sqing wrote: Let $ AB$ be the diameter of semicircle $O$ , $C, D $ be points on the arc $AB$, $P, Q$ be respectively the circumcenter of $\triangle OAC $ and $\triangle OBD $ . Prove that:$CP\cdot CQ=DP \cdot DQ$. This is not a good proof but it's quite direct! Suppose $ \angle AOC=2y, \angle BOD=2x $ then there are two cases. Case 1: $ x+y \le \pi $ We have $ CP=\frac{1}{2\cos(y)}, DQ=\frac{1}{2\cos(x)} $ by using Law of Sines on $ \triangle AOC, \triangle BOD $, respectively. And now, using Law of Cosines on $ \triangle COQ, \triangle DOP $ then we get $ CQ^2=1+\frac{1}{4\cos(x)^2}+\frac{\cos(x+2y)}{\cos(x)}, DP^2=1+\frac{1}{4\cos(y)^2}+\frac{\cos(2x+y)}{\cos(y)} $ Finally, the problem becomes $ \frac{1}{4\cos(y)^2}[1+\frac{1}{4\cos(x)^2}+\frac{\cos(x+2y)}{\cos(x)}]=\frac{1}{4\cos(x)^2}[1+\frac{1}{4\cos(y)^2}+\frac{\cos(2x+y)}{\cos(y)}] $ , which is not hard! Case 2: $ x+y > \pi $ It's similar to case 1~((I just too lazy to type

Let $M,N$ be the midpoints of $AC,AO$. Then $AP\cdot CB=\frac{1}{2}OP\cdot OM= \frac{1}{2}ON\cdot OA=\frac{1}{4}OA^2$, hence $AP\cdot CB=BQ\cdot AD$. Thus $\frac{AP}{AD}=\frac{BQ}{BC}$ and also by anglechasing we can get $\angle APD=\angle BQC$. Thus we conclude that since $\triangle APD\sim\triangle BQC$, $\frac{AP}{PC}=\frac{BQ}{QD}$, which is equivalent to $CP\cdot CQ=DP\cdot DQ$.

WLOG arc $BD$ be larger than $AC$ so if $\angle{CAB} = \alpha$ and $\angle{DAB} = \beta$ then $\alpha > \beta$. We have $\angle{COB} = \angle{OCA}+\angle{OAC} = 2\alpha = \angle{OPC}$ since they are both isosceles, we get $\triangle{OPC} \sim \triangle{BOC}$. From similar reasoning, we get $\triangle{OAD} \sim \triangle{QOD}$. Hence, $\frac{OP}{BO} = \frac{CO}{CB}$ and $\frac{OQ}{AO} = \frac{DO}{DA}$ then multiply to get $\frac{OP}{DA} \cdot \frac{DO}{BO} = \frac{CO}{CB} \cdot \frac{OQ}{AO}$ so $\frac{AP}{AD} = \frac{OP}{AD} = \frac{OQ}{BC} = \frac{BQ}{BC}$. But, angle chase gives : \[\angle{PAD} = \angle{DAB} - \angle{PAO} = 90^{\circ} - \beta - (90^{\circ}-\alpha) = \alpha - \beta\] \[\angle{QBC} = \angle{QBO} - \angle{CBA} = 90^{\circ} - \beta - (90^{\circ}-\alpha) = \alpha - \beta\] Thus, $\angle{PAD} = \angle{QBC}$ and $\triangle{PAD} \sim \triangle{QBC}$ so, $\frac{PA}{QB} = \frac{PD}{CQ} \longrightarrow CQ \cdot CP = DQ \cdot DP$. Done.

