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: 35

In a trapezoid $ABCD$ with $AD$ parallel to $BC$ points $E, F$ are on sides $AB, CD$ respectively. $A_1, C_1$ are on $AD,BC$ such that $A_1, E, F, A$ lie on a circle and so do $C_1, E, F, C$. Prove that lines $A_1C_1, BD, EF$ are concurrent.
On the sides $AB$ and $AC$ of the triangle $ABC$, the points $P$ and $Q$ are chosen, respectively, so that $PQ\parallel BC$. Segments $BQ$ and $CP$ intersect at point $O$. Point $A'$ is symmetric to point $A$ relative to line $BC$. The segment $A'O$ intersects the circumcircle $w$ of the triangle $APQ$ at the point $S$. Prove that circumcircle of $BSC$ is tangent to the circle $w$.
Let $ABC$ be a non-equilateral triangle and let $M_a$, $M_b$, $M_c$ be the midpoints of the sides $BC$, $CA$, $AB$, respectively. Let $S$ be a point lying on the Euler line. Denote by $X$, $Y$, $Z$ the second intersections of $M_aS$, $M_bS$, $M_cS$ with the nine-point circle. Prove that $AX$, $BY$, $CZ$ are concurrent.
$AH$ is the altitude of triangle $ABC$ and $H^\prime$ is the reflection of $H$ trough the midpoint of $BC$. If the tangent lines to the circumcircle of $ABC$ at $B$ and $C$, intersect each other at $X$ and the perpendicular line to $XH^\prime$ at $H^\prime$, intersects $AB$ and $AC$ at $Y$ and $Z$ respectively, prove that $\angle ZXC=\angle YXB$.
Let $ABCD$ be a cyclic quadrilateral such that $AD=BD=AC$. A point $P$ moves along the circumcircle $\omega$ of triangle $ABCD$. The lined $AP$ and $DP$ meet the lines $CD$ and $AB$ at points $E$ and $F$ respectively. The lines $BE$ and $CF$ meet point $Q$. Find the locus of $Q$.
$AH$ is the altitude of triangle $ABC$ and $H^\prime$ is the reflection of $H$ trough the midpoint of $BC$. If the tangent lines to the circumcircle of $ABC$ at $B$ and $C$, intersect each other at $X$ and the perpendicular line to $XH^\prime$ at $H^\prime$, intersects $AB$ and $AC$ at $Y$ and $Z$ respectively, prove that $\angle ZXC=\angle YXB$.
In a triangle $ABC$ with $AB=AC$, $\omega$ is the circumcircle and $O$ its center. Let $D$ be a point on the extension of $BA$ beyond $A$. The circumcircle $\omega_{1}$ of triangle $OAD$ intersects the line $AC$ and the circle $\omega$ again at points $E$ and $G$, respectively. Point $H$ is such that $DAEH$ is a parallelogram. Line $EH$ meets circle $\omega_{1}$ again at point $J$. The line through $G$ perpendicular to $GB$ meets $\omega_{1}$ again at point $N$ and the line through $G$ perpendicular to $GJ$ meets $\omega$ again at point $L$. Prove that the points $L, N, H, G$ lie on a circle.
Let $ABCD$ be a cyclic quadrilateral whose diagonals $AC$ and $BD$ meet at $E$. The extensions of the sides $AD$ and $BC$ beyond $A$ and $B$ meet at $F$. Let $G$ be the point such that $ECGD$ is a parallelogram, and let $H$ be the image of $E$ under reflection in $AD$. Prove that $D,H,F,G$ are concyclic.
Let $BC$ be a fixed segment in the plane, and let $A$ be a variable point in the plane not on the line $BC$. Distinct points $X$ and $Y$ are chosen on the rays $CA^\to$ and $BA^\to$, respectively, such that $\angle CBX = \angle YCB = \angle BAC$. Assume that the tangents to the circumcircle of $ABC$ at $B$ and $C$ meet line $XY$ at $P$ and $Q$, respectively, such that the points $X$, $P$, $Y$ and $Q$ are pairwise distinct and lie on the same side of $BC$. Let $\Omega_1$ be the circle through $X$ and $P$ centred on $BC$. Similarly, let $\Omega_2$ be the circle through $Y$ and $Q$ centred on $BC$. Prove that $\Omega_1$ and $\Omega_2$ intersect at two fixed points as $A$ varies. [i]Daniel Pham Nguyen, Denmark[/i]
Let $ABC$ be a triangle and let $P$ be a point not lying on any of the three lines $AB$, $BC$, or $CA$. Distinct points $D$, $E$, and $F$ lie on lines $BC$, $AC$, and $AB$, respectively, such that $\overline{DE}\parallel \overline{CP}$ and $\overline{DF}\parallel \overline{BP}$. Show that there exists a point $Q$ on the circumcircle of $\triangle AEF$ such that $\triangle BAQ$ is similar to $\triangle PAC$. [i]Andrew Gu[/i]