Given a circle $\omega$, a point $A$ lying inside $\omega$, and point $B$ ($B \ne A$). All possible triangles $BXY$ are considered, such that the points $X$ and $Y$ lie on $\omega$ and the chord $XY$ passes through the point $A$. Prove that the centers of the circumcircles of the triangles $BXY$ lie on the same straight line.
Problem
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Tags: geometry, collinear
geonnwwoo
26.09.2024 04:02
Let $D$, $E$ be the midpoints of $BX$, $BY$ respectively. Let the circumcenter of a random triangle for $BXY$ be $O$. Note that $BDOE$ cyclic due to $\angle{BDO} = \angle{BEO} = 90$. Let $C = AB\cap DE$, and let $Z = (BDOE)\cap AB$. $\textbf{\textcolor{blue}{Claim: }}$$C$ and $Z$ stays constant with respect to $\omega$ (Note $A$, $B$ constant with respect to $\omega$) $\overline{CDE}$ collinear by definition of $C$ and $BD=DX$, $BE=EY$. Thus, $AC = BC$ and so $C$ stays constant with respect to $\omega$. Since $A$ is constant with respect to the circle $\omega$, $(XA)(YA)$ remains constant due to Power of a Point $\implies (DC)(CE)$ remains constant $\implies (BC)(CZ)$ remains constant. This means that $Z$ is also constant with respect to $\omega$. $\textbf{\textcolor{blue}{Claim:}}$ Circumcenters of triangles $BXY$ lie on the line perpendicular to $AB$ intersecting $Z$, a straight line. $BDOE$ cyclic, so $\angle{BZO} = \angle{BEO} = 90$. This means that $O$ lies on the line perpendicular to $AB$ intersecting $Z$. Since the location of $Z$ remains constant, we've the statement is true.