Found problems: 3585
Let $\omega_1$ and $\omega_2$ be two different circles that intersect at two different points, $X$ and $Y$. Let lines $l_1$ and $l_2$ be common tangent lines of these circles such that $l_1$ is tangent $\omega_1$ at $A$ and $\omega_2$ at $C$ and $l_2$ is tangent $\omega_1$ at $B$ and $\omega_2$ at $D$. Let $Z$ be the reflection of $Y$ respect to $l_1$ and let $BC$ and $\omega_1$ meet at $K$ for the second time. Let $AD$ and $\omega_2$ meet at $L$ for the second time. Prove that the line tangent to $\omega_1$ and passes through $K$ and the line tangent to $\omega_2$ and passes through $L$ meet on the line $XZ$.
Let $ABC$ be an acute scalene triangle, $H$ its orthocenter and $G$ its geocenter. The circumference with diameter $AH$ cuts the circumcircle of $BHC$ in $A'$ ($A' \neq H$). Points $B'$ and $C'$ are defined similarly. Show that the points $A'$, $B'$, $C'$, and $G$ lie in one circumference.
Let $ABC$ be an acute,non-isosceles triangle with $AB<AC<BC$.Let $D,E,Z$ be the midpoints of $BC,AC,AB$ respectively and segments $BK,CL$ are altitudes.In the extension of $DZ$ we take a point $M$ such that the parallel from $M$ to $KL$ crosses the extensions of $CA,BA,DE$ at $S,T,N$ respectively (we extend $CA$ to $A$-side and $BA$ to $A$-side and $DE$ to $E$-side).If the circumcirle $(c_{1})$ of $\triangle{MBD}$ crosses the line $DN$ at $R$ and the circumcirle $(c_{2})$ of $\triangle{NCD}$ crosses the line $DM$ at $P$ prove that $ST\parallel PR$.
Points $ A_{1}$, $ B_{1}$, $ C_{1}$ are chosen on the sides $ BC$, $ CA$, $ AB$ of a triangle $ ABC$ respectively. The circumcircles of triangles $ AB_{1}C_{1}$, $ BC_{1}A_{1}$, $ CA_{1}B_{1}$ intersect the circumcircle of triangle $ ABC$ again at points $ A_{2}$, $ B_{2}$, $ C_{2}$ respectively ($ A_{2}\neq A, B_{2}\neq B, C_{2}\neq C$). Points $ A_{3}$, $ B_{3}$, $ C_{3}$ are symmetric to $ A_{1}$, $ B_{1}$, $ C_{1}$ with respect to the midpoints of the sides $ BC$, $ CA$, $ AB$ respectively. Prove that the triangles $ A_{2}B_{2}C_{2}$ and $ A_{3}B_{3}C_{3}$ are similar.
Let $(O)$ be a fixed circle with the radius $R$. Let $A$ and $B$ be fixed points in $(O)$ such that $A,B,O$ are not collinear. Consider a variable point $C$ lying on $(O)$ ($C\neq A,B$). Construct two circles $(O_1),(O_2)$ passing through $A,B$ and tangent to $BC,AC$ at $C$, respectively. The circle $(O_1)$ intersects the circle $(O_2)$ in $D$ ($D\neq C$). Prove that:
a) \[ CD\leq R \]
b) The line $CD$ passes through a point independent of $C$ (i.e. there exists a fixed point on the line $CD$ when $C$ lies on $(O)$).
In triangle $ABC$, $P$ is a point on altitude $AD$. $Q,R$ are the feet of the perpendiculars from $P$ to $AB,AC$, and $QP,RP$ meet $BC$ at $S$ and $T$ respectively. the circumcircles of $BQS$ and $CRT$ meet $QR$ at $X,Y$.
a) Prove $SX,TY, AD$ are concurrent at a point $Z$.
b) Prove $Z$ is on $QR$ iff $Z=H$, where $H$ is the orthocenter of $ABC$.
[i]Ray Li.[/i]
In a scalene triangle $ABC$, the points $E$ and $F$ are the foot of altitudes drawn from $B$ and $C$, respectively. The points $X$ and $Y$ are the reflections of the vertices $B$ and $C$ to the line $EF$, respectively. Let the circumcircles of the $\triangle ABC$ and $\triangle AEF$ intersect at $T$ for the second time. Show that the four points $A, X, Y, T$ lie on a single circle.
