); label("$H$",((100/3-1)/2,(100/3-1)/2),SW); label("$I$",(35,25),NE); label("$J$",(((100/3-1)/2+35)/2,((100/3-1)/2+25)/2),S); draw((0,100/3-1)--(100/3-1,0)); draw((20,50)--(50,0)); draw((100/6-1/2,100/6-1/2)--(35,25)); draw((((100/3-1)/2+35)/2,((100/3-1)/2+25)/2)--(20,50)); [/asy]

Let $ABC$ be an acute triangle and let $\omega$ be its circumcircle. Let the tangents to $\omega$ through $B,C$ meet each other at point $P$. Prove that the perpendicular bisector of $AB$ and the parallel to $AB$ through $P$ meet at line $AC$.

As shown in the following figure, point $ P$ lies on the circumcicle of triangle $ ABC.$ Lines $ AB$ and $ CP$ meet at $ E,$ and lines $ AC$ and $ BP$ meet at $ F.$ The perpendicular bisector of line segment $ AB$ meets line segment $ AC$ at $ K,$ and the perpendicular bisector of line segment $ AC$ meets line segment $ AB$ at $ J.$ Prove that \[ \left(\frac{CE}{BF} \right)^2 \equal{} \frac{AJ \cdot JE}{AK \cdot KF}.\]

Let $F=A_0A_1...A_n$ be a convex polygon in the plane. Define for all $1 \leq k \leq n-1$ the operation $f_k$ which replaces $F$ with a new polygon $f_k(F)=A_0A_1..A_{k-1}A_k^\prime A_{k+1}...A_n$ where $A_k^\prime$ is the symmetric of $A_k$ with respect to the perpendicular bisector of $A_{k-1}A_{k+1}$. Prove that $(f_1\circ f_2 \circ f_3 \circ...\circ f_{n-1})^n(F)=F$.

Let two chords $AC$ and $BD$ of a circle $k$ meet at the point $K$, and let $O$ be the center of $k$. Let $M$ and $N$ be the circumcenters of triangles $AKB$ and $CKD$. Show that the quadrilateral $OMKN$ is a parallelogram.

$A,B,C$ are points that three complex numbers $z_0=a\text{i},z_1=\frac{1}{2}+b\text{i},z_2=1+c\text{i}(a,b,c\in\mathbb{R})$ refer to on complex plane (not collinear). Prove that curve $Z=Z_0\cos^4t+2Z_1\cos^2t\sin^2t+Z_2\sin^4t(t\in\mathbb{R})$ has only one common point with the perpendicular bisector of $AC$, and find the point.

Points $A, B$ move with equal speeds along two equal circles. Prove that the perpendicular bisector of $AB$ passes through a fixed point.

For every $n\geq 3$, determine all the configurations of $n$ distinct points $X_1,X_2,\ldots,X_n$ in the plane, with the property that for any pair of distinct points $X_i$, $X_j$ there exists a permutation $\sigma$ of the integers $\{1,\ldots,n\}$, such that $\textrm{d}(X_i,X_k) = \textrm{d}(X_j,X_{\sigma(k)})$ for all $1\leq k \leq n$. (We write $\textrm{d}(X,Y)$ to denote the distance between points $X$ and $Y$.) [i](United Kingdom) Luke Betts[/i]

Let $ABC$ be a triangle in which $\angle B$ is obtuse.Let $\Gamma$ be its circumcircle and $O$ be the centre of $\Gamma$.Let the tangent to $\Gamma$ at $C$ intersect the line $AB$ in $B_1$.Let $O_1$ be the circumcentre of the circumcircle $\Gamma_1$ of $\triangle AB_1 C$.Take any point $B_2$ on the segment $BB_1$ different from $B,B_1$.Let $C_1$ be the point of contact of the tangent from $B_2$ to $\Gamma$ which is closer to $C$.Let $O_2$ be the circumcentre of $\triangle AB_2 C_1$.Prove that $O,O_2,O_1,C_1,C$ are concyclic if $OO_2\perp AO_1$.

Equations of non-intersecting curves are $y = ax^2 + bx + c$ and $y = dx^2 + ex + f$ where $ad < 0.$ Prove that there is a line of the plane which does not meet either of the curves.

