This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 429

Given two circles ${\omega _1},{\omega _2}$, $S$ denotes all $\Delta ABC$ satisfies that ${\omega _1}$ is the circumcircle of $\Delta ABC$, ${\omega _2}$ is the $A$- excircle of $\Delta ABC$ , ${\omega _2}$ touches $BC,CA,AB$ at $D,E,F$. $S$ is not empty, prove that the centroid of $\Delta DEF$ is a fixed point.
Let $\Gamma $ be the circumcircle of triangle $ABC$. A circle $\Omega$ is tangent to the line segment $AB$ and is tangent to $\Gamma$ at a point lying on the same side of the line $AB$ as $C$. The angle bisector of $\angle BCA$ intersects $\Omega$ at two different points $P$ and $Q$. Prove that $\angle ABP = \angle QBC$.
Given an acute scalene triangle $ABC$ inscribed in circle $(O)$. Let $H$ be its orthocenter and $M$ be the midpoint of $BC$. Let $D$ lie on the opposite rays of $HA$ so that $BC=2DM$. Let $D'$ be the reflection of $D$ through line $BC$ and $X$ be the intersection of $AO$ and $MD$. a) Show that $AM$ bisects $D'X$. b) Similarly, we define the points $E,F$ like $D$ and $Y,Z$ like $X$. Let $S$ be the intersection of tangent lines from $B,C$ with respect to $(O)$. Let $G$ be the projection of the midpoint of $AS$ to the line $AO$. Show that there exists a point with the same power to all the circles $(BEY),(CFZ),(SGO)$ and $(O)$.
Let $ABC$ be a triangle and $P$ be an interior point. Let $\omega_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ internally and tangent to the circumcircle of $ABC$ at $A_1$ internally and let $\Gamma_A$ be the circle that is tangent to the circumcircle of $BPC$ at $P$ externally and tangent to the circumcircle of $ABC$ at $A_2$ internally. Define $B_1$, $B_2$, $C_1$, $C_2$ analogously. Let $O$ be the circumcentre of $ABC$. Prove that the lines $A_1A_2$, $B_1B_2$, $C_1C_2$ and $OP$ are concurrent.
let $\omega_1$ be a circle with $O_1$ as its center , let $\omega_2$ be a circle passing through $O_1$ with center $O_2$ let $A$ be one of the intersection of $\omega_1$ and $\omega_2$ let $x$ be a line tangent line to $\omega_1$ passing from $A$ let $\omega_3$ be a circle passing through $O_1,O_2$ with its center on the line $x$ and intersect $\omega_2$ at $P$ (not $O_1$) prove that the reflection of $P$ through $x$ is on $\omega_1$
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 $ A_1A_2A_3$ be a non-isosceles triangle with incenter $ I.$ Let $ C_i,$ $ i \equal{} 1, 2, 3,$ be the smaller circle through $ I$ tangent to $ A_iA_{i\plus{}1}$ and $ A_iA_{i\plus{}2}$ (the addition of indices being mod 3). Let $ B_i, i \equal{} 1, 2, 3,$ be the second point of intersection of $ C_{i\plus{}1}$ and $ C_{i\plus{}2}.$ Prove that the circumcentres of the triangles $ A_1 B_1I,A_2B_2I,A_3B_3I$ are collinear.
Let $I$ be the incenter of $ABC$ and $I_A$ the excenter of the side $BC$, let $M$ be the midpoint of $CB$ and $N$ the midpoint of arc $BC$(with the point $A$). If $T$ is the symmetric of the point $N$ by the point $A$, prove that the quadrilateral $I_AMIT$ is cyclic.
Let $ P$ be a point inside a triangle $ ABC$ such that \[ \angle APB \minus{} \angle ACB \equal{} \angle APC \minus{} \angle ABC. \] Let $ D$, $ E$ be the incenters of triangles $ APB$, $ APC$, respectively. Show that the lines $ AP$, $ BD$, $ CE$ meet at a point.
Let $ABC$ be a scalene triangle with incenter $I$. The incircle of $ABC$ touches $\overline{BC},\overline{CA},\overline{AB}$ at points $D,E,F$, respectively. Let $P$ be the foot of the altitude from $D$ to $\overline{EF}$, and let $M$ be the midpoint of $\overline{BC}$. The rays $AP$ and $IP$ intersect the circumcircle of triangle $ABC$ again at points $G$ and $Q$, respectively. Show that the incenter of triangle $GQM$ coincides with $D$. [i]Zack Chroman and Daniel Liu[/i]
Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
Let $ABC$ be a triangle with $AB = AC \neq BC$ and let $I$ be its incentre. The line $BI$ meets $AC$ at $D$, and the line through $D$ perpendicular to $AC$ meets $AI$ at $E$. Prove that the reflection of $I$ in $AC$ lies on the circumcircle of triangle $BDE$.
In the plane we are given two circles intersecting at $ X$ and $ Y$. Prove that there exist four points with the following property: (P) For every circle touching the two given circles at $ A$ and $ B$, and meeting the line $ XY$ at $ C$ and $ D$, each of the lines $ AC$, $ AD$, $ BC$, $ BD$ passes through one of these points.
