National Math Camp Geometry Exam Problem-3

Discussion on Bangladesh National Math Camp
User avatar
Mahiir
Posts:9
Joined:Sun Dec 13, 2020 1:50 pm
Contact:
National Math Camp Geometry Exam Problem-3

Unread post by Mahiir » 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.

~Aurn0b~
Posts:46
Joined:Thu Dec 03, 2020 8:30 pm

Re: National Math Camp Geometry Exam Problem-3

Unread post by ~Aurn0b~ » Sun May 09, 2021 5:00 pm

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.
$\textbf{Solution}$
Let $\omega_1$ be the circle centered at $Q$ with radius $BQ$.
Let $\omega_2$ be the circle centered at $R$ with radius $CR$.
And let the circumcircle of $\triangle ABC$ be $\Gamma$ .
Let $X=\omega_1 \cap \omega_2$

$\textbf{Claim}:$ $X$ lies on $\Gamma$ and the circumcircle of $\triangle AQR$
Proof : $\angle QPR=180-\angle QPB-\angle RPC$.
$=180-\angle CBA-\angle ACB$
$=\angle BAC$
So, $\angle QXR=\angle QPR=\angle BAC=\angle QAR \Rightarrow A,Q,X,R$ concyclic.
Now, As $X,P,C$ lie on $\omega_2$, $\angle XCP=\angle XRQ$
Therefore, $\measuredangle XAB=\measuredangle XAQ=\measuredangle XRQ=\measuredangle XCP=\measuredangle XCB\Rightarrow X\in\Gamma$ (Proved)

Now, it suffices to prove that $O$ lies on the circumircle of points $A,X,Q,R$.
$\measuredangle OXQ=\measuredangle OXB-\measuredangle QXB$
$=(90-\measuredangle BCX)-\measuredangle XBQ$
$=90-\measuredangle BCX -\measuredangle XCA$
$=90-\measuredangle BCA$
$=\measuredangle OAB$
$=\measuredangle OAQ\Rightarrow A,X,O,Q,R$ are concyclic.$\blacksquare$

Nayer_Sharar
Posts:16
Joined:Mon Dec 21, 2020 9:26 pm

Re: National Math Camp Geometry Exam Problem-3

Unread post by Nayer_Sharar » Wed Jun 16, 2021 10:35 pm

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.

$Q.E.D$

Post Reply