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

Tags were heavily modified to better represent problems.

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Found problems: 847

Cevians $AP$ and $AQ$ of a triangle $ABC$ are symmetric with respect to its bisector. Let $X$, $Y$ be the projections of $B$ to $AP$ and $AQ$ respectively, and $N$, $M$ be the projections of $C$ to $AP$ and $AQ$ respectively. Prove that $XM$ and $NY$ meet on $BC$.
Given a circle $(O)$ with center $O$ and $P$ a point outside $(O)$. $A$ and $B$ are points on $(O)$ such that $PA$ and $PB$ are tangents to $(O)$. The line $\ell$ through $P$ intersects $(O)$ at points $C$ and $D$, respectively ($C$ lies between $P$ and $D$). Line $BF$ is parallel to line $PA$ and intersects line $AC$ and line $AD$ at $E$ and $F$, respectively. Prove that $BE = BF$.
In a triangle $ABC$ points $A_1,B_1,C_1$ are the midpoints of $BC,CA,AB$ respectively,and $A_2,B_2,C_2$ are the midpoints of the altitudes from $A,B,C$ respectively. Show that the lines $A_1A_2,B_1B_2,C_1,C_2$ are concurrent.
Let $\triangle ABC$ be a triangle with circumcenter $O$ satisfying $AB=13$, $BC = 15$, and $AC = 14$. Suppose there is a point $P$ such that $PB \perp BC$ and $PA \perp AB$. Let $X$ be a point on $AC$ such that $BX \perp OP$. What is the ratio $AX/XC$? [i]Proposed by Thomas Lam[/i]
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 a triangle $ABC$ satisfy $AC = BC$; in other words, let $ABC$ be an isosceles triangle with base $AB$. Let $P$ be a point inside the triangle $ABC$ such that $\angle PAB = \angle PBC$. Denote by $M$ the midpoint of the segment $AB$. Show that $\angle APM + \angle BPC = 180^{\circ}$.
Let $ABCD$ be a cyclic quadrilateral satisfying $AD^2 + BC^2 = AB^2$. The diagonals of $ABCD$ intersect at $E$. Let $P$ be a point on side $\overline{AB}$ satisfying $\angle APD = \angle BPC$. Show that line $PE$ bisects $\overline{CD}$. [i]Proposed by Ankan Bhattacharya[/i]
Let $O$ be the circumcenter of an acute triangle $ABC$. Line $OA$ intersects the altitudes of $ABC$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $PQH$ lies on a median of triangle $ABC$.
Let $ABC$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $AC$ and $AB$, respectively, and let $M$ be the midpoint of $EF$. Let the perpendicular bisector of $EF$ intersect the line $BC$ at $K$, and let the perpendicular bisector of $MK$ intersect the lines $AC$ and $AB$ at $S$ and $T$, respectively. If the quadrilateral $KSAT$ is cycle, prove that $\angle{KEF}=\angle{KFE}=\angle{A}$.
Let $\triangle ABC$ be an acute-scalene triangle, and let $N$ be the center of the circle wich pass trough the feet of altitudes. Let $D$ be the intersection of tangents to the circumcircle of $\triangle ABC$ at $B$ and $C$. Prove that $A$, $D$ and $N$ are collinear iff $\measuredangle BAC = 45º$.
Let $ABC$ be an isosceles triangle ($AB=AC$) with incenter $I$. Circle $\omega$ passes through $C$ and $I$ and is tangent to $AI$. $\omega$ intersects $AC$ and circumcircle of $ABC$ at $Q$ and $D$, respectively. Let $M$ be the midpoint of $AB$ and $N$ be the midpoint of $CQ$. Prove that $AD$, $MN$ and $BC$ are concurrent. [i]Proposed by Alireza Dadgarnia[/i]
Let $ABC$ be an acute angled triangle with circumcircle $O$. Line $AO$ intersects the circumcircle of triangle $ABC$ again at point $D$. Let $P$ be a point on the side $BC$. Line passing through $P$ perpendicular to $AP$ intersects lines $DB$ and $DC$ at $E$ and $F$ respectively . Line passing through $D$ perpendicular to $BC$ intersects $EF$ at point $Q$. Prove that $EQ = FQ$ if and only if $BP = CP$.
Given triangle $ABC$. $D$ is a point on $BC$. $AC$ meets $(ABD)$ again at $E$,and $AB$ meets $(ACD)$ again at $F$. $M$ is the midpoint of $EF$. $BC$ meets $(DEF)$ again at $P$. Prove that $\angle BAP = \angle MAC$.
