Let ${ABC}$ be an acute angled triangle, let ${O}$ be its circumcentre, and let ${D,E,F}$ be points on the sides ${BC,CA,AB}$, respectively. The circle ${(c_1)}$ of radius ${FA}$, centered at ${F}$, crosses the segment ${OA}$ at ${A'}$ and the circumcircle ${(c)}$ of the triangle ${ABC}$again at ${K}$. Similarly, the circle ${(c_2)}$ of radius $DB$, centered at $D$, crosses the segment $\left( OB \right)$ at ${B}'$ and the circle ${(c)}$ again at ${L}$. Finally, the circle ${(c_3)}$ of radius $EC$, centered at $E$, crosses the segment $\left( OC \right)$at ${C}'$ and the circle ${(c)}$ again at ${M}$. Prove that the quadrilaterals $BKF{A}',CLD{B}'$ and $AME{C}'$ are all cyclic, and their circumcircles share a common point. Evangelos Psychas (Greece)
Problem
Source: 2016 JBMO Shortlist G1
Tags: geometry, JBMO
08.10.2017 12:56
FA'A=FAA'=ABO => BFA'O cyclic. so BFA'=2BAA'=180-AOB=180-2C=180-C-AFA'=AKB-AKA'=BKA' => BKFA'O cyclic. so as the other two circles, all meet at O.
13.08.2018 05:27
Because $FA=FA'=FK, \angle FAA'=\angle FA'A, \angle FKA'=\angle FA'K,\angle FAK=\angle FKA,$ and \begin{align} 2(\angle FAA'+\angle FA'K+\angle FKA)=180^{\circ}. \end{align}. By construction, $A, A', O$ are collinear and $AO=BO$ so $\angle OAB=\angle ABO.$ Hence, $\angle FBO=\angle FAA'=\angle FA'A$ so $FA'OB$ is cyclic. Also, note that $\angle KFA'=180-2\angle FKA'$ and $\angle AOK $ $=\angle AOB-\angle KOB $ $=(180^{\circ}-2\angle FAA')-2\angle KAB$ $=2\angle FKA'$ (by $(1))$ so $KFA'O$ is cyclic. Since three points on a circle determine a unique circle, then $BKFA'O$ is concyclic. Similarly, $CLDB'O$ and $AMEC'O$ are also concyclic and concur at $O$, as desired. $\blacksquare$
26.08.2018 09:56
Nice problem! Solved it exactly as @above.
26.08.2018 10:19
In fact, if you mark the second intersection of line AB with circle C1 as K1, it turns out that K1 is the incenter of triangle A1BK.
29.01.2021 22:19
GREAT PROBLEM!!! Well done @Evangelos Psychas! I solved it exactly like navi_092210114