Let $AB$ be a chord of circle $\Gamma$. Let $O$ be the centre of a circle which is tangent to $AB$ at $C$ and internally tangent to $\Gamma$ at $P$. Point $C$ lies between $A$ and $B$. Let the circumcircle of triangle $POC$ intersect $\Gamma$ at distinct points $P$ and $Q$. Prove that $\angle{AQP}=\angle{CQB}$.
Problem
Source: NZMO 2
Tags: geometry
BarisKoyuncu
07.10.2021 09:45
Let $J$ be the center of $\Gamma$. We have $P-O-J$. Let $PJ\cap \Gamma=\{P,P'\}$ and $PC\cap \Gamma=\{P,M\}$. We know that $M$ is the midpoint of the arc $AB$ which doesn't contain $P$. Let $\overarc{AP}$$=2\alpha$ and $\overarc{BP}$$=2\beta$. (Minor $\overarc{AP}$ and $\overarc{BP}$). We have $\angle AQP=\alpha$. Also, $\angle PQB=\beta$. Hence, we need to show that $\angle CQB=\alpha-\beta$. $\angle PJA=2\alpha\Rightarrow \angle AJP'=180-2\alpha$. Also, $\angle AJM=180-\alpha-\beta$. Hence, $\angle P'JM=\alpha-\beta$. Then, $\angle P'PM=\frac 12\cdot (\alpha-\beta)$. Thus, $\angle OCP=\frac 12\cdot (\alpha-\beta)$ and $\angle COP=180-(\alpha-\beta)$. $OPQC$ is cyclic so $\angle CQB=180-\angle COP=\alpha-\beta$. Hence completes the proof.
Attachments:
TheCollatzConjecture
07.10.2021 19:46
A really nice problem which inspires inversion(because of the two tangent circles) but it is just an angle chase in fact
Firstly let $PC$ meet $\Gamma$ at $D$, and we know by the curvilinear incircle/circles inscribed in segments lemma that $\triangle ADC$ is similar to $\triangle PDA$. Now let $\angle OCP=\angle OPC =x$, so $\angle CQP=2x$. Further, let $\angle PAB= \angle PQB=y$. Now since $\angle OCA=90^{\circ}$, we get that $\angle CPA =\angle DPA=90-x-y=\angle DAC$. Further since $\angle DCA = 90-x$, we get that $\angle ADC= \angle ADP=\angle AQP=2x+y$, and we know that $\angle CQB= \angle CQP+\angle PQB=2x+y$, and we're done $\blacksquare$
JAnatolGT_00
08.10.2021 20:17
As a fact this is the converse of P4 Japan MO Finals 2009. Ray $QO$ bisects angle $PQC,$ since $OPQC$ is cyclic. Consider inversion with power $|QP|\cdot |QC|$ wrt $P,$ followed by symmetry wrt $QO;$ clearly $(O)$ is fixed, hence $\Gamma, AB$ are swapped, and so $A,B.$ The conclusion follows.