Let $BE$ and $CF$ be the medians of an acute triangle $ABC.$ On the line $BC,$ points $K \ne B$ and $L \ne C$ are chosen such that $BE = EK$ and $CF = FL.$ Prove that $AK = AL.$ Proposed by Heorhii Zhilinskyi
2024 Yasinsky Geometry Olympiad
VIII
Let $I$ be the incenter and $O$ be the circumcenter of triangle $ABC,$ where $\angle A < \angle B < \angle C.$ Points $P$ and $Q$ are such that $AIOP$ and $BIOQ$ are isosceles trapezoids ($AI \parallel OP,$ $BI \parallel OQ$). Prove that $CP = CQ.$ Proposed by Volodymyr Brayman and Matthew Kurskyi
Let \( W \) be the midpoint of the arc \( BC \) of the circumcircle of triangle \( ABC \), such that \( W \) and \( A \) lie on opposite sides of line \( BC \). On sides \( AB \) and \( AC \), points \( P \) and \( Q \) are chosen respectively so that \( APWQ \) is a parallelogram, and on side \( BC \), points \( K \) and \( L \) are chosen such that \( BK = KW \) and \( CL = LW \). Prove that the lines \( AW \), \( KQ \), and \( LP \) are concurrent. Proposed by Matthew Kurskyi
On side \( AB \) of an isosceles trapezoid \( ABCD \) (\( AD \parallel BC \)), points \( E \) and \( F \) are chosen such that a circle can be inscribed in quadrilateral \( CDEF \). Prove that the circumcircles of triangles \( ADE \) and \( BCF \) are tangent to each other. Proposed by Matthew Kurskyi
On side \( AC \) of triangle \( ABC \), a point \( P \) is chosen such that \( AP = \frac{1}{3} AC \), and on segment \( BP \), a point \( S \) is chosen such that \( CS \perp BP \). A point \( T \) is such that \( BCST \) is a parallelogram. Prove that \( AB = AT \). Proposed by Bohdan Zheliabovskyi
IX
Inside triangle \( ABC \), a point \( D \) is chosen such that \( \angle ADB = \angle ADC \). The rays \( BD \) and \( CD \) intersect the circumcircle of triangle \( ABC \) at points \( E \) and \( F \), respectively. On segment \( EF \), points \( K \) and \( L \) are chosen such that \linebreak \( \angle AKD = 180^\circ - \angle ACB \) and \( \angle ALD = 180^\circ - \angle ABC \), with segments \( EL \) and \( FK \) \linebreak not intersecting line \( AD \). Prove that \( AK = AL \). Proposed by Matthew Kurskyi
Let \( M \) be the midpoint of side \( BC \) of triangle \( ABC \), and let \( D \) be an arbitrary point on the arc \( BC \) of the circumcircle that does not contain \( A \). Let \( N \) be the midpoint of \( AD \). A circle passing through points \( A \), \( N \), and tangent to \( AB \) intersects side \( AC \) at point \( E \). Prove that points \( C \), \( D \), \( E \), and \( M \) are concyclic. Proposed by Matthew Kurskyi
Let \( H \) be the orthocenter of an acute triangle \( ABC \), and let \( AT \) be the diameter of the circumcircle of this triangle. Points \( X \) and \( Y \) are chosen on sides \( AC \) and \( AB \), respectively, such that \( TX = TY \) and \( \angle XTY + \angle XAY = 90^\circ \). Prove that \( \angle XHY = 90^\circ \). Proposed by Matthew Kurskyi
Let \( \omega \) be the circumcircle of triangle \( ABC \), where \( AB > AC \). Let \( N \) be the midpoint of arc \( \smile\!BAC \), and \( D \) a point on the circle \( \omega \) such that \( ND \perp AB \). Let \( I \) be the incenter of triangle \( ABC \). Reconstruct triangle \( ABC \), given the marked points \( A, D, \) and \( I \). Proposed by Oleksii Karlyuchenko and Hryhorii Filippovskyi
Let \( AL \) be the bisector of triangle \( ABC \), \( O \) the center of its circumcircle, and \( D \) and \( E \) the midpoints of \( BL \) and \( CL \), respectively. Points \( P \) and \( Q \) are chosen on segments \( AD \) and \( AE \) such that \( APLQ \) is a parallelogram. Prove that \( PQ \perp AO \). Proposed by Mykhailo Plotnikov
X-XI
Let \( I \) and \( O \) be the incenter and circumcenter of the right triangle \( ABC \) (\( \angle C = 90^\circ \)), and let \( K \) be the tangency point of the incircle with \( AC \). Let \( P \) and \( Q \) be the points where the circumcircle of triangle \( AOK \) intersects \( OC \) and the circumcircle of triangle \( ABC \), respectively. Prove that points \( C, I, P, \) and \( Q \) are concyclic. Proposed by Mykhailo Sydorenko
Let \( O \) and \( H \) be the circumcenter and orthocenter of the acute triangle \( ABC \). On sides \( AC \) and \( AB \), points \( D \) and \( E \) are chosen respectively such that segment \( DE \) passes through point \( O \) and \( DE \parallel BC \). On side \( BC \), points \( X \) and \( Y \) are chosen such that \( BX = OD \) and \( CY = OE \). Prove that \( \angle XHY + 2\angle BAC = 180^\circ \). Proposed by Matthew Kurskyi
Inside triangle \( ABC \), points \( D \) and \( E \) are chosen such that \( \angle ABD = \angle CBE \) and \( \angle ACD = \angle BCE \). Point \( F \) on side \( AB \) is such that \( DF \parallel AC \), and point \( G \) on side \( AC \) is such that \( EG \parallel AB \). Prove that \( \angle BFG = \angle BDC \). Proposed by Anton Trygub
Let \( I \) and \( M \) be the incenter and the centroid of a scalene triangle \( ABC \), respectively. A line passing through point \( I \) parallel to \( BC \) intersects \( AC \) and \( AB \) at points \( E \) and \( F \), respectively. Reconstruct triangle \( ABC \) given only the marked points \( E, F, I, \) and \( M \). Proposed by Hryhorii Filippovskyi
Let \( ABCDEF \) be a cyclic hexagon such that \( AD \parallel EF \). Points \( X \) and \( Y \) are marked on diagonals \( AE \) and \( DF \), respectively, such that \( CX = EX \) and \( BY = FY \). Let \( O \) be the intersection point of \( AE \) and \( FD \), \( P \) the intersection point of \( CX \) and \( BY \), and \( Q \) the intersection point of \( BF \) and \( CE \). Prove that points \( O, P, \) and \( Q \) are collinear. Proposed by Matthew Kurskyi