If you could copy the code and edit this in to replace the attached image: [asy][asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy][/asy]

Solution: We will say $\angle{OAC}=A$ and $\angle{OBD}=B$.Note that $\frac{AP}{BQ}=\frac{2AP}{2BQ}=\frac{\frac{OC}{sinA}}{\frac{OB}{sinB}}=\frac{sinB}{sinA}=\frac{\frac{AD}{AB}}{\frac{BC}{AB}}=\frac{AD}{BC}$.So $\frac{AP}{AD}=\frac{BQ}{BC}$. We also see that $\angle{PAO}=90-A$ and $\angle{DAB}=90-B$ so $\angle{PAD}=|A-B|$.Similarly $\angle{CBQ}=|A-B|$.Combining these details we see that $\triangle{PAD} \sim \triangle{QBC}$.Thus $\frac{CP}{PD}=\frac{AP}{PD}=\frac{BQ}{QC}=\frac{QD}{QC} \implies CP \cdot CQ=DP \cdot DQ$ as desired.

Note $\angle OCB=90^{\circ}-\angle OCA=\angle POA\implies \triangle OBC\sim \triangle PAO$. Similarly $\triangle OAD\sim \triangle QOB$. Let $OA=R$, then $\frac{PA}{OB}=\frac{OA}{BC}\implies AP\times BC=R^2= AD\times QB\implies \frac{AP}{AD}=\frac{BQ}{BC}$. Also $\angle PAD=\angle PAO-\angle OAD=\angle OBC-\angle QBO=\angle QBC\implies \triangle PAD\sim \triangle QBC\implies \frac{AP}{PD}=\frac{BQ}{QC}\implies CP\cdot CQ=DP\cdot DQ.$

Let $PD$ cuts $CQ$ at $M$. Prove that radical axis of circles $(MCP)$ and $(MDQ)$ passes through intersection of $AC$ and $BD$.

buratinogigle wrote: Let $PD$ cuts $CQ$ at $M$. Prove that radical axis of circles $(MCP)$ and $(MDQ)$ passes through intersection of $AC$ and $BD$. Let $ E $ $ \equiv $ $ AC $ $ \cap $ $ \odot (CMP), $ $ F $ $ \equiv $ $ BD $ $ \cap $ $ \odot (DMQ). $ From the proof at post #7 we get $ \triangle PAD $ $ \stackrel{+}{\sim} $ $ \triangle QBC, $ so $ \measuredangle (PE,AC) $ $ = $ $ \measuredangle (PD,QC) $ $ = $ $ \measuredangle (AD,BC) $ $ \Longrightarrow $ $ \measuredangle (PE,AD) $ $ = $ $ \measuredangle (AC,BC) $ $ = $ $ 90^{\circ}, $ hence $ PE $ is parallel to $ BD. $ Similarly, we can prove $ QF $ $ \parallel $ $ AC, $ so $ \measuredangle EPA $ $ = $ $ \measuredangle FQB $ $ \Longrightarrow $ $ \triangle AEP $ and $ \triangle BFQ $ are pseudo-similar, hence $ \tfrac{AE}{BF} $ $ = $ $ \tfrac{AP}{BQ} $ $ = $ $ \tfrac{AD}{BC} $ $ = $ $ \tfrac{RA}{RB} $ where $ R $ $ \equiv $ $ AC $ $ \cap $ $ BD $ $ \Longrightarrow $ $ AB $ $ \parallel $ $ EF. $ From Reim's theorem we get $ C, $ $ D, $ $ E, $ $ F $ are concyclic, so $ RC $ $ \cdot $ $ RE $ $ = $ $ RD $ $ \cdot $ $ RF. $

We use complex numbers.

We claim that DPA ~ CQB. This is obvious because if E = AC \cap BD then by extended law of sines on AOC, BOD, AP/BQ = EA/EB which is equal to AD/BC. Thus, it suffices to show that <DAP = <CBQ or equivalently AP and BQ meet at an angle equal to <E, which is obvious because <PAB + <QBA = 180 - <E.

Only need to calculate ! Suppose $ \angle AOC=2\alpha, \angle BOD=2\beta $. in fact, $CP^2\cdot CQ^2=DP^2\cdot DQ^2=\frac{1}{4}(\frac{1}{4{\cos}^2{\alpha}{\cos}^2{\beta}}+2\tan\alpha\tan\beta+2).$