ISL 2003

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shehab ahmed
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ISL 2003

Unread post by shehab ahmed » Sun Apr 08, 2012 8:03 pm

Three distinct points $A,B,C$ are fixed on a line in this order.Denote by P the intersection of the tangents to a circle at $A$ and $C$ where the centre of the circle doesn't lie on the line $AC$.Suppose, $Q$ is intersection point of the segment $PB$ and the circle.Prove that the intersection of the bisector of $\angle AQC$ and the line $AC$ doesn't depend on the choice of the circle.
Last edited by sourav das on Mon Apr 09, 2012 8:30 pm, edited 1 time in total.
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*Mahi*
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Re: ISL 2003

Unread post by *Mahi* » Sun Apr 08, 2012 9:25 pm

This has quite the same main idea like this year's APMO 4.
If the tangents at $B,C$ of the circumcircle of triangle $ABC$ meets at $P$, then $AP$ is a symmedian of $ABC$. What are the properties of a symmedian?
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shehab ahmed
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Re: ISL 2003

Unread post by shehab ahmed » Sun Apr 08, 2012 9:43 pm

Can u please send me a msg about the question of APMO 4?I have forgot the question

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*Mahi*
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Re: ISL 2003

Unread post by *Mahi* » Sun Apr 08, 2012 10:30 pm

APMO questions are published in the official site, and so we have no problem to discuss them publicly.
Here is the link to APMO official site - http://www.mmjp.or.jp/competitions/APMO/ . You can collect the questions from there.
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zadid xcalibured
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Re: ISL 2003

Unread post by zadid xcalibured » Mon Apr 09, 2012 12:49 pm

Hint:
after finding the symmedian we can use a little trig(just sine) too.

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*Mahi*
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Re: ISL 2003

Unread post by *Mahi* » Mon Apr 09, 2012 1:58 pm

zadid xcalibured wrote:Hint:
after finding the symmedian we can use a little trig(just sine) too.
Or, we can use basic algerbra ;)
But, the symmedian part is the main crux move here :)
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zadid xcalibured
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Re: ISL 2003

Unread post by zadid xcalibured » Mon Apr 09, 2012 2:26 pm

after using this kind of lemma,this problem doesn't seem to be crux.

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Re: ISL 2003

Unread post by zadid xcalibured » Mon Apr 09, 2012 2:44 pm

my solution:
$\frac{AB}{BC}=\frac{AQsinx}{CQsiny}$,here $x$ and $y$ are the angles subtended by segments $AB$ & $BC$ at vertex $Q$.let the midpoint of $AC$ be $M$.then $\frac{AM}{MC}= \frac{AQsiny}{CQsinx}$.so $\frac {AK}{CK}=\frac{AQ}{CQ}$.$\rightarrow \frac{AK^2}{CK^2}=\frac{AB}{BC}$,here $K$ is the foot of angle bisector.
Mahi,what did u mean by basic algebra?

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Re: ISL 2003

Unread post by *Mahi* » Mon Apr 09, 2012 3:59 pm

I meant
$\frac mn = \frac ij $ and $\frac {m-x}{n-x} =\frac {i+x} {j+x}$ implies $x=0$.
Do not be afraid, it is not complex ;)
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Re: ISL 2003

Unread post by zadid xcalibured » Mon Apr 09, 2012 4:31 pm

কমপ্লেক্স হলে ভয় পাওয়ার কি আছে ?

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