Let $ABC$ be an acute-angled triangle with circumscribed circle $k$ and centre of the circumscribed circle $O$. A line through $O$ intersects the sides $AB$ and $AC$ at $D$ and $E$.Denote by $B'$ and $C'$ the reflections of $B$ and $C$ over $O$, respectively. Prove that the circumscribed circles of $ODC'$ and $OEB'$ concur on $k$.
Problem
Source: 2022 Nordic Mathematical Olympiad
Tags: geometry, circumcircle
TheStrayCat
16.04.2022 06:17
By Pascal, $B'E$ and $C'D$ intersect on $k$. The rest is angle chasing.
Jalil_Huseynov
16.04.2022 09:41
Let $(ADE)\cap (ABC)=F$. Since $F$ is center of spiral similarity that sends $DE$ to $BC$, we get $\angle FDO=\angle FDE=\angle FBC=\angle FC'C=\angle FC'O \implies DOFC'$ is cyclic, similarly $EOB'F$ is cyclic. Hence $(DOC')\cap (EOB')=F$ which lies on $k$. So we are done.
Mahdi_Mashayekhi
16.04.2022 10:43
Let $ADE$ meet $k$ at $S$. we have $\angle SDO = \angle SDE = \angle 180 - \angle SAE = \angle 180 - \angle SAC = \angle SBC = \angle SC'C = \angle SC'O \implies SC'DO$ is cyclic. $\angle SEO = \angle SED = \angle SAD = \angle SAB = \angle SB'B = \angle SB'O \implies SB'EO$ is cyclic so $ODC'$ and $OEB'$ meet at $S$ on $k$.
ohiorizzler1434
15.08.2024 04:45
what the skibidi rizzler? draw the miquel point of DECB which is on the circumcircle, then it is a trivial angle chase to demonstrate that it lies on both circles