Olympiad problems

2016 Sharygin Geometry Olympiad, P22

Let $M_A, M_B, M_C$ be the midpoints of the sides $BC, CA, AB$ respectively of a non-isosceles triangle $ABC$. Points $H_A, H_B, H_C$ lie on the corresponding sides, different from $M_A, M_B, M_C$ such that $M_AH_B=M_AH_C, $ $M_BH_A=M_BH_C,$ and $M_CH_A=M_CH_B$. Prove that $H_A, H_B, H_C$ are the feet of the corresponding altitudes.

2001 Saint Petersburg Mathematical Olympiad, 9.5

Tags: geometry , isogonal conjugates , midline , isogonal lines , sine chasing , angle bisector

Points $A_1$, $B_1$, $C_1$ are midpoints of sides $BC$, $AC$, $AB$ of triangle $ABC$. On midlines $C_1B_1$ and $A_1B_1$ points $E$ and $F$ are chosen such that $BE$ is the angle bisector of $AEB_1$ and $BF$ is the angle bisector of $CFB_1$. Prove that bisectors of angles $ABC$ and $FBE$ coincide. [I]Proposed by F. Baharev[/i]

2018 IMO Shortlist, G6

Tags: IMO , imo 2018 , geometry , IMO Shortlist , isogonal conjugates , symmedian

A convex quadrilateral $ABCD$ satisfies $AB\cdot CD = BC\cdot DA$. Point $X$ lies inside $ABCD$ so that \[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\] Prove that $\angle{BXA} + \angle{DXC} = 180^\circ$. [i]Proposed by Tomasz Ciesla, Poland[/i]

2011 USAMO, 5

Tags: AMC , USA(J)MO , USAMO , geometry , trigonometry , isogonal conjugates

Let $P$ be a given point inside quadrilateral $ABCD$. Points $Q_1$ and $Q_2$ are located within $ABCD$ such that \[\angle Q_1BC=\angle ABP,\quad\angle Q_1CB=\angle DCP,\quad\angle Q_2AD=\angle BAP,\quad\angle Q_2DA=\angle CDP.\] Prove that $\overline{Q_1Q_2}\parallel\overline{AB}$ if and only if $\overline{Q_1Q_2}\parallel\overline{CD}$.

Kvant 2019, M2577

Tags: geometry , isogonal conjugates , Kvant

Inside the acute-angled triangle $ABC$ we take $P$ and $Q$ two isogonal conjugate points. The perpendicular lines on the interior angle-bisector of $\angle BAC$ passing through $P$ and $Q$ intersect the segments $AC$ and $AB$ at the points $B_p\in AC$, $B_q\in AC$, $C_p\in AB$ and $C_q\in AB$, respectively. Let $W$ be the midpoint of the arc $BAC$ of the circle $(ABC)$. The line $WP$ intersects the circle $(ABC)$ again at $P_1$ and the line $WQ$ intersects the circle $(ABC)$ again at $Q_1$. Prove that the points $P_1$, $Q_1$, $B_p$, $B_q$, $C_p$ and $C_q$ lie on a circle. [i]Proposed by P. Bibikov[/i]

2018 IMO, 6

Tags: IMO , imo 2018 , geometry , IMO Shortlist , isogonal conjugates , symmedian

A convex quadrilateral $ABCD$ satisfies $AB\cdot CD = BC\cdot DA$. Point $X$ lies inside $ABCD$ so that \[\angle{XAB} = \angle{XCD}\quad\,\,\text{and}\quad\,\,\angle{XBC} = \angle{XDA}.\] Prove that $\angle{BXA} + \angle{DXC} = 180^\circ$. [i]Proposed by Tomasz Ciesla, Poland[/i]

Russian TST 2019, P3

Tags: geometry , isogonal conjugates , Kvant

Inside the acute-angled triangle $ABC$ we take $P$ and $Q$ two isogonal conjugate points. The perpendicular lines on the interior angle-bisector of $\angle BAC$ passing through $P$ and $Q$ intersect the segments $AC$ and $AB$ at the points $B_p\in AC$, $B_q\in AC$, $C_p\in AB$ and $C_q\in AB$, respectively. Let $W$ be the midpoint of the arc $BAC$ of the circle $(ABC)$. The line $WP$ intersects the circle $(ABC)$ again at $P_1$ and the line $WQ$ intersects the circle $(ABC)$ again at $Q_1$. Prove that the points $P_1$, $Q_1$, $B_p$, $B_q$, $C_p$ and $C_q$ lie on a circle. [i]Proposed by P. Bibikov[/i]