Let $PQRS$ be a cyclic quadrilateral such that the segments $PQ$ and $RS$ are not parallel. Consider the set of circles through $P$ and $Q$, and the set of circles through $R$ and $S$. Determine the set $A$ of points of tangency of circles in these two sets.
Let $ABC$ be a triangle with $AB < AC$ and let $D{}$ be the other intersection point of the angle bisector of $\angle A$ with the circumcircle of the triangle $ABC$. Let $E{}$ and $F{}$ be points on the sides $AB$ and $AC$ respectively, such that $AE = AF$ and let $P{}$ be the point of intersection of $AD$ and $EF$. Let $M{}$ be the midpoint of $BC{}$. Prove that $AM$ and the circumcircles of the triangles $AEF$ and $PMD$ pass through a common point.
In $\triangle ABC$, $H$ is the orthocenter, and $AD,BE$ are arbitrary cevians. Let $\omega_1, \omega_2$ denote the circles with diameters $AD$ and $BE$, respectively. $HD,HE$ meet $\omega_1,\omega_2$ again at $F,G$. $DE$ meets $\omega_1,\omega_2$ again at $P_1,P_2$ respectively. $FG$ meets $\omega_1,\omega_2$ again $Q_1,Q_2$ respectively. $P_1H,Q_1H$ meet $\omega_1$ at $R_1,S_1$ respectively. $P_2H,Q_2H$ meet $\omega_2$ at $R_2,S_2$ respectively. Let $P_1Q_1\cap P_2Q_2 = X$, and $R_1S_1\cap R_2S_2=Y$. Prove that $X,Y,H$ are collinear.
[i]Ray Li.[/i]
Let $ABCD$ be a rectangle inscribed in circle $\Gamma$, and let $P$ be a point on minor arc $AB$ of $\Gamma$. Suppose that $P A \cdot P B = 2$, $P C \cdot P D = 18$, and $P B \cdot P C = 9$. The area of rectangle $ABCD$ can be expressed as $\frac{a\sqrt{b}}{c}$ , where $a$ and $c$ are relatively prime positive integers and $b$ is a squarefree positive integer. Compute $100a + 10b + c$.
$ABC$ is an acute triangle with orthocenter $H$. Point $P$ is in triangle $BHC$ that $\angle HPC = 3 \angle HBC $ and $\angle HPB =3 \angle HCB $. Reflection of point $P$ through $BH,CH$ is $X,Y$. if $S$ is the center of circumcircle of $AXY$ , Prove that:
$$\angle BAS = \angle CAP$$
[i]Proposed by Pouria Mahmoudkhan Shirazi [/i]
Let $ABC$ be a triangle with $\angle{C} = 90^{\circ}$, and let $H$ be the foot of the altitude from $C$. A point $D$ is chosen inside the triangle $CBH$ so that $CH$ bisects $AD$. Let $P$ be the intersection point of the lines $BD$ and $CH$. Let $\omega$ be the semicircle with diameter $BD$ that meets the segment $CB$ at an interior point. A line through $P$ is tangent to $\omega$ at $Q$. Prove that the lines $CQ$ and $AD$ meet on $\omega$.
Let $\omega$ be the circumcircle of isosceles triangle $ABC$ ($AB=AC$). Points $P$ and $Q$ lie on $\omega$ and $BC$ respectively such that $AP=AQ$ .$AP$ and $BC$ intersect at $R$. Prove that the tangents from $B$ and $C$ to the incircle of $\triangle AQR$ (different from $BC$) are concurrent on $\omega$.
[i]Proposed by Ali Zamani, Hooman Fattahi[/i]
Let $ABC$ be an acute-angled triangle with $AB<AC$ and let $O$ be the centre of its circumcircle $\omega$. Let $D$ be a point on the line segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M$, $N$ and $P$ are the midpoints of the line segments $BE$, $OD$ and $AC$, respectively, show that the points $M$, $N$ and $P$ are collinear.
