Olympiad problems

2009 Germany Team Selection Test, 1

Tags: geometry , trapezoid , circumcircle , IMO Shortlist , Charles Leytem

Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]

2009 Belarus Team Selection Test, 2

Tags: geometry , trapezoid , circumcircle , IMO Shortlist , Charles Leytem

Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]

2009 Germany Team Selection Test, 1

Tags: geometry , trapezoid , circumcircle , IMO Shortlist , Charles Leytem

Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]

2009 Ukraine Team Selection Test, 1

Tags: geometry , trapezoid , circumcircle , IMO Shortlist , Charles Leytem

Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]

2007 IMO Shortlist, 4

Tags: geometry , parallelogram , circumcircle , IMO , IMO 2007 , Charles Leytem , homothety

Consider five points $ A$, $ B$, $ C$, $ D$ and $ E$ such that $ ABCD$ is a parallelogram and $ BCED$ is a cyclic quadrilateral. Let $ \ell$ be a line passing through $ A$. Suppose that $ \ell$ intersects the interior of the segment $ DC$ at $ F$ and intersects line $ BC$ at $ G$. Suppose also that $ EF \equal{} EG \equal{} EC$. Prove that $ \ell$ is the bisector of angle $ DAB$. [i]Author: Charles Leytem, Luxembourg[/i]

2016 EGMO, 4

Tags: geometry , EGMO , Charles Leytem , EGMO 2016 , circles , marvio

Two circles $\omega_1$ and $\omega_2$, of equal radius intersect at different points $X_1$ and $X_2$. Consider a circle $\omega$ externally tangent to $\omega_1$ at $T_1$ and internally tangent to $\omega_2$ at point $T_2$. Prove that lines $X_1T_1$ and $X_2T_2$ intersect at a point lying on $\omega$.

2007 IMO, 2

Tags: geometry , parallelogram , circumcircle , IMO , IMO 2007 , Charles Leytem , homothety

Consider five points $ A$, $ B$, $ C$, $ D$ and $ E$ such that $ ABCD$ is a parallelogram and $ BCED$ is a cyclic quadrilateral. Let $ \ell$ be a line passing through $ A$. Suppose that $ \ell$ intersects the interior of the segment $ DC$ at $ F$ and intersects line $ BC$ at $ G$. Suppose also that $ EF \equal{} EG \equal{} EC$. Prove that $ \ell$ is the bisector of angle $ DAB$. [i]Author: Charles Leytem, Luxembourg[/i]

1998 IMO, 1

Tags: geometry , circumcircle , quadrilateral , perpendicular bisector , Charles Leytem , IMO , IMO 1998

A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas.

2009 Brazil Team Selection Test, 2

Tags: geometry , trapezoid , circumcircle , IMO Shortlist , Charles Leytem

Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]

2017 IMO, 4

Tags: geometry , IMO , IMO 2017 , circumcircle , tangent , Charles Leytem

Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$. [i]Proposed by Charles Leytem, Luxembourg[/i]

2017 IMO Shortlist, G2

Tags: geometry , IMO , IMO 2017 , circumcircle , tangent , Charles Leytem

Let $R$ and $S$ be different points on a circle $\Omega$ such that $RS$ is not a diameter. Let $\ell$ be the tangent line to $\Omega$ at $R$. Point $T$ is such that $S$ is the midpoint of the line segment $RT$. Point $J$ is chosen on the shorter arc $RS$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $JST$ intersects $\ell$ at two distinct points. Let $A$ be the common point of $\Gamma$ and $\ell$ that is closer to $R$. Line $AJ$ meets $\Omega$ again at $K$. Prove that the line $KT$ is tangent to $\Gamma$. [i]Proposed by Charles Leytem, Luxembourg[/i]

1998 IMO Shortlist, 1

Tags: geometry , circumcircle , quadrilateral , perpendicular bisector , Charles Leytem , IMO , IMO 1998

A convex quadrilateral $ABCD$ has perpendicular diagonals. The perpendicular bisectors of the sides $AB$ and $CD$ meet at a unique point $P$ inside $ABCD$. Prove that the quadrilateral $ABCD$ is cyclic if and only if triangles $ABP$ and $CDP$ have equal areas.

2008 IMO Shortlist, 2

Tags: geometry , trapezoid , circumcircle , IMO Shortlist , Charles Leytem

Given trapezoid $ ABCD$ with parallel sides $ AB$ and $ CD$, assume that there exist points $ E$ on line $ BC$ outside segment $ BC$, and $ F$ inside segment $ AD$ such that $ \angle DAE \equal{} \angle CBF$. Denote by $ I$ the point of intersection of $ CD$ and $ EF$, and by $ J$ the point of intersection of $ AB$ and $ EF$. Let $ K$ be the midpoint of segment $ EF$, assume it does not lie on line $ AB$. Prove that $ I$ belongs to the circumcircle of $ ABK$ if and only if $ K$ belongs to the circumcircle of $ CDJ$. [i]Proposed by Charles Leytem, Luxembourg[/i]