Points $B,C$ vary on two fixed rays emanating from point $A$ such that $AB+AC$ is constant. Show that there is a point $D$, other than $A$, such that the circumcircle of triangle $ABC$ passes through $D$ for all possible choices of $B, C$.

Let $AL$ and $BK$ be angle bisectors in the non-isosceles triangle $ABC$ ($L$ lies on the side $BC$, $K$ lies on the side $AC$). The perpendicular bisector of $BK$ intersects the line $AL$ at point $M$. Point $N$ lies on the line $BK$ such that $LN$ is parallel to $MK$. Prove that $LN = NA$.

Consider a line segment $[AB]$ with midpoint $M$ and perpendicular bisector $m$. For each point$ X \ne M$ on m consider we are the intersection point $Y$ of the line $BX$ with the bisector from the angle $\angle BAX$. As $X$ approaches $M$, then approaches $Y$ to a point of $[AB]$. Which? [img]https://cdn.artofproblemsolving.com/attachments/a/3/17d72a23011a9ec22deb20184717cc9c020a2b.png[/img] [hide=original wording]Beschouw een lijnstuk [AB] met midden M en middelloodlijn m. Voor elk punt X 6= M op m beschouwenwe het snijpunt Y van de rechte BX met de bissectrice van de hoek < BAX . Als X tot M nadert, dan nadert Y tot een punt van [AB]. Welk? [/hide]

Two circles $ (O_1)$ and $ (O_2)$ touch externally at the point $C$ and internally at the points $A$ and $B$ respectively with another circle $(O)$. Suppose that the common tangent of $ (O_1)$ and $ (O_2)$ at $C$ meets $(O)$ at $P$ such that $PA=PB$. Prove that $PO$ is perpendicular to $AB$.

Let $\alpha, \beta$ and $\gamma$ denote the angles of the triangle $ABC$. The perpendicular bisector of $AB$ intersects $BC$ at the point $X$, the perpendicular bisector of $AC$ intersects it at $Y$. Prove that $\tan(\beta) \cdot \tan(\gamma) = 3$ implies $BC= XY$ (or in other words: Prove that a sufficient condition for $BC = XY$ is $\tan(\beta) \cdot \tan(\gamma) = 3$). Show that this condition is not necessary, and give a necessary and sufficient condition for $BC = XY$.

The perpendicular bisectors of the sides AB and BC of a triangle ABC meet the lines BC and AB at the points X and Z, respectively. The angle bisectors of the angles XAC and ZCA intersect at a point B'. Similarly, define two points C' and A'. Prove that the points A', B', C' lie on one line through the incenter I of triangle ABC. [i]Extension:[/i] Prove that the points A', B', C' lie on the line OI, where O is the circumcenter and I is the incenter of triangle ABC. Darij

Let $ABC$ be an acute, non-isosceles triangle with the circumcircle $(O)$. Denote $D, E$ as the midpoints of $AB,AC$ respectively. Two circles $(ABE)$ and $(ACD)$ intersect at $K$ differs from $A$. Suppose that the ray $AK$ intersects $(O)$ at $L$. The line $LB$ meets $(ABE)$ at the second point $M$ and the line $LC$ meets $(ACD)$ at the second point $N$. a) Prove that $M, K, N$ collinear and $MN$ perpendicular to $OL$. b) Prove that $K$ is the midpoint of $MN$