Let $ABC$ be a triangle with $CA \neq CB$. Let $D$, $F$, and $G$ be the midpoints of the sides $AB$, $AC$, and $BC$ respectively. A circle $\Gamma$ passing through $C$ and tangent to $AB$ at $D$ meets the segments $AF$ and $BG$ at $H$ and $I$, respectively. The points $H'$ and $I'$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H'I'$ meets $CD$ and $FG$ at $Q$ and $M$, respectively. The line $CM$ meets $\Gamma$ again at $P$. Prove that $CQ = QP$.
[i]Proposed by El Salvador[/i]
Let $ABCD$ be a cyclic quadrilateral with $\angle BAD < \angle ADC$. Let $M$ be the midpoint of the arc $CD$ not containing $A$. Suppose there is a point $P$ inside $ABCD$ such that $\angle ADB = \angle CPD$ and $\angle ADP = \angle PCB$.
Prove that lines $AD, PM$, and $BC$ are concurrent.
Let $P$ be a point inside triangle $ABC$, and suppose lines $AP$, $BP$, $CP$ meet the circumcircle again at $T$, $S$, $R$ (here $T \neq A$, $S \neq B$, $R \neq C$). Let $U$ be any point in the interior of $PT$. A line through $U$ parallel to $AB$ meets $CR$ at $W$, and the line through $U$ parallel to $AC$ meets $BS$ again at $V$. Finally, the line through $B$ parallel to $CP$ and the line through $C$ parallel to $BP$ intersect at point $Q$. Given that $RS$ and $VW$ are parallel, prove that $\angle CAP = \angle BAQ$.
Let $ABC$ be an acute-angled triangle and $\Gamma$ be a circle with $AB$ as diameter intersecting $BC$ and $CA$ at $F ( \not= B)$ and $E (\not= A)$ respectively. Tangents are drawn at $E$ and $F$ to $\Gamma$ intersect at $P$. Show that the ratio of the circumcentre of triangle $ABC$ to that if $EFP$ is a rational number.
On a circle there are $2n+1$ points, dividing it into equal arcs ($n\ge 2$). Two players take turns to erase one point. If after one player's turn, it turned out that all the triangles formed by the remaining points on the circle were obtuse, then the player wins and the game ends.
Who has a winning strategy: the starting player or his opponent?
Let there be an acute triangle $ABC$ with orthocenter $H$. Let $BH, CH$ hit the circumcircle of $\triangle ABC$ at $D, E$. Let $P$ be a point on $AB$, between $B$ and the foot of the perpendicular from $C$ to $AB$. Let $PH \cap AC = Q$. Now $\triangle AEP$'s circumcircle hits $CH$ at $S$, $\triangle ADQ$'s circumcircle hits $BH$ at $R$, and $\triangle AEP$'s circumcircle hits $\triangle ADQ$'s circumcircle at $J (\not=A)$. Prove that $RS$ is the perpendicular bisector of $HJ$.
Let O be the center of the circumscribed circle of the triangle ABC. Let AH be the altitude in this triangle, and let P be the base of the perpendicular drawn from point A to the line CO. Prove that the line HP passes through the midpoint of the side AB. (6 points)
Egor Bakaev
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$ and circumcentre $O$. Points $D\neq B$ and $E\neq C$ lie on $\omega$ such that $BD\perp AC$ and $CE\perp AB$. Let $CO$ meet $AB$ at $X$, and $BO$ meet $AC$ at $Y$.
Prove that the circumcircles of triangles $BXD$ and $CYE$ have an intersection lie on line $AO$.
[i]Ivan Chan Kai Chin, Malaysia[/i]
Given a triangle $ABC$, points $X,Y,$ and $Z$ are the midpoints of $BC,CA,$ and $AB$ respectively. The perpendicular bisector of $AB$ intersects line $XY$ and line $AC$ at $Z_1$ and $Z_2$ respectively. The perpendicular bisector of $AC$ intersects line $XZ$ and line $AB$ at $Y_1$ and $Y_2$ respectively. Let $K$ be a point such that $KZ_1 = KZ_2$ and $KY_1 = KY_2$. Prove that $KB=KC$.
$AD$ and $BE$ are altitudes of an acute triangle $ABC$. From $D$, perpendiculars are dropped
to $AB$ at $G$ and $AC$ at $K$. From $E$, perpendiculars are dropped to $AB$ at $F$ and $BC$ at $H$.
Prove that $FG$ is parallel to $HK$ and $FK = GH$.