Circles $\omega_1$ and $\omega_2$ intersect each other at points $A$ and $B$. Point $C$ lies on the tangent line from $A$ to $\omega_1$ such that $\angle ABC = 90^\circ$. Arbitrary line $\ell$ passes through $C$ and cuts $\omega_2$ at points $P$ and $Q$. Lines $AP$ and $AQ$ cut $\omega_1$ for the second time at points $X$ and $Z$ respectively. Let $Y$ be the foot of altitude from $A$ to $\ell$. Prove that points $X, Y$ and $Z$ are collinear. [i]Proposed by Iman Maghsoudi[/i]

Figure shows a semicircle with diameter $AD$. The chords $AC$ and $BD$ meet at $P$. $Q$ is the foot of the perpendicular from $P$ to $AD$. find $\angle BCQ$ in terms of $\theta$ and $\phi$ . [img]https://cdn.artofproblemsolving.com/attachments/a/2/2781050e842b2dd01b72d246187f4ed434ff69.png[/img]

Let $ABC$ be a triangle with $\angle B > \angle C$. Let $P$ and $Q$ be two different points on line $AC$ such that $\angle PBA = \angle QBA = \angle ACB $ and $A$ is located between $P$ and $C$. Suppose that there exists an interior point $D$ of segment $BQ$ for which $PD=PB$. Let the ray $AD$ intersect the circle $ABC$ at $R \neq A$. Prove that $QB = QR$.

Points $P$ and $Q$ lie respectively on sides $AB$ and $AC$ of a triangle $ABC$ and $BP=CQ$. Segments $BQ$ and $CP$ cross at $R$. Circumscribed circles of triangles $BPR$ and $CQR$ cross again at point $S$ different from $R$. Prove that point $S$ lies on the bisector of angle $BAC$.

Let $ABC$ be a triangle with altitude $AD$ , angle bisectors $AE$ and $BZ$ that intersecting at point $I$. From point $I$ let $IT$ be a perpendicular on $AC$. Also let line $(e)$ be perpendicular on $AC$ at point $A$. Extension of $ET$ intersects line $(e)$ at point $K$. Prove that $AK=AD$.

Square $ABCD$ is rotated $20^\circ$ clockwise about its center to obtain square $EFGH$, as shown below. What is the degree measure of $\angle EAB$? [asy] size(170); defaultpen(linewidth(0.6)); real r = 25; draw(dir(135)--dir(45)--dir(315)--dir(225)--cycle); draw(dir(135-r)--dir(45-r)--dir(315-r)--dir(225-r)--cycle); label("$A$",dir(135),NW); label("$B$",dir(45),NE); label("$C$",dir(315),SE); label("$D$",dir(225),SW); label("$E$",dir(135-r),N); label("$F$",dir(45-r),E); label("$G$",dir(315-r),S); label("$H$",dir(225-r),W); [/asy] $\textbf{(A) }20^\circ\qquad\textbf{(B) }30^\circ\qquad\textbf{(C) }32^\circ\qquad\textbf{(D) }35^\circ\qquad\textbf{(E) }45^\circ$

The diagram below shows the first three figures of a sequence of figures. The first figure shows an equilateral triangle $ABC$ with side length $1$. The leading edge of the triangle going in a clockwise direction around $A$ is labeled $\overline{AB}$ and is darkened in on the figure. The second figure shows the same equilateral triangle with a square with side length $1$ attached to the leading clockwise edge of the triangle. The third figure shows the same triangle and square with a regular pentagon with side length $1$ attached to the leading clockwise edge of the square. The fourth figure in the sequence will be formed by attaching a regular hexagon with side length $1$ to the leading clockwise edge of the pentagon. The hexagon will overlap the triangle. Continue this sequence through the eighth figure. After attaching the last regular figure (a regular decagon), its leading clockwise edge will form an angle of less than $180^\circ$ with the side $\overline{AC}$ of the equilateral triangle. The degree measure of that angle can be written in the form $\tfrac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. [asy] size(250); defaultpen(linewidth(0.7)+fontsize(10)); pair x[],y[],z[]; x[0]=origin; x[1]=(5,0); x[2]=rotate(60,x[0])*x[1]; draw(x[0]--x[1]--x[2]--cycle); for(int i=0;i<=2;i=i+1) { y[i]=x[i]+(15,0); } y[3]=rotate(90,y[0])*y[2]; y[4]=rotate(-90,y[2])*y[0]; draw(y[0]--y[1]--y[2]--y[0]--y[3]--y[4]--y[2]); for(int i=0;i<=4;i=i+1) { z[i]=y[i]+(15,0); } z[5]=rotate(108,z[4])*z[2]; z[6]=rotate(108,z[5])*z[4]; z[7]=rotate(108,z[6])*z[5]; draw(z[0]--z[1]--z[2]--z[0]--z[3]--z[4]--z[2]--z[7]--z[6]--z[5]--z[4]); dot(x[2]^^y[2]^^z[2],linewidth(3)); draw(x[2]--x[0]^^y[2]--y[4]^^z[2]--z[7],linewidth(1)); label("A",(x[2].x,x[2].y-.3),S); label("B",origin,S); label("C",x[1],S);[/asy]

Let $ABC$ be a triangle with circumcircle $k$. The tangents at $A$ and $C$ intersect at $T$. The circumcircle of triangle $ABT$ intersects the line $CT$ at $X$ and $Y$ is the midpoint of $CX$. Prove that the lines $AX$ and $BY$ intersect on $k$.

Let $G, H$ be the centroid and orthocentre of $\vartriangle ABC$ which has an obtuse angle at $\angle B$. Let $\omega$ be the circle with diameter $AG$. $\omega$ intersects $\odot(ABC)$ again at $L \ne A$. The tangent to $\omega$ at $L$ intersects $\odot(ABC)$ at $K \ne L$. Given that $AG = GH$, prove $\angle HKG = 90^o$ . [i]Sam Bealing, United Kingdom[/i]

In triangle $ABC$ $AD$ and $BE$ are altitudes. Let $L$ be a point on $ED$ such that $ED$ is orthogonal to $BL$. If $LB^2=LD\cdot LE$ prove that triangle $ABC$ is isosceles