Given is an acute triangle $ABC$ $\left(AB<AC<BC\right)$,inscribed in circle $c(O,R)$.The perpendicular bisector of the angle bisector $AD$ $\left(D\in BC\right)$ intersects $c$ at $K,L$ ($K$ lies on the small arc $\overarc{AB}$).The circle $c_1(K,KA)$ intersects $c$ at $T$ and the circle $c_2(L,LA)$ intersects $c$ at $S$.Prove that $\angle{BAT}=\angle{CAS}$. [hide=Diagram][asy]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 = -6.94236331697463, xmax = 15.849400903703716, ymin = -5.002235438802758, ymax = 7.893104843949444; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen qqqqtt = rgb(0.,0.,0.2); draw((1.8318261909633622,3.572783369254345)--(0.,0.)--(6.,0.)--cycle, aqaqaq); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-117.14497824050169,-101.88970202103212)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); draw(arc((1.8318261909633622,3.572783369254345),0.6426249310341638,-55.85706977865775,-40.60179355918817)--(1.8318261909633622,3.572783369254345)--cycle, qqqqtt); /* draw figures */ draw((1.8318261909633622,3.572783369254345)--(0.,0.), uququq); draw((0.,0.)--(6.,0.), uququq); draw((6.,0.)--(1.8318261909633622,3.572783369254345), uququq); draw(circle((3.,0.7178452373968209), 3.0846882800136055)); draw((2.5345020274407277,0.)--(1.8318261909633622,3.572783369254345)); draw(circle((-0.01850947366601585,1.3533783539547308), 2.889550258039566)); draw(circle((5.553011501106743,2.4491551634556963), 3.887127532933951)); draw((-0.01850947366601585,1.3533783539547308)--(5.553011501106743,2.4491551634556963), linetype("2 2")); draw((1.8318261909633622,3.572783369254345)--(0.7798408954511686,-1.423695174396108)); draw((1.8318261909633622,3.572783369254345)--(5.22015910454883,-1.4236951743961088)); /* dots and labels */ dot((1.8318261909633622,3.572783369254345),linewidth(3.pt) + dotstyle); label("$A$", (1.5831274347452782,3.951671933606579), NE * labelscalefactor); dot((0.,0.),linewidth(3.pt) + dotstyle); label("$B$", (-0.6,0.05), NE * labelscalefactor); dot((6.,0.),linewidth(3.pt) + dotstyle); label("$C$", (6.188606107156787,0.07450151636712989), NE * labelscalefactor); dot((2.5345020274407277,0.),linewidth(3.pt) + dotstyle); label("$D$", (2.3,-0.7), NE * labelscalefactor); dot((-0.01850947366601585,1.3533783539547308),linewidth(3.pt) + dotstyle); label("$K$", (-0.3447473583572136,1.6382221818835927), NE * labelscalefactor); dot((5.553011501106743,2.4491551634556963),linewidth(3.pt) + dotstyle); label("$L$", (5.631664500260511,2.580738747400365), NE * labelscalefactor); dot((0.7798408954511686,-1.423695174396108),linewidth(3.pt) + dotstyle); label("$T$", (0.5977692071595602,-1.960477431907719), NE * labelscalefactor); dot((5.22015910454883,-1.4236951743961088),linewidth(3.pt) + dotstyle); label("$S$", (5.160406217502124,-1.8747941077698307), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/hide]

Lines tangent to circle $O$ in points $A$ and $B$, intersect in point $P$. Point $Z$ is the center of $O$. On the minor arc $AB$, point $C$ is chosen not on the midpoint of the arc. Lines $AC$ and $PB$ intersect at point $D$. Lines $BC$ and $AP$ intersect at point $E$. Prove that the circumcentres of triangles $ACE$, $BCD$, and $PCZ$ are collinear.

Let $ABCD$ be a rectangle. Determine the set of all points $P$ from the region between the parallel lines $AB$ and $CD$ such that $\angle APB=\angle CPD$.

Let $ ABC$ be a triangle in which $ AB \equal{} AC$ and let $ I$ be its in-centre. Suppose $ BC \equal{} AB \plus{} AI$. Find $ \angle{BAC}$

At most how many points with integer coordinates are there over a circle with center of $(\sqrt{20}, \sqrt{10})$ in the $xy$-plane? $ \textbf{(A)}\ 8 \qquad\textbf{(B)}\ 4 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ \text{None} $

$ABCD$ is a cyclic quadrilateral, with diagonals $AC,BD$ perpendicular to each other. Let point $F$ be on side $BC$, the parallel line $EF$ to $AC$ intersect $AB$ at point $E$, line $FG$ parallel to $BD$ intersect $CD$ at $G$. Let the projection of $E$ onto $CD$ be $P$, projection of $F$ onto $DA$ be $Q$, projection of $G$ onto $AB$ be $R$. Prove that $QF$ bisects $\angle PQR$.

Let $ABC$ be an acute-angled triangle such that $AB<AC$. Let $D$ be the point of intersection of the perpendicular bisector of the side $BC$ with the side $AC$. Let $P$ be a point on the shorter arc $AC$ of the circumcircle of the triangle $ABC$ such that $DP \parallel BC$. Finally, let $M$ be the midpoint of the side $AB$. Prove that $\angle APD=\angle MPB$. [i]Proposed by Dominik Burek, Poland[/i]