Let there be a circle with center $O$, and three distinct points $A, B, X$ on the circle, where $A, B, O$ are not collinear. Let $\Omega$ be the circumcircle of triangle $ABO$. Segments $AX, BX$ intersect $\Omega$ at points $C(\neq A), D(\neq B)$, respectively. Prove that $O$ is the orthocenter of triangle $CXD$.
Problem
Source: KMO 2024 P1
Tags: geometry
MathLuis
09.11.2024 13:16
From Anti-Parallels it's trivial that $XO \perp CD$, but also notice that $\angle ODB=180-(90-\angle AXB)$ meaning that if you let $M,N$ midpoints of $AX,BX$ respectively then $M,O,D$ are colinear, similarily $N,O,C$ are colinear and clearly now from perpendicular bisectors we have $DO \perp XA$ and $CO \perp XB$, therefore $X$ is the orthocenter of $\triangle XCD$ as desired thus we are done .
seoneo
14.11.2024 17:45
Let $E = DO \cap AX$, $F=CO \cap XB$. By inscribed angle properties, we have: \[ \angle XCO = \alpha = \angle ODA = \angle OBA. \]Since \(O\) is the circumcenter of \(\triangle XAB\), it follows that: \[ \angle CXF = 90^\circ - \alpha. \]Similarly $\angle XED = 90^\circ$ which complete proof. - solved by my collegue This was posted https://artofproblemsolving.com/community/u281710h3441800p33180428
Davud29_09
31.12.2024 12:05
We note that <XBO=α=<BXO and <OAX=<OXA=β.We get <OBA=OAB=90-α-β from angle-chasing.ABDOC is cyclic.From this cyclic we get <XDC=90-α and we get from this XO perpendicular to CD.From cyclic <DBO=<DCO=α .From angle chasing we get CO perpendicular to XD.Then O is orthocenter in triangle XCD. We are done.
Attachments:
