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: 41

The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that \[ {AM\over AC}={CN\over CE}=r. \] Determine $r$ if $B,M$ and $N$ are collinear.
The circle inscribed in a triangle $ABC$ touches the sides $BC,CA,AB$ in $D,E, F$, respectively, and $X, Y,Z$ are the midpoints of $EF, FD,DE$, respectively. Prove that the centers of the inscribed circle and of the circles around $XYZ$ and $ABC$ are collinear.
A bicentric quadrilateral is one that is both inscribable in and circumscribable about a circle, i.e. both the incircle and circumcircle exists. Show that for such a quadrilateral, the centers of the two associated circles are collinear with the point of intersection of the diagonals.
Let $ABC$ be a triangle with circumcircle $\Omega$, incircle $\omega$, and $A$-excircle $\omega_A$. Let $X$ and $Y$ be the tangency points of $\omega_A$ with $AB$ and $AC$. Lines $XY$ and $BC$ intersect in $T$. The tangent from $T$ to $\omega$ different from $BC$ intersects $\omega$ at $K$. The radical axis of $\omega_A$ and $\Omega$ intersects $BC$ in $S$. The tangent from $S$ to $\omega_A$ different from $BC$ intersects $\omega_A$ at $L$. Prove that $A$, $K$ and $L$ are collinear. [i]Proposed by Ana Boiangiu[/i]
Circle $\omega$ is inscribed in quadrilateral $ABCD$ such that $AB$ and $CD$ are not parallel and intersect at point $O.$ Circle $\omega_1$ touches the side $BC$ at $K$ and touches line $AB$ and $CD$ at points which are located outside quadrilateral $ABCD;$ circle $\omega_2$ touches side $AD$ at $L$ and touches line $AB$ and $CD$ at points which are located outside quadrilateral $ABCD.$ If $O,K,$ and $L$ are collinear$,$ then show that the midpoint of side $BC,AD,$ and the center of circle $\omega$ are also collinear.
Let $ABC$ be an acute triangle such that $AB < AC$. Let $\omega$ be the circumcircle of $ABC$ and assume that the tangent to $\omega$ at $A$ intersects the line $BC$ at $D$. Let $\Omega$ be the circle with center $D$ and radius $AD$. Denote by $E$ the second intersection point of $\omega$ and $\Omega$. Let $M$ be the midpoint of $BC$. If the line $BE$ meets $\Omega$ again at $X$, and the line $CX$ meets $\Omega$ for the second time at $Y$, show that $A, Y$, and $M$ are collinear. [i]Proposed by Nikola Velov, North Macedonia[/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 $ABC$ be a triangle with incenter $I$ whose incircle is tangent to $\overline{BC}$, $\overline{CA}$, $\overline{AB}$ at $D$, $E$, $F$. Point $P$ lies on $\overline{EF}$ such that $\overline{DP} \perp \overline{EF}$. Ray $BP$ meets $\overline{AC}$ at $Y$ and ray $CP$ meets $\overline{AB}$ at $Z$. Point $Q$ is selected on the circumcircle of $\triangle AYZ$ so that $\overline{AQ} \perp \overline{BC}$. Prove that $P$, $I$, $Q$ are collinear.
The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that \[ {AM\over AC}={CN\over CE}=r. \] Determine $r$ if $B,M$ and $N$ are collinear.
Circle $\omega$ is inscribed in quadrilateral $ABCD$ such that $AB$ and $CD$ are not parallel and intersect at point $O.$ Circle $\omega_1$ touches the side $BC$ at $K$ and touches line $AB$ and $CD$ at points which are located outside quadrilateral $ABCD;$ circle $\omega_2$ touches side $AD$ at $L$ and touches line $AB$ and $CD$ at points which are located outside quadrilateral $ABCD.$ If $O,K,$ and $L$ are collinear$,$ then show that the midpoint of side $BC,AD,$ and the center of circle $\omega$ are also collinear.
Let $ABC$ be a triangle, $H$ its orthocenter, and $\Gamma$ its circumcircle. Let $J$ be the point diametrically opposite to $A$ on $\Gamma$. The points $D$, $E$ and $F$ are the feet of the altitudes from $A$, $B$ and $C$, respectively. The line $AD$ intersects $\Gamma$ again at $P$. The circumcircle of $EFP$ intersects $\Gamma$ again at $Q$. Let $K$ be the second point of intersection of $JH$ with $\Gamma$. Prove that $K$, $D$ and $Q$ are collinear.
In an acute-angled triangle $ ABC,$ let $ AD,BE$ be altitudes and $ AP,BQ$ internal bisectors. Denote by $ I$ and $ O$ the incenter and the circumcentre of the triangle, respectively. Prove that the points $ D, E,$ and $ I$ are collinear if and only if the points $ P, Q,$ and $ O$ are collinear.
Let $ABC$ be an isosceles triangle with base $AC$. Points $D$ and $E$ are chosen on the sides $AC$ and $BC$, respectively, such that $CD = DE$. Let $H, J,$ and $K$ be the midpoints of $DE, AE,$ and $BD$, respectively. The circumcircle of triangle $DHK$ intersects $AD$ at point $F$, whereas the circumcircle of triangle $HEJ$ intersects $BE$ at $G$. The line through $K$ parallel to $AC$ intersects $AB$ at $I$. Let $IH \cap GF =$ {$M$}. Prove that $J, M,$ and $K$ are collinear points.
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 $ABCD$ be a rhombus. Eight additional points $X_1$, $X_2$, $Y_1$, $Y_2$, $Z_1$, $Z_2$, $W_1$, $W_2$ were chosen so that the quadrilaterals $AX_1BX_2$, $BY_1CY_2$, $CZ_1DZ_2$, $DW_1AW_2$ are squares. Prove that the eight new points lie on two straight lines.
The diagonals $AC$ and $CE$ of the regular hexagon $ABCDEF$ are divided by inner points $M$ and $N$ respectively, so that \[ {AM\over AC}={CN\over CE}=r. \] Determine $r$ if $B,M$ and $N$ are collinear.