BdMO National Higher Secondary 2009/5

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Moon
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BdMO National Higher Secondary 2009/5

Unread post by Moon » Sun Feb 06, 2011 11:33 pm

Problem 5:
In triangle $ABC,\ \angle A = 90\circ$. $M$ is the midpoint of $BC$. Choose $D$ on $AC$ such that $AD=AM$. The circumcircles of triangles $AMC$ and $BDC$ intersect at $C$ and at a point $P$. What is the ratio: \[\frac {\angle ACB}{\angle PCB}=?\]
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Ashfaq Uday
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Re: BdMO National Higher Secondary 2009/5

Unread post by Ashfaq Uday » Wed Oct 19, 2011 1:01 am

I think the answer is 2. P is the incenter of ABC. Can somebody please post the proof?? I am missing the easier solution and keeping messing around with many things. Proved a lot of unnecessary contents.

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Re: BdMO National Higher Secondary 2009/5

Unread post by Labib » Sun Dec 25, 2011 5:57 pm

Would love some hints for this one!!
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Re: BdMO National Higher Secondary 2009/5

Unread post by sourav das » Sun Dec 25, 2011 7:53 pm

Try to work backward and see if you can find any congruent triangles.....
You spin my head right round right round,
When you go down, when you go down down......
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Re: BdMO National Higher Secondary 2009/5

Unread post by Ashfaq Uday » Tue Dec 27, 2011 12:53 am

let\[O_{1}=\]circumcenter of \[\triangle AMC\]\[X=O_{1}P\cap AB,Y=BA\cap CO_{1}\]\[O_{1}AXM\]is a rhombus. \[\Rightarrow AE=EM\Rightarrow AP=PM\Rightarrow \angle PYA=\angle PCA=\angle PAM=\angle PCM\]

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Re: BdMO National Higher Secondary 2009/5

Unread post by sourav das » Wed Dec 28, 2011 11:40 am

@Ashfaq Uday
My questions

i) Reason of claiming $O_1AXM$ a rohmbus.
ii) Define point $E$.

Please make your solution much more detailed so that everyone can understand and learn the way of your solution. It helps both the reader and solver.

I'm giving my solution:
$M$ is the midpoint of $BC$ and so circumcenter of $\triangle ABC$ as $\angle A= 90$. So $BM=AM=AD$
$CAPM$ and $PBDC$ are cyclic. So $\angle PAD=180-\angle PMC=\angle PMB$ . Same way $\angle PBM= \angle PDA$.

But these imply $\triangle APD \cong \triangle MPB$ . So, $AP=PM$ and it implies $P$ is the midpoint of arc $AM$ of the circumcircle of $AMC$. So, $\angle ACP = \angle BCP = \frac{1}{2} \angle ACB$. So the ratio is 2.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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