China Western Mathematical Olympiad 2014 ,Problem 2

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China Western Mathematical Olympiad 2014 ,Problem 2

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Source: China Zongqing 16 Aug 2014

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41782 posts

sqing

#1 h Aug 16, 2014, 8:43 AM • 4 Y

Y by phantranhuongth, Adventure10, Mango247, GeoKing

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: $CP\cdot CQ=DP \cdot DQ$ . $[asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy]$

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641 posts

BSJL

#2 h Aug 16, 2014, 1:22 PM • 2 Y

Y by Adventure10, Mango247

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

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341 posts

#3 h Aug 18, 2014, 11:25 AM • 3 Y

Y by ZacPower123, Adventure10, and 1 other user

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 $CP\cdot CQ=DP\cdot DQ$ .

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191 posts

MrRTI

#4 h Aug 19, 2014, 2:54 PM • 2 Y

Y by Adventure10, Mango247

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.

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TheMaskedMagician

2955 posts

TheMaskedMagician

#5 h Aug 19, 2014, 10:30 PM • 2 Y

Y by Adventure10, Mango247

If you could copy the code and edit this in to replace the attached image:

$[asy] import cse5; import olympiad; unitsize(3.5cm); dotfactor=4; pathpen=black; real h=sqrt(55/64); pair A=(-1,0), O=origin, B=(1,0),C=shift(-3/8,h)*O,D=shift(4/5,3/5)*O,P=circumcenter(O,A,C), Q=circumcenter(O,D,B); D(arc(O,1,0,180),darkgreen); D(MP("A",A,W)--MP("C",C,N)--MP("P",P,SE)--MP("D",D,E)--MP("Q",Q,E)--C--MP("O",O,S)--D--MP("B",B,E)--cycle,deepblue); D(O); [/asy]$

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sayantanchakraborty

505 posts

sayantanchakraborty

#6 h Sep 28, 2014, 3:57 AM • 4 Y

Y by strawberry_circle, mijail, Adventure10, Mango247

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.

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#7 h Mar 27, 2016, 11:56 AM • 2 Y

Y by Adventure10, Mango247

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 $\angle PAD=\angle PAO-\angle OAD=\angle OBC-\angle QBO=\angle QBC\implies \triangle PAD\sim \triangle QBC\implies \frac{AP}{PD}=\frac{BQ}{QC}\implies CP\cdot CQ=DP\cdot DQ.$

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2344 posts

#8 h Apr 11, 2016, 3:41 AM • 4 Y

Y by baopbc, hoangA1K44PBC, mijail, Adventure10

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

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2312 posts

#9 h Apr 11, 2016, 5:12 AM • 6 Y

Y by buratinogigle, baopbc, hoangA1K44PBC, enhanced, Adventure10, Mango247

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 $RC$ $\cdot$ $RE$ $=$ $RD$ $\cdot$ $RF.$

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#10 h Jan 19, 2023, 2:48 PM

Y by

We use complex numbers.
bash

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1650 posts

v4913

#11 h Jan 19, 2023, 3:36 PM

Y by

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.

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pud

#12 h May 16, 2024, 1:52 PM • 1 Y

Y by GeoKing

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

:roll:

This post has been edited 1 time. Last edited by pud, May 16, 2024, 1:53 PM

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