Let $\vartriangle ABC$ be an acute triangle. $R$ is a point on arc $BC$. Choose two points $P, Q$ on $AR$ such that $B,P,C,Q$ are concyclic. Let the second intersection of $BP$, $CP$, $BQ$, $CQ$ and the circumcircle of $\vartriangle ABC$ is $P_B$, $P_C$, $Q_B$, $Q_C$, respectively. Let the circumcenter of $\vartriangle P P_BP_C$ and $\vartriangle QQ_BQ_C$ are $O_P$ and $O_Q$, respectively. Prove that $A,O_P,O_Q,R$ are concylic. proposed by andychang
Problem
Source: 2022 IMOC G3 https://artofproblemsolving.com/community/c6h2918249p26069468
Tags: geometry, Concyclic
parmenides51
11.09.2022 22:15
I had a typo $O_Q$, is circumcenter of $\vartriangle QQ_QB_Q$ and not $\vartriangle Q_QB_QC$ , my fault , sorry
Prod55
14.09.2022 14:35
parmenides51 wrote: I had a typo $O_Q$, is circumcenter of $\vartriangle QQ_QB_Q$ and not $\vartriangle Q_QB_QC$ , my fault , sorry Points $Q_Q,B_Q$ are not specified in the problem statement
parmenides51
14.09.2022 18:24
I had a second typo $O_Q$, is circumcenter of $\vartriangle QQ_BQ_C$ and not $\vartriangle QQ_QB_Q$ , again sorry here is the official wording
Attachments:
CrazyInMath
24.09.2022 16:36
Oof, my bad. R is a point on $\overarc{BC}$ and not $BC$. My file didn't display the \overarc assignment, I don't know why.
CrazyInMath
24.09.2022 18:07
First, we have $AR\parallel P_BQ_C\parallel P_CQ_B\parallel O_PO_Q$.
So we only need to prove that $P_BO_P=Q_CO_Q$ because $AP_BQ_CR$ is an isosceles trapezoid.
we have $\measuredangle P_BPP_C=\measuredangle BPC=\measuredangle BQC=\measuredangle Q_BQQ_C$.
Meaning that $(PP_BP_C)$ and $(QQ_BQ_C)$ have the same radius length, which completes the proof.