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: \[\frac {\angle ACB}{\angle PCB}=?\]
BdMO National Higher Secondary 2009/5
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Re: BdMO National Higher Secondary 2009/5
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.
Re: BdMO National Higher Secondary 2009/5
Would love some hints for this one!!
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"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
Learn how to write equations, and don't forget to read Forum Guide and Rules.
"When you have eliminated the impossible, whatever remains, however improbable, must be the truth." - Sherlock Holmes
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Re: BdMO National Higher Secondary 2009/5
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......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
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Re: BdMO National Higher Secondary 2009/5
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
@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:
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:
You spin my head right round right round,
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )
When you go down, when you go down down......(-$from$ "$THE$ $UGLY$ $TRUTH$" )