$A,B,C$ and $D$ are four points in that order on the circumference of a circle $K$. $AB$ is perpendicular to $BC$ and $BC$ is perpendicular to $CD$. $X$ is a point on the circumference of the circle between $A$ and $D$. $AX$ extended meets $CD$ extended at $E$ and $DX$ extended meets $BA$ extended at $F$. Prove that the circumcircle of triangle $AXF$ is tangent to the circumcircle of triangle $DXE$ and that the common tangent line passes through the center of the circle $K$.

In an acute triangle $ABC$, points $D,E,F$ are the feet of the altitudes from $A,B,C$, respectively. A line through $D$ parallel to $EF$ meets $AC$ at $Q$ and $AB$ at $R$. Lines $BC$ and $EF$ intersect at $P$. Prove that the circumcircle of triangle $PQR$ passes through the midpoint of $BC$.

Let $\overline{AH_1}, \overline{BH_2}$, and $\overline{CH_3}$ be the altitudes of an acute scalene triangle $ABC$. The incircle of triangle $ABC$ is tangent to $\overline{BC}, \overline{CA},$ and $\overline{AB}$ at $T_1, T_2,$ and $T_3$, respectively. For $k = 1, 2, 3$, let $P_i$ be the point on line $H_iH_{i+1}$ (where $H_4 = H_1$) such that $H_iT_iP_i$ is an acute isosceles triangle with $H_iT_i = H_iP_i$. Prove that the circumcircles of triangles $T_1P_1T_2$, $T_2P_2T_3$, $T_3P_3T_1$ pass through a common point.

Form a pentagon by taking a square of side length $1$ and an equilateral triangle of side length $1$ and placing the triangle so that one of its sides coincides with a side of the square. Then "circumscribe" a circle around the pentagon, passing through three of its vertices, so that the circle passes through exactly one vertex of the equilateral triangle, and exactly two vertices of the square. What is the radius of the circle? $\textbf{(A) }\dfrac23\hspace{14.4em}\textbf{(B) }\dfrac34\hspace{14.4em}\textbf{(C) }1$ $\textbf{(D) }\dfrac54\hspace{14.4em}\textbf{(E) }\dfrac43\hspace{14.4em}\textbf{(F) }\dfrac{\sqrt2}2$ $\textbf{(G) }\dfrac{\sqrt3}2\hspace{13.5em}\textbf{(H) }\sqrt2\hspace{13.8em}\textbf{(I) }\sqrt3$ $\textbf{(J) }\dfrac{1+\sqrt3}2\hspace{12em}\textbf{(K) }\dfrac{2+\sqrt6}2\hspace{11.9em}\textbf{(L) }\dfrac76$ $\textbf{(M) }\dfrac{2+\sqrt6}4\hspace{11.5em}\textbf{(N) }\dfrac45\hspace{14.4em}\textbf{(O) }2007$

In triangle $ABC$ $(b \geq c)$, point $E$ is the midpoint of shorter arc $BC$. If $D$ is the point such that $ED$ is the diameter of circumcircle $ABC$, prove that $\angle DEA = \frac{1}{2}(\beta-\gamma)$

Let $P$ be a point inside a triangle $ABC$ such that $\angle PAC= \angle PCB$. Let the projections of $P$ onto $BC$, $CA$, and $AB$ be $X,Y,Z$ respectively. Let $O$ be the circumcenter of $\triangle XYZ$, $H$ be the foot of the altitude from $B$ to $AC$, $N$ be the midpoint of $AC$, and $T$ be the point such that $TYPO$ is a parallelogram. Show that $\triangle THN$ is similar to $\triangle PBC$. [i]Proposed by Sammy Luo[/i]

Let $ABC$ be a triangle, and points $D,E$ are on $BA,CA$ respectively such that $DB=BC=CE$. Let $O,I$ be the circumcenter, incenter of $\triangle ABC$. Prove that the circumradius of $\triangle ADE$ is equal to $OI$.

