Given is a triangle $ABC$ and a circle through $A, B$. The perpendicular bisector of $AB$ meets the circle at $P, Q$, such that $AP>AQ$. Let $M$ be a point on the segment $AB$. The lines through $M$, parallel to $QA, QB$ meet $PB, PA$ at $R, S$. Prove that $MQ$ bisects $RS$.
Problem
Source: Bulgarian Autumn Tournament 2022 11.2
Tags: geometry
GuvercinciHoca
21.11.2022 16:21
$RM,SM$ meet $AP,BP$ at $E,F$ respectively. $K$ be the mid-point of $RS$. Let $QM\cap (APQ)=T$ we have $\angle MEP=\angle MFP=\angle MTP= 90$ since $PQ$ is the diameter of $(ABP)$. Hence $P,T,E,M,F$ are all lie on a circle. Let $PT\cap RS=L$. By Miquel's theorem on $\triangle PRL$ we have $L\in (TES)$ and $T\in (PRS) \implies L\in EF$. Hence $PL$ is the polar of $M$ WRT $(EFRS) \implies KM\perp PL$ which means $K\in MQ$.
SevroMyName
11.04.2023 17:22
Neat synthetic solution, although I want to propose a trig bash. MQ bisects RS is equivalent to (sin(SMQ)/sin(RSM))=(sin(RMQ)/sin (MRS)) Reorganizing we get: (sin(SMQ)/sin(RMQ))=(sin(RSM)/sin(MRS)) RHS: with sin theorem is equal to RM/MS. Note that EMS~BMR (equal angles with simple angle chase). RHS = RM/MS = MB/EM Denote EQ intersect SM = X Denote BQ intersect MR = Y MYQX is a parallelogram. <SMQ = <MQB <RMQ = <EQM LHS = sin(MQB)/sin(MQE) LHS = (sin(MQB)/sin(MBQ))/(sin(MEQ)/sin(MQE)) As MBQ=MEQ because Q is on the perpendicular bisector. LHS = (MB/MQ)/(MQ/ME) LHS = MB/ME LHS = RHS As the equality we formed is equivalent to MQ bisecting RS and the equality is true MQ does bisect RS.