Let $ABC$ be a triangle and let $O$ be its circumcentre. The internal and external bisectrices of the angle $BAC$ meet the line $BC$ at points $D$ and $E$, respectively. Let further $M$ and $L$ respectively denote the midpoints of the segments $BC$ and $DE$. The circles $ABC$ and $ALO$ meet again at point $N$. Show that the angles $BAN$ and $CAM$ are equal.
Let $ABCD$ be a convex quadrilateral with $AB=BD=DC$ and $AB \perp BD \perp DC$. Let $M$ be the midpoint of segment $BC$. Show that $\angle BAM+\angle DCA=45^\circ$.
The circle $ \omega $ inscribed in triangle $ABC$ touches its sides $AB, BC, CA$ at points $K, L, M$ respectively. In the arc $KL$ of the circle $ \omega $ that does not contain the point $M$, we select point $S$. Denote by $P, Q, R, T$ the intersection points of straight $AS$ and $KM, ML$ and $SC, LP$ and $KQ, AQ$ and $PC$ respectively. It turned out that the points $R, S$ and $M$ are collinear. Prove that the point $T$ also lies on the line $SM$.
To two circles $r_1$ and $r_2$, intersecting at points $A$ and $B$, their common tangent $CD$ is drawn ($C$ and $D$ are tangency points, respectively, point $B$ is closer to line $CB$ than $A$). Line passing through $A$ , intersects $r_1$ and $r_2$ for second time at points $K$ and $L$, respectively ($A$ lies between $K$ and $L$). Lines $KC$ and $LD$ intersect at point $P$. Prove that $PB$ is the symmedian of triangle $KPL$. (Yu. Blinkov)
Let $ABCD$ be a parallelogram such that $AB = 35$ and $BC = 28$. Suppose that $BD \perp BC$. Let $\ell_1$ be the reflection of $AC$ across the angle bisector of $\angle BAD$, and let $\ell_2$ be the line through $B$ perpendicular to $CD$. $\ell_1$ and $\ell_2$ intersect at a point $P$. If $PD$ can be expressed in simplest form as $\frac{m}{n}$, find $m + n$.
Given $\triangle ABC$, let $I,O,\Gamma$ denote its incenter, circumcenter and circumcircle respecitvely. Let $AI$ intersect $\Gamma$ at $M(\neq A)$. Circle $\omega$ is tangent to $AB$, $AC$ and $\Gamma$ internally at $T$ (i.e. the mixtilinear incircle opposite $A$). Let the tangents at $A$ and $T$ to $\Gamma$ meet at $P$, and let $PI$ and $TM$ intersect at $Q$. Prove that $QA$ and $MO$ intersect at a point on $\Gamma$.
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 $m+n$.
Let $H$ be the orthocenter of a nonisosceles triangle $ABC$. The bisector of angle $BHC$ meets $AB$ and $AC$ at points $P$ and $Q$ respectively. The perpendiculars to $AB$ and $AC$ from $P$ and $Q$ meet at $K$. Prove that $KH$ bisects the segment $BC$.
For a triangle $\triangle ABC$ with an obtuse angle $\angle A$ , let $E , F$ be feet of altitudes from $B , C$ on sides $AC , AB$ respectively. The tangents from $B , C$ to circumcircle of triangle $\triangle ABC$ intersect line $EF$ at points $K , L$ respectively and we know that $\angle CLB=135$. Point $R$ lies on segment $BK$ in such a way that $KR=KL$ and let $S$ be a point on line $BK$ such that $K$ is between $B , S$ and $\angle BLS=135$. Prove that the circle with diameter $RS$ is tangent to circumcircle of triangle $\triangle ABC$. [i]Proposed by Mehran Talaei[/i]
We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear. [i]Proposed by Sammy Luo[/i]
Let $ABCD$ to be a quadrilateral inscribed in a circle $\Gamma$. Let $r$ and $s$ to be the tangents to $\Gamma$ through $B$ and $C$, respectively, $M$ the intersection between the lines $r$ and $AD$ and $N$ the intersection between the lines $s$ and $AD$. After all, let $E$ to be the intersection between the lines $BN$ and $CM$, $F$ the intersection between the lines $AE$ and $BC$ and $L$ the midpoint of $BC$. Prove that the circuncircle of the triangle $DLF$ is tangent to $\Gamma$.
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.