Let $ABC$ be a triangle with $\angle B=90$. The incircle of $ABC$ touches the side $BC$ at $D$. The incenters of triangles $ABD$ and $ADC$ are $X$ and $Z$ , respectively. The lines $XZ$ and $AD$ are intersecting at the point $K$. $XZ$ and circumcircle of $ABC$ are intersecting at $U$ and $V$. Let $M$ be the midpoint of line segment $[UV]$ . $AD$ intersects the circumcircle of $ABC$ at $Y$ other than $A$. Prove that $|CY|=2|MK|$ .

Let $M$ be the circumcenter of an acute-angled triangle $ABC$. The circumcircle of triangle $BMA$ intersects $BC$ at $P$ and $AC$ at $Q$. Show that $CM \perp PQ$.

Given an acute triangle $ABC$ with altituties AD and BE. O circumcinter of $ABC$.If o lies on the segment DE then find the value of $sinAsinBcosC$

$L$ is a Line that intersect with the side $AB,BC,AC$ of triangle $ABC$ at $F,D,E$ The line perpendicular to $BC$ from $D$ intersect $AB,AC$ at $A_{1},A_{2}$ respectively Name $B_{1},B_{2},C_{1},C_{2}$ similarly Prove that the circumcenters of $AA_{1}A_{2},BB_{1}B_{2},CC_{1}C_{2}$ are collinear

In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ and odd integer). Let $\alpha$ be the acute angel subtending, from $A$, that segment which contains the mdipoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse fo the triangle. Prove that: \[ \tan{\alpha}=\dfrac{4nh}{(n^2-1)a}. \]

Let $ABC$ be an isosceles triangle with $AC=BC$, let $M$ be the midpoint of its side $AC$, and let $Z$ be the line through $C$ perpendicular to $AB$. The circle through the points $B$, $C$, and $M$ intersects the line $Z$ at the points $C$ and $Q$. Find the radius of the circumcircle of the triangle $ABC$ in terms of $m = CQ$.

It is known that $\odot O$ is the circumcircle of $\vartriangle ABC$ wwith diameter $AB$. The tangents of $\odot O$ at points $B$ and $C$ intersect at $P$ . The line perpendicular to $PA$ at point $A$ intersects the extension of $BC$ at point $D$. Extend $DP$ at length $PE = PB$. If $\angle ADP = 40^o$ , find the measure of $\angle E$.

In acute triangle $ABC$ angle $B$ is greater than$C$. Let $M$ is midpoint of $BC$. $D$ and $E$ are the feet of the altitude from $C$ and $B$ respectively. $K$ and $L$ are midpoint of $ME$ and $MD$ respectively. If $KL$ intersect the line through $A$ parallel to $BC$ in $T$, prove that $TA=TM$.

Three wooden equilateral triangles of side length $18$ inches are placed on axles as shown in the diagram to the right. Each axle is $30$ inches from the other two axles. A $144$-inch leather band is wrapped around the wooden triangles, and a dot at the top corner is painted as shown. The three triangles are then rotated at the same speed and the band rotates without slipping or stretching. Compute the length of the path that the dot travels before it returns to its initial position at the top corner. [asy] size(150); defaultpen(linewidth(0.8)+fontsize(10)); pair A=origin,B=(48,0),C=rotate(60,A)*B; path equi=(0,0)--(18,0)--(9,9*sqrt(3))--cycle,circ=circle(centroid(A,B,C)*18/48,1/3); picture a; fill(a,equi,grey); fill(a,circ,white); add(a); add(shift(15,15*sqrt(3))*a); add(shift(30,0)*a); draw(A--B--C--cycle,linewidth(1)); path top = circle(C,2/3); unfill(top); draw(top); real r=-5/2; draw((9,r+1)--(9,r-1)^^(9,r)--(39,r)^^(39,r-1)--(39,r+1)); label("$30$",(24,r),S); [/asy]

In a triangle $ABC$, with $\widehat{A} > 90^\circ$, let $O$ and $H$ denote its circumcenter and orthocenter, respectively. Let $K$ be the reflection of $H$ with respect to $A$. Prove that $K, O$ and $C$ are collinear if and only if $\widehat{A} - \widehat{B} = 90^\circ$.

