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: 41

1997 IMO Shortlist, 16
Tags: geometry , incenter , circumcircle , trigonometry , angle bisector , collinearity
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.

1997 IMO Shortlist, 9
Tags: geometry , incenter , circumcircle , collinearity
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.

2020 Junior Macedonian National Olympiad, 4
Tags: nice , geometry , collinearity , circumcircle , parallelogram
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.

2018 Peru Iberoamerican Team Selection Test, P9
Tags: geometry , circumcircle , collinear , collinearity
Let $\Gamma$ be the circumcircle of a triangle $ABC$ with $AB <BC$, and let $M$ be the midpoint from the side $AC$ . The median of side $AC$ cuts $\Gamma$ at points $X$ and $Y$ ($X$ in the arc $ABC$). The circumcircle of the triangle $BMY$ cuts the line $AB$ at $P$ ($P \ne B$) and the line $BC$ in $Q$ ($Q \ne B$). The circumcircles of the triangles $PBC$ and $QBA$ are cut in $R$ ($R \ne B$). Prove that points $X, B$ and $R$ are collinear.

1982 IMO, 2
Tags: geometry , hexagon , collinearity , golden ratio
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.

2022 Oral Moscow Geometry Olympiad, 6
Tags: geometry , collinear , collinearity
In an acute non-isosceles triangle $ABC$, the inscribed circle touches side $BC$ at point $T, Q$ is the midpoint of altitude $AK$, $P$ is the orthocenter of the triangle formed by the bisectors of angles $B$ and $C$ and line $AK$. Prove that the points $P, Q$ and $T$ lie on the same line. (D. Prokopenko)

Mathematical Minds 2024, P8
Tags: olympiad geometry , excircle , incircle , collinearity , geometry
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]

1986 IMO Longlists, 44
Tags: geometry , incenter , circumcircle , collinearity
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.

2003 Junior Balkan Team Selection Tests - Romania, 1
Tags: geometry , rectangle , collinear , collinearity
Suppose $ABCD$ and $AEFG$ are rectangles such that the points $B,E,D,G$ are collinear (in this order). Let the lines $BC$ and $GF$ intersect at point $T$ and let the lines $DC$ and $EF$ intersect at point $H$. Prove that points $A, H$ and $T$ are collinear.

1989 IMO Shortlist, 14
Tags: geometry , circumcircle , quadrilateral , collinearity
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.

2016 Harvard-MIT Mathematics Tournament, 10
Tags: incircle , collinearity
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.

2016 Sharygin Geometry Olympiad, 6
Tags: collinearity , circumcircle , geometry
A triangle $ABC$ is given. The point $K$ is the base of the external bisector of angle $A$. The point $M$ is the midpoint of the arc $AC$ of the circumcircle. The point $N$ on the bisector of angle $C$ is such that $AN \parallel BM$. Prove that the points $M,N,K$ are collinear. [i](Proposed by Ilya Bogdanov)[/i]

2023 Sharygin Geometry Olympiad, 23
Tags: geometry , ellipse , de longchamps point , collinearity
An ellipse $\Gamma_1$ with foci at the midpoints of sides $AB$ and $AC$ of a triangle $ABC$ passes through $A$, and an ellipse $\Gamma_2$ with foci at the midpoints of $AC$ and $BC$ passes through $C$. Prove that the common points of these ellipses and the orthocenter of triangle $ABC$ are collinear.

STEMS 2023 Math Cat A, 5
Tags: geometry , radical axis , collinearity
A convex quadrilateral $ABCD$ is such that $\angle B = \angle D$ and are both acute angles. $E$ is on $AB$ such that $CB = CE$ and $F$ is on $AD$ such that $CF = CD$. If the circumcenter of $CEF$ is $O_1$ and the circumcenter of $ABD$ is $O_2$. Prove that $C,O_1,O_2$ are collinear. [i]Proposed by Kapil Pause[/i]

2020 Turkey Junior National Olympiad, 3
Tags: geometry , collinearity
The circumcenter of an acute-triangle $ABC$ with $|AB|<|BC|$ is $O$, $D$ and $E$ are midpoints of $|AB|$ and $|AC|$, respectively. $OE$ intersects $BC$ at $K$, the circumcircle of $OKB$ intersects $OD$ second time at $L$. $F$ is the foot of altitude from $A$ to line $KL$. Show that the point $F$ lies on the line $DE$

2019 Thailand Mathematical Olympiad, 8
Tags: geometry , incenter , circumcircle , collinear , collinearity
Let $ABC$ be a triangle such that $AB\ne AC$ and $\omega$ be the circumcircle of this triangle. Let $I$ be the center of the inscribed circle of $ABC$ which touches $BC$ at $D$. Let the circle with diameter $AI$ meets $\omega$ again at $K$. If the line $AI$ intersects $\omega$ again at $M$, show that $K, D, M$ are collinear.

1987 IMO Longlists, 52
Tags: geometry , circumcircle , incenter , collinearity
Given a nonequilateral triangle $ABC$, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $A'B'C'$ (the vertices listed counterclockwise) for which the triples of points $A,B', C'; A',B, C';$ and $A',B', C$ are collinear. [i]Proposed by Poland.[/i]

1982 IMO Longlists, 37
Tags: geometry , hexagon , collinearity , golden ratio
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.

1986 IMO Shortlist, 14
Tags: geometry , incenter , circumcircle , collinearity
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.

1989 IMO Longlists, 48
Tags: geometry , circumcircle , quadrilateral , collinearity
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.

1987 IMO Shortlist, 12
Tags: geometry , circumcircle , incenter , collinearity
Given a nonequilateral triangle $ABC$, the vertices listed counterclockwise, find the locus of the centroids of the equilateral triangles $A'B'C'$ (the vertices listed counterclockwise) for which the triples of points $A,B', C'; A',B, C';$ and $A',B', C$ are collinear. [i]Proposed by Poland.[/i]

2022 Macedonian Mathematical Olympiad, Problem 5
Tags: geometry , incenter , circumcircle , collinearity , excenter
An acute $\triangle ABC$ with circumcircle $\Gamma$ is given. $I$ and $I_a$ are the incenter and $A-$excenter of $\triangle ABC$ respectively. The line $AI$ intersects $\Gamma$ again at $D$ and $A'$ is the antipode of $A$ with respect to $\Gamma$. $X$ and $Y$ are point on $\Gamma$ such that $\angle IXD = \angle I_aYD = 90^\circ$. The tangents to $\Gamma$ at $X$ and $Y$ intersect in point $Z$. Prove that $A', D$ and $Z$ are collinear. [i]Proposed by Nikola Velov[/i]

2017 Hanoi Open Mathematics Competitions, 14
Tags: geometry , trapezoid , collinear , collinearity , perpendicular
Given trapezoid $ABCD$ with bases $AB \parallel CD$ ($AB < CD$). Let $O$ be the intersection of $AC$ and $BD$. Two straight lines from $D$ and $C$ are perpendicular to $AC$ and $BD$ intersect at $E$ , i.e. $CE \perp BD$ and $DE \perp AC$ . By analogy, $AF \perp BD$ and $BF \perp AC$ . Are three points $E , O, F$ located on the same line?

1982 IMO Shortlist, 5
Tags: geometry , hexagon , collinearity , golden ratio
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.

1998 Belarus Team Selection Test, 3
Tags: geometry , incenter , circumcircle , collinearity
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.