Let be given a semicircle with diameter $AB$ and center $O$, and a line intersecting the semicircle at $C$ and $D$ and the line $AB$ at $M$ ($MB < MA$, $MD < MC$). The circumcircles of the triangles $AOC$ and $DOB$ meet again at $L$. Prove that $\angle MKO$ is right. [i]L. Kuptsov[/i]
Let $ABC$ be a triangle with circumcircle $\omega$. Let $l_B$ and $l_C$ be two lines through the points $B$ and $C$, respectively, such that $l_B \parallel l_C$. The second intersections of $l_B$ and $l_C$ with $\omega$ are $D$ and $E$, respectively. Assume that $D$ and $E$ are on the same side of $BC$ as $A$. Let $DA$ intersect $l_C$ at $F$ and let $EA$ intersect $l_B$ at $G$. If $O$, $O_1$ and $O_2$ are circumcenters of the triangles $ABC$, $ADG$ and $AEF$, respectively, and $P$ is the circumcenter of the triangle $OO_1O_2$, prove that $l_B \parallel OP \parallel l_C$. [i]Proposed by Stefan Lozanovski, Macedonia[/i]
There are arbitrary 7 points in the plane. Circles are drawn through every 4 possible concyclic points. Find the maximum number of circles that can be drawn.
We say a finite set $S$ of points in the plane is [i]very[/i] if for every point $X$ in $S$, there exists an inversion with center $X$ mapping every point in $S$ other than $X$ to another point in $S$ (possibly the same point). (a) Fix an integer $n$. Prove that if $n \ge 2$, then any line segment $\overline{AB}$ contains a unique very set $S$ of size $n$ such that $A, B \in S$. (b) Find the largest possible size of a very set not contained in any line. (Here, an [i]inversion[/i] with center $O$ and radius $r$ sends every point $P$ other than $O$ to the point $P'$ along ray $OP$ such that $OP\cdot OP' = r^2$.) [i]Proposed by Sammy Luo[/i]
Let $ABC$ be an acute-angled triangle with circumcircle $\omega$. A circle $\Gamma$ is internally tangent to $\omega$ at $A$ and also tangent to $BC$ at $D$. Let $AB$ and $AC$ intersect $\Gamma$ at $P$ and $Q$ respectively. Let $M$ and $N$ be points on line $BC$ such that $B$ is the midpoint of $DM$ and $C$ is the midpoint of $DN$. Lines $MP$ and $NQ$ meet at $K$ and intersect $\Gamma$ again at $I$ and $J$ respectively. The ray $KA$ meets the circumcircle of triangle $IJK$ again at $X\neq K$. Prove that $\angle BXP = \angle CXQ$. [i]Kian Moshiri, United Kingdom[/i]
Let $H$ be the orthocenter of acute triangle $ABC$, let $F$ be the foot of the altitude from $C$ to $AB$, and let $P$ be the reflection of $H$ across $BC$. Suppose that the circumcircle of triangle $AFP$ intersects line $BC$ at two distinct points $X$ and $Y$. Prove that $C$ is the midpoint of $XY$.
Point $O$ is a center of circumcircle of acute triangle $ABC$, bisector of angle $BAC$ cuts side $BC$ in point $D$. Let $M$ be a point such that, $MC \perp BC$ and $MA \perp AD$. Lines $BM$ and $OA$ intersect in point $P$. Show that circle of center in point $P$ passing through a point $A$ is tangent to line $BC$.
Let $\overline{AB}$ be a chord of a circle $\omega$, and let $P$ be a point on the chord $\overline{AB}$. Circle $\omega_1$ passes through $A$ and $P$ and is internally tangent to $\omega$. Circle $\omega_2$ passes through $B$ and $P$ and is internally tangent to $\omega$. Circles $\omega_1$ and $\omega_2$ intersect at points $P$ and $Q$. Line $PQ$ intersects $\omega$ at $X$ and $Y$. Assume that $AP=5$, $PB=3$, $XY=11$, and $PQ^2 = \tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Given a triangle $ABC$ satisfying $AC+BC=3\cdot AB$. The incircle of triangle $ABC$ has center $I$ and touches the sides $BC$ and $CA$ at the points $D$ and $E$, respectively. Let $K$ and $L$ be the reflections of the points $D$ and $E$ with respect to $I$. Prove that the points $A$, $B$, $K$, $L$ lie on one circle. [i]Proposed by Dimitris Kontogiannis, Greece[/i]
Trapezoid $ ABCD$, with $ \overline{AB}\parallel{}\overline{CD}$, is inscribed in circle $ \omega$ and point $ G$ lies inside triangle $ BCD$. Rays $ AG$ and $ BG$ meet $ \omega$ again at points $ P$ and $ Q$, respectively. Let the line through $ G$ parallel to $ \overline{AB}$ intersects $ \overline{BD}$ and $ \overline{BC}$ at points $ R$ and $ S$, respectively. Prove that quadrilateral $ PQRS$ is cyclic if and only if $ \overline{BG}$ bisects $ \angle CBD$.
Convex quadrilateral $ABCD$ has $\angle ABC = \angle CDA = 90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $BD$. Points $S$ and $T$ lie on sides $AB$ and $AD$, respectively, such that $H$ lies inside triangle $SCT$ and \[ \angle CHS - \angle CSB = 90^{\circ}, \quad \angle THC - \angle DTC = 90^{\circ}. \] Prove that line $BD$ is tangent to the circumcircle of triangle $TSH$.
There is a convex quadrilateral $ ABCD $ which satisfies $ \angle A=\angle D $. Let the midpoints of $ AB, AD, CD $ be $ L,M,N $. Let's say the intersection point of $ AC, BD $ be $ E $ . Let's say point $ F $ which lies on $ \overrightarrow{ME} $ satisfies $ \overline{ME}\times \overline{MF}=\overline{MA}^{2} $. Prove that $ \angle LFM=\angle MFN $. :)