In the $ ABC$ triangle, the bisector of $A $ intersects the $ [BC] $ at the point $ A_ {1} $ , and the circle circumscribed to the triangle $ ABC $ at the point $ A_ {2} $. Similarly are defined $ B_ {1} $ and $ B_ {2} $ , as well as $ C_ {1} $ and $ C_ {2} $. Prove that $$ \frac {A_{1}A_{2}}{BA_{2} + A_{2}C} + \frac {B_{1}B_{2}}{CB_{2} + B_{2}A} + \frac {C_{1}C_{2}}{AC_{2} + C_{2}B} \geq \frac {3}{4}$$

For a triangle $ABC$, $O$ is the circumcircle and $D$ is a point on ray $BA$. $E$ and $F$ are points on $O$ so that $DE$ and $DF$ are tangent to $O$ and $EF$ cuts $AC$ at $T(\neq C)$. $P(\neq B,C)$ is a point on the arc $BC$ not containing $A$, and $DP$ cuts $O$ at $Q (\neq P)$. Let $BQ$ and $DT$ meets on $X (\neq Q)$, and $PT$ cuts $O$ at $Y (\neq P)$. Prove that $C,X,Y$ are collinear.

There is a $\triangle ABC$ which $\angle C=90$, and $\bar{AB}=36$ On the circumcircle of $\triangle ABC$, there is $\overarc{BC}$ which does not include point $A$. D is on $\overarc{BC}$. It satisfies $2\times\angle CAD = \angle BAD $ $E: \bar{AD}\cap\bar{BC} $ $ \bar{AE}=20 $ Find $ \bar{BD}^2 $

Given a scalene triangle $ \triangle ABC $. $ B', C' $ are points lie on the rays $ \overrightarrow{AB}, \overrightarrow{AC} $ such that $ \overline{AB'} = \overline{AC}, \overline{AC'} = \overline{AB} $. Now, for an arbitrary point $ P $ in the plane. Let $ Q $ be the reflection point of $ P $ w.r.t $ \overline{BC} $. The intersections of $ \odot{\left(BB'P\right)} $ and $ \odot{\left(CC'P\right)} $ is $ P' $ and the intersections of $ \odot{\left(BB'Q\right)} $ and $ \odot{\left(CC'Q\right)} $ is $ Q' $. Suppose that $ O, O' $ are circumcenters of $ \triangle{ABC}, \triangle{AB'C'} $ Show that 1. $ O', P', Q' $ are colinear 2. $ \overline{O'P'} \cdot \overline{O'Q'} = \overline{OA}^{2} $

Let $ABC$ be an acute scalene triangle with $O$ as its circumcenter. Point $P$ lies inside triangle $ABC$ with $\angle PAB = \angle PBC$ and $\angle PAC = \angle PCB$. Point $Q$ lies on line $BC$ with $QA = QP$. Prove that $\angle AQP = 2\angle OQB$.

Let $I$ be the incenter of the triangle $ABC$, et let $A',B',C'$ be the symmetric of $I$ with respect to the lines $BC,CA,AB$ respectively. It is known that $B$ belongs to the circumcircle of $A'B'C'$. Find $\widehat {ABC}$. Pierre.

The incircle of $ABC$ touches $BC$, $CA$, $AB$ at $K$, $L$, $M$ respectively. The line through $A$ parallel to $LK$ meets $MK$ at $P$, and the line through $A$ parallel to $MK$ meets $LK$ at $Q$. Show that the line $PQ$ bisects $AB$ and bisects $AC$.

Let the points $O$ and $H$ be the circumcenter and orthocenter of an acute angled triangle $ABC.$ Let $D$ be the midpoint of $BC.$ Let $E$ be the point on the angle bisector of $\angle BAC$ such that $AE\perp HE.$ Let $F$ be the point such that $AEHF$ is a rectangle. Prove that $D,E,F$ are collinear.