$Problem$ $3$.
Let $ABC$ be a triangle with $BC$ being the longest side. Let $O$ be the circumcenter
of $ABC$. $P$ is an arbitrary point on $BC$. The perpendicular bisector of $BP$ meet $AB$ at $Q$ and the
perpendicular bisector of $PC$ meet $AC$ at $R$. Prove that $AQOR$ is cyclic.
National Math Camp Geometry Exam Problem-3
Re: National Math Camp Geometry Exam Problem-3
$\textbf{Solution}$Mahiir wrote: ↑Sun May 09, 2021 3:13 pm$Problem$ $3$.
Let $ABC$ be a triangle with $BC$ being the longest side. Let $O$ be the circumcenter
of $ABC$. $P$ is an arbitrary point on $BC$. The perpendicular bisector of $BP$ meet $AB$ at $Q$ and the
perpendicular bisector of $PC$ meet $AC$ at $R$. Prove that $AQOR$ is cyclic.
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Re: National Math Camp Geometry Exam Problem-3
Restate the problem as : Let $P$ be a moving point on side $BC$ of $\triangle ABC$. Let the rotation of $PB$ with angle $B$ and centre $P$ meet $AB$ at $Q$. Define $R$ similarly.It suffices to show $R,Q,A,O$ is cyclic.
Degree of $P =1$. So degree of line $PQ = d(PB)+d(p)=0+1=1$ since $PB(BC)$ is a fixed line. So $ d(Q)=d(PQ)+d(PB)=1+0=1$ .Simlarly $d(R)=1$
Now $AQOR$ is cyclic has degree at most $d(AQ)+d(AR)+d(OQ)+d(OQ)=1+1+0+0=2$ Since $AR,AQ$ are fixed lines
$\therefore $ It suffices to check $3$ special positions of $P$
$ 1) P=B \Longrightarrow Q=B \Longrightarrow \angle ARQ =\angle ARP=2C = \angle AOB=\angle AOQ$
$ 2) P=C$ is similar.
$3) P= AH \cap BC $ Then $Q,R$ are the midpoints of $AB,AC$ respectively.Then $O,R,Q,A$ are indeed cyclic.
$Q.E.D$
Degree of $P =1$. So degree of line $PQ = d(PB)+d(p)=0+1=1$ since $PB(BC)$ is a fixed line. So $ d(Q)=d(PQ)+d(PB)=1+0=1$ .Simlarly $d(R)=1$
Now $AQOR$ is cyclic has degree at most $d(AQ)+d(AR)+d(OQ)+d(OQ)=1+1+0+0=2$ Since $AR,AQ$ are fixed lines
$\therefore $ It suffices to check $3$ special positions of $P$
$ 1) P=B \Longrightarrow Q=B \Longrightarrow \angle ARQ =\angle ARP=2C = \angle AOB=\angle AOQ$
$ 2) P=C$ is similar.
$3) P= AH \cap BC $ Then $Q,R$ are the midpoints of $AB,AC$ respectively.Then $O,R,Q,A$ are indeed cyclic.
